toaz.info-117-polynomial-problems-pr 56664cbb273bd94814cfe5423987a7a8

xyzt~
~
Polynomial
Problems
from the AwesomeMath
Summer Program
117 Polynomial Problems
From the AwesomeMath
Summer Program
Titu Andreescu
Navid Safaei
Alessandro Ventullo
Library of Congress
Control Number: · 2018915280
ISBN-13: 978-0-9993428-4-8
ISBN-10: 0-9993428-4-3
© 2019 XYZ Press, LLC
All rights reserved . This work may not be translated or copied in whole or in
part without the written permission of the publisher (XYZ Pres s, LLC , 3425
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www .awesomemath .org
Cover design by lury Ulzutuev
Th e authors dedicat e this book as follo ws:
Titu Andr eescu: to Alina , my lovely wij e
Navid Safaei: to my moth er Shirin and my brother M ehr egan
Al essandro Ventullo: to Ra.ffaella, my love
Preface
Polynomial s form the cornerstone of mod ern mathematics
and ot her discrete field s. Apart from the direct link to Algebra. and sub seq uentl y, Ca.lculus: the y also appear frequently in many branch es of sciences . Th e ubiquity
of polynomials and their ability to characterize complex patt erns let us bctt er
understand generalizations, theorems. and elegant paths to solu t ions that th ey
provide . \\ e strive to showcase the true beau ty of polynomials t hrough a wellthought collection of probl ems from mathemati cs compet it ions and int ui t ive
lect ures that follow the sub-topics. Thus. ,ve pres ent a view of polynomials
that incorporates various techniques paired with the favorit e th emes that show
up in math contests.
First , the two introductory chapte rs detail th e ma chin ery and not at ion
used to characterize pol ynomial s. Factorization identi ties, the notion of GCD ,
composition, types of roots , and the Intermediat e Value Th eorem, to name
some , lay the foundational groundwork that the rea.der will use to in te rnaliz e
the contents of the later chapters. To give the student a sense of dir ection
or relation to real world problem s, polynomials and th eir properti es are di scussed based on the number of variable s. sta r t ing with second degree and up
to fourth degree represen tat ions. Two additional chapt ers t hat deal with key
theorems and some applications of polynomials in Numb er Th eor y follow. Th e
book concludes with the Introdu ct ory and Advanced pr ob lem sectio ns , along
with their respective solution segments. There a.re significantly mor e than 117
problems that are selected with st rict considerations , with an ab undan ce of
great problems in the first seven chapters. It actually features more than 180
problems in the theory component as well. l\,Iost of the hard er and classic
problems have more than one solution to familiariz e the rea.der with a variety
of approaches. We have also added problems proposed recent ly in journals
and competitions for the student to better consolidate and assimilate thes e
techniques.
Since mathematics competitions almost always includ e algebra.ie problems ,
polynomials can form a large subset of the necessar y materiał one should be
acquainted with in order to succeed. The authors consider pol ynomials a
dear and classic art. Through the problems , lecture , and theory, we do our
best to transfer most of the knowledge , strategies, and tricks explored to our
reade rs. This book is best suited for AMC 10/ 12, AIME, and USAMO/JMO
competitors , and is th e first volume in a three-book series.
Preface
Vlll
Many thanks to Richard Stong and Adrian Andreescu for significant improvement of the text. We would also like to thank Alexandr Khrabrov, Peter
Boyvalenkov, Gregor Dolinar, and Bayasagalan Banzarach for their contribution.
Enjoy the problems!
The authors
Contents
Pre face .....
......
...
...
. .. .
1
Basic properties of polynomials - Part I
1.1
Identities . . . . . . . . . . . . .
1.2
The coefficients of xd in polynomial produ cts
1.3
Factoring and its implications .
1.4
Values of polynomials . . . . . .
1.5
Division , GCD of polynomials ..
1.6
Th e composition of polynomials
1.7
Odd and even polynomials
2
Basic properties of polynomials - Part II
2.1
Polynomial roots . . . . . . . . .
2.2
Int eger and rational roots of polynomials
2.3
Interm ediate value theorem, increasing and
decreasing polynomials .
3
Second degree polynomials . .
3.1
The form ax 2 + bx + c
3.2
The discriminant
3.3
Root s .. . ....
.
3.4
Vieta's formulas . .
3.5
Solving inequalit ies .
3.6
Miscellaneous problems
3. 7
More advanced problems
Third degree polynomials
4
4.1
Roots and graph . . . . .
4.2
Vieta 's formulas . . . . .
4.3
More advanced problems
Fourth degree polynomials .
5
5.1
Solving equations ...
..
5.2
Vieta' s formulas . . . . .
5.3
Number of real roots and graph.
5.4
Miscellaneous . . . . . . . . . . .
6
On roots of polynomials - elementary problems
Vll
1
2
6
11
18
26
35
37
39
39
44
49
59
59
61
66
73
75
78
80
85
85
90
93
101
101
101
104
107
111
Contents
X
6.1
6.2
6.3
7
8
9
10
11
12
Vieta's formulas in the generał case . . . .
Inequalities between coefficients and roots
Miscellaneous problems . . . . . . . . . .
N umb er theory and polynomials . . .. .. ...
7.1
Number theory and low degree polynomials
7.2
P(a) - P(b) ..
Introductory problems . . ....
.
Advanced problems . . . . . . . . .
Solutions to introductory problems
Solutions to advanced problems
Other Books from XYZ Pre ss . . .
111
115
121
125
125
127
135
145
155
197
261
1
Bas ic properties of polynomials - Part I
1
Basic properties
of polynomials
- Part I
A polynomial is an expression of the form
n
P( x ) _
- anx n+ an-IX n-1 + . . . + a 1x + a0 = ~
L..t aix i .
i=O
Th e numbers ai ar e said the coefficients of the polynomial P(x). Usually, we
consider ai in Z, Q, JR,C and we say that the polynomial has integer, rational,
real or complex coefficients, respectively. We denote by Z[x], Q[x], IR[x ], C[x)
the set of polynomials with integer, rational, real or complex coefficients , respectively.
The coefficient ao is said the constant term .
From the definition, it follows that two polynomials
P(x)
n
m
i=O
j=O
= L aixi and Q(x) = L bjxi
are equal if and only if ai = bi for all i (if m > n , then bn+I = ... = bm = O).
n
We define the degree of polynomial P(x)
=
L aixi as the greatest integer
i =O
i such that ai # O and we denote the degree by <legP(x). If i is the greatest
integer i such that ai # O, we say that ai is the leading coefficient of P(x).
lf the leading coefficient is equal to 1, we say that the polynomial is monie.
Notice that the degree of a constant polynomial P(x) = ao # O is zero . We
don 't give any degree to the zero polynomial P( x )
O (i.e. the polynomial
=
1
whose coefficients are all zeros) . Usually, we omit the terms having zero as a
coefficient. For example , we write the polynomial Ox3 + lx 2 + 2x + O as x 2 + 2x.
This is clearly an example of a monie polynomial of degree 2.
We can perform some operations on polynomials.
For example, if P(x) =
n
m
i=O
j =O
L aixi and Q(x) = L bjxi are two polynomials
and m 2: n, then the sum of P(x) and Q(x) is defined by
m
P(x)
+ Q(x) = L(ah
+ bh)xh
h=O
and the product of P(x) and Q(x) is defined by
1
By convention , we can also assign to the zero polynomial the degr ee -oo.
Basic properties of polynornials - Part J
2
We have the following.
Theoren1
Let P(x) and Q(x) be two polynomials and let k be a positive int eger.
Th en ,
1. deg(P(x)Q( x)) = deg P( x) + deg Q(x )
2. deg(P( x) + Q(x)) < max(d eg P( x ), deg Q(x))
3. deg[(P( x )l]
1. 1
= k · degP(x).
ldentities
The term identity designates an equality that holds for all allowed values of
the unknowns it contains (usually all real numbers or all complex neumbers).
Wh en it is elear from context one often omits explicitly specifying the allowed
range for th e unknowns. For example, the following equations are identities:
a2 - b2 = (a - b) (a + b),
a 3 - b3 = (a - b) (a 2 +ab+ b2 ) ,
(a+ b + c)2 = a2 + b2 + c2 + 2(ab +ac+ be),
a2
b2
b2
c2
a2
a+b
b+c
a+c·
c2
--+--+--=--+--+-a+b
b+c
c+a
The first three of these hold for all real (or all complex) values of a, b, c, whereas
the last only holds if none of a + b, b + c, and c + a vanishes. However, the
equalities below do not meet this criterion since they do not hold universally :
2x + 1 = 5,
1
X-
1
+ -X = 3,
2
a 3 + b3 + c3 = 3abc.
Identities are the pillars of our mathematical computations.
They are
commonly encountered in mathematics competitions, where many problems
require such knowledge .
Here we record some of the most important identities. It is important that,
in order to enhance your strength in working with algebraic expressions , you
force yourself to learn these identities.
- Part J
[Ja:,ic properties of polynomials
3
Useful ldentities
Conjugat e ld enti ty:
a2 - b2 = (a - b)(a + b)
Square Id entity I:
(a + b)
2
= a 2 + 2ab + b2
Square Identity II:
(a+ b + c)2 = a
2
+ b2 + c2 + 2(ab +b e + ca)
Power Difference:
Power Sum:
if n is odd.
Euler' s I dentity:
a 3 + b3 + c3 - 3abc =(a+
b + c)(a 2 + b2 + c2 - ab - be - ca)
Binomial Identity:
(a+ bt =an+
(;)a n-lb + ... +(;)an-kbk+
... + bn.
Comment I. Euler 's Id entity impli es that if a+ b + c = O, then
a 3 + b3 + c3 = 3abc.
Comment II. The generał case of the Square Identity is of special interest .
That is, let x 1 , ... , Xn be n numbers. Then ,
(x1 + · · · + Xn ) 2 = Xi2 + · · · + Xn2 + 2
For example, when n = 4, we have,
(a + b + c + d) 2 = a
2
+ b2 + c2 + d 2 + 2 (ab + ac + ad + be + bd + cd) .
Comment III. The generał case of Binomial Id entit y is of special int erest.
Basic properties of polynomials - Part I
4
Let x 1 , . .. , Xn be n numbers and let rn be a positive int eger. Th en,
For example, when
The above identity is called the Multinomial Id entity.
n = m = 3 we have
(a+ b + c) 3 = a3 + b3 + c3 + 3 (a 2 b + b2 c + c2 a + b2 a + c2 b + a 2 c) + 6abc.
Many elementary algebra problems are greatly simplified by the above formulas. Nevertheless , we will also see some of th eir creative applications in more
complex solution s.
Example 1.1.
(Saint-Petersburg Mathemat ical Olympiad 1999) Let n> 3
be a positive integer. Prove that n 12 can be represented as a sum of cubes of
three natural numbers.
Solution. Note that
(a - 3b)3 = a3 - 9a2 b + 27ab2 - 27b3 .
Multiplying both sides by a, we have
a(a - 3b)3 = a4 - 9b(a - b)3 - 9b4 .
Now, set a= n 3 and b = 3m 3 • Then,
This is equivalent to
Example 1.2.
(Serbian Mathematical Olympiad 2012) If x+y+ z = O and
2
2
2
x + y + z = 6 what is the maximal value of
l(x - y)(y - z )(z - x)I?
Solution . Set z = --x-y. We find that 2x 2 +2xy+2y 2 = 6, i.e., x2+xy+y2 · 3.
· y 2 + yz + z 2 = z 2 + zx + x 2 = 3, thus xy + yz + zx = -3. Now,
L 1.kew1se,
/
2
(x - y ) = x 2 + xy + y 2 - 3xy = 3 - 3xy,
thus
((x - y)(y - z)(x - z))2 = (3 - 3xy)(3 - 3yz)(3 - 3zx)
5
Basi c properties of polynom ials - Part I
Th e right-h and sid e is equal t o
2
27 (4 - x 2y 2z )
27 (1 - x y - y z - zx + x y z(x + y + z ) - x 2y 2z 2)
< 27 · 4.
Ther efore, l(x - y)(y - z )( z - x )I ::; 6)3. Since thi s value is attained when
{x, y ,z } = {O, J3 , -v'3}, this is th e maximum value.
Example
equations
1.3.
(Math emati cs and Youth Journ al 2003) Solve the syst em of
x 2(y + z)2
y2( z + x)2
+ x + 1)y 2 z 2
(4y 2 + y + l) z 2 x 2
z2(x + y)2
(5z2 + z + 1) x 2 y 2 .
(3x
2
Solution. If x = O, then all thr ee equations redu ce to y2z 2 = O and we find
the solutions (x , y, z) = (O, t , O) or (O, O, t) for any real t. Similarly , the cases
y = O or z = O add one mor e family of solutions (x , y , z ) = (t , O, O).
Now assume xyz =/O. Setting , x = ¼, y =
z = ¾, we obt ain the following
system of equations
i,
(b + c)2
(c + a) 2
(a+ b)2
3 +a + a 2
2
4+b+b
5 +C+c
2
By summing them, we find that (a+ b + c)2 - (a + b + c) - 12 = O. We
substitute t = a+ b + c, and simplify to t 2 - t - 12 = O, which gives t = 4
or t = -3. If t = 4, then a + b + c = 4, and by substituting into the abov e
equations, we find (4 - a) 2 = 3 +a+ a 2, i.e. , 9a = 13, so x = 9/ 13. By the
same argumenty=
3/4 and z = 9/11. If t = 3, then a+ b + c = -3 , and by
the same argument we find the solution (x , y ,z ) = (-5 / 6,-1 , -5 / 4).
Example 1.4.
(Ivan Tonov - Bulgarian Mathematical Olympiad 2008)
If the following equation is an identity
(x + y)2n+l _ x2n+l _ y2n+l = (2n + l)xy(x
+ y)( x 2 + x y + y2r-1,
find the value of n.
Solution. Let x = y = l. Then 22n+l - 2 = 2(2n + 1)3n- 1, thus
22n - 1 = (2n + 1)3n-l .
We prove that the equality does not occur for n> 3. Write t he equality as
i )n 2n + 1 1
(3
3
+ 3n·
Basic properties of polynomials - Part I
6
Then, for n > 3 we have
Gr
(l+~r
n n(n - 1)
1
l + 3 + 2 · 32 + · · · + 3n
n n(n - 1)
1
> l + 3 + 2 · 32 + 3n ·
Now
2n+l
3
n n(n-1)
> l + 3 + 2 · 32
'
which leads to the inequality n 2 - 7n + 12 = (n - 3)(n - 4) < O, which is false
for n > 4. If n E {1, 2, 3}, the equation is indeed an identity.
Note. We can also use a number theory argument to refute the identity case:
22n -1 = (2n+ 1)3n- 1 . Let v 3 (N) denote the exact number of times the prime
3 divides N . Since v3(22n - 1) = v3(4n - 1) 2: n - 1, we find that 3n- 2 I n ,
which is false for n > 3.
1.2
The coefficients
of xd in polynomial
products
Suppose that we want to compute the coefficient of x 50 in the following product
(1 + 2x + 3x2 + · · · + 10lx 100 )(1 + x + x 2 + · ·. + x 25 ) .
For sake of this, we need to study in which ways a monomial from first factor
and a monomial from second factor generate the term x 50 . That is,
x5o = x5o . 1 = x49 . x = .. . = x25 . x25.
Hence, the coefficient of x 50 is the sum of coefficients of the constructed monomials and is equal to
51 + 50 + ... + 26 = 1001.
Example 1.5.
(Navid Safaei) Let k be a positive integer such that
2
1 + xk + x k = (1 + a1x + x 2 )(1 + a2x + x 2 ) · ...
ay
· (1
+ akx + x 2 ).
ai.
Find the value of
+ ... +
Solution. If k = 1, then a1 = 1 and the desired value is 1. If k = 2, then
2
1+x + x
4
= (1 + a1x + x 2 )(1 + a2x + x 2 ).
Basic properties of polynomials - Part J
7
Comparing the coefficients of x 2 and x in both sides, we find that
i.e.
aI + a~ = (a1 + a2) 2 - 2a 1a2 = 2 .
Moreover, one can find t hat {a 1, a 2 } = {1, -1}. Hence,
1 + X 2 + X 4 = (1 - X+
X
2
)(1· +X+
2
X ).
Assume now k > 3. Then , the coefficients of x and x 2 in the product
must be zero. Examining the aforementioned coeflicients, one can find that
a1 + ... + ak = O
and
k+
L aiaj = O,
1$ i<j$k
i.e.,
L aiaj = -k. Hence , by the generalization of the square identity,
1$i < j $ k
we find that
ar + ... + a~ = (a1 + ... + ak)
2
- 2
L aiaj = 2k .
1$i< j $ k
Hence, the answer is k if k = 1, 2 and 2k if k > 3.
Example 1.6.
Find the coeflicient of x 100in the expressio n:
(1 + x + x 2 + ... + x 100)3.
Solution. Note that
(l+x+x2+
... +x100)3 = (l+x+ ... +xlOO)(l+x+ ... +xlOO)(l+x+ .. . +x100).
Thus, a term x 100 can arise from a product of the form xaxbxc from the
respective factors where a+ b + c = 100 and a, b, c > O. Let a = O. Then
b + c = 100 and there are 101 cases. If a= 1, then b + c = 99 and there are
100 cases. Similarly when a= 100, then b + c = O, and there is only one case.
Hence the total number of cases is
1 + 2 + ... + 101 = 5151.
Basic properties of polynomials - Part I
8
Example 1. 7.
(Federico Poloni - Italian Mathematical Olympiad 2013, Local Round) Let P(x) and Q(x) be two trinomials. How many non-zero monomials does their product P(x)Q(x) have at least?
Solution. Assume that
P(x) = AxR + Bx 8 + Cxr,
Q(x) = axr + bx 8 + ext
a,
where , A , B , C, b, c =/=O and R, S , T, r, s, t > O are integers, R =/=S =/=T and
r =/=s =/=t. Without loss of generality assume that R > S > T and r > s > t.
Then,
P(x)Q(x) = AaAxR+r + ... + CcxT+t,
a product with 9 terms. It is elear that the monomials xR+r and xT+t cannot
cancel out because of minimality and maximality. Thus, the product has at
least two terms. Now we provide an example with exactly two terms, that is,
consider the polynomial x 4 + 4. If we factor it, we find that
x 4 + 4 = (x 2 - 2x + 2)(x 2 + 2x + 2).
Thus, the answer is 2.
Example
1.8.
Define a family of polynomials recursively by
Po(x)
x - 2,
Pk(x) = Pf_ 1(x) - 2 if k > 1.
Find the coefficient of x 2 in Pk (x) in terms of k.
Solution. Note that for all k > 1 we have Pk(0) = 2, thus
pk (X) = 2 + akx + bkx 2 + .. . ,
then Pk+1(x) = 2 + ak+lx + bk+1x 2 + ... = (2 + akx + bkx2 + ... )2 _ 2.
An
. easy calculation shows that ak+l = 4ak and bk+l = ak2 + 4bk·
Smee
a1 = -4, we find that ak = _4k, so
bk = 42k-2
+ ... + 4k-l
= 4k-l(l
+ 4 + ... + 4k-l) = 42k-1 - 4k-l
3
Example
1.9.
(AIME 2016) Let P(x) = 1 _ x + x
3
2
and define
6
i=O
50
Find
L jail•
i=O
Basic properties of polynomials
- Part I
9
Soluti_on. N~te that al_lthe coefficients of the polynomial P( -x) are nonnegative, mdeed its coefficients are the absolute values of the coefficients of the
polynomial P (x) · Thus all the coefficients of the polynomial
Q(-x)
= P(-x)P(-x
3
)P(-x
7 )P(-x
9)
are nonnegative and are the absolute values of the coefficients of the polynomial
Q(x). Hence
Example 1.10. (V.A. Senderov - Russian Mathematical Olympiad 2008)
Find all positive integers n such that there exist nonzero a, b, c, d such that
(ax + b)1000 - (ex+ d) 1000 has exactly n nonzero coeffi.cients.
Solution. The answer is n E {500, 1000, 1001}. Indeed for n= 1001 consider
the expression (2x+2) 1000 -(x+1) 1000 and for n= 1000 consider the expression
(2x + 1) 1000 - ( x + l) 1000 . If we have more than one zero coeffi.cient the n there
exists two coefficients, say the coefficients of xr and xt such that
Thus
so ad
be
~ = l which gives also ,~, = 1. Now it is elear that if we replace
b
'
c
the polynomial ax + b with -ax - b, the condition does not change. Thus
without loss of generality we assume that a/c = 1 and we have two cases. If
d/b = l, then we have (ax+b) 1000 - (ax +b) 1000 which has only zero coeffi.cients,
a contradiction. If d/b = -1, then we have (ax + b) 1000 - (ax - b) 1000 , which
has exactly 500 nonzero coeffi.cients.
Example 1.11.
(Tournament of Towns 2012) Let P(O) = 1 and
P(x) 2 = 1 + x + x
100
Q(x) .
100
Find the coefficients of x 99 in the expansion of (1 + P(x))
.
100
Solution. Note that in the expansion of (1 + P( x))
+ (1 - P(x)) 100 we
have only even powers of P( x) . Indeed , the above expression is a polynomial
in P(x)2 = 1 + x + x 100 Q(x) of degree 50. Taking the equation modulo
x 100, the highest coefficient of the expression is 2(1 + x) 50 ; thus there is no
x99 term in the expansion. Moreover, since P(O) = 1, one ca n find that the
polynomial P(x) - 1 is divisible by x. Hence, (1 - P(x)) 100 = x 100 R( x ), for
Basic properties of polynomials - Part I
10
5
· t h expans ion.
·
some polynomial R(x), which implies that there is no x 99 m
1
0
Thu s there doesn 't exist any x 99 in the expansion of ( 1 + P( x))
too.
Example
1.12.
(Moscow Mathematical
Olympiad 1997) Let
1 + x + x2 + ... + xn-l = F(x)G(x),
where n> 1 and where F and G are polynomials, whose coefficients are zeros
and ones. Prove that one of the polynomials F and G can be represented in
the form (1 + x + x 2 + ... + xk- 1 )T(x), where k > 1 and where T is also a
polynomial whose coefficients are zeros and ones.
Solution. Set F(x) = a0 + a 1x + ••• and G(x) = bo+ b1x + · · · • From _the
constant term we get aobo = 1, which gives a0 = bo = 1. From the x coeffic1ent
we therefore get a 1 + b1 = 1. Without loss of generality, assume that a1 = 1
and b1 = O. If G(x) = 1, then we are done, so we may assume there is some
least non-zero monomial in G (x), say xk, so that G (x) = 1 + xk + · · · . Looking
at the coefficients of xi for i = 2, ... , k, we conclude that ao = a1 = a2 = · · · =
ak-l = 1 and ak = O.
Now we will show that every monomial in G is of the form xkr for some r (or in
terms of polynomials G(x) = Q(xk) for some polynomial Q with coefficients
zero and one) and that every non-zero monomial in F occurs in a run of
nonzero monomials of the form xkr + xkr+l + xkr+2 + • • • + xkr +(k-l) (or in
terms of polynomials F(x) = (l+x+x 2 +• · •+xk-l )P(xk) for some polynomial
P(x) with coefficients zero and one). Note that this will solve the problem by
setting T(x) = P(xk) .
Suppose on the contrary that this is not the case. Then there is some lowest
degree monomial where it fails. There are two ways this can happen.
If the first bad monomial is in G, then there is some monomial xkr+s in G
where O < s < k. Since the product F(x)G(x) contains the monomial xkr
there must be some monomial xa in F and some monomial xb in G with
a + b = kr. Since a, b < kr + s and the monomial xkr+s is the first deviation
from our proposed pattern, it follows that b = kr' for some r' < r and hence
a = k(r - r'). Again since this is xkr+s was the first bad monomial this
must begin a run of monomials xk(r-r') + xk(r-r')+l + ... + xk(r-r')+(k-1)
' in
F. But then the coefficient of xkr+s in F(x)G(x) gets a contribution of 1
from xk(r- r')+s · xkr' and another contribution of 1 form x 0 . xkr+s. Since any
additional contributions would only be positive, the coefficient of xkr+s in the
product will be at least 2, a contradiction .
If the first bad monomial is in F, then there is some r such that F contains
only a proper subset of the monomials xkr, xkr+I, . .. , xkr+k-I _ Say it contains
xkr+i but not xkr+J for some O< i, j < k. The monomial xkr+J must occur in
F(x)G(x), say as xa · xb. Since the run xkr, .. . ,xrk+k - I contains the smallest
degree bad monomial, we must have b = kr' for some 1 < r' < r and hence
11
Basic properties of polynomials - Part I
a= k(r - r') + j. Since this is smaller, F must contain the entire run
Xk(r-r')
+ Xk(r-r')+l
+ ... + Xk(r-r')+k-1
_
Now look at how many times the monomial xkr+i occurs in F(x)G(x) . lt
occurs once as xk(r-r')+i · xkr' and once as xkr+i . x 0 . Thus it occurs at least
twice, a contradiction.
Example 1.13. (Moscow Mathematical Olympiad 1994) Is there a polynomial P(x) with a negative coefficient while all the coefficients of any power
(P( x )r are positive for n> l?
Solution.
The answer is yes. Let P (x) = adxd + ... + a 0 have positive
coefficients. Then all of its powers have positive coefficients.
Assume J (x) = x 4 + x 3 + x + 1 and set g ( x) = f (x) - cx 2 for some c > O
sufficiently small. Then all the coefficients of (g(x)) 2 and (g(x)) 3 are close to
the coefficients of
(f(x)) 2 = x 8 + 2x7 + x 6 + 2x5 + 4x 4 + 2x3 + x 2 + 2x + 1
and
(J(x)) 3 = x 12 +3x 11 +3x 10 +4x 9 +9x 8 +9x 7 +6x 6 +9x 5 +9x 4 +4x 3 +3x 2 +3x+l.
The coefficients of (f(x)) 2 and (f(x)) 3 are all positive and thus the coefficients
of (g(x)) 2 and (g(x)) 3 must be positive. Then, since all positive · integers n can
be written as n = 2a + 3b for some nonnegative integers a, b, all the powers
(g(x)r have only positive coefficients.
1.3
Factoring and its implications
Factoring of an algebraic expression means the decomposition of the original
expression into a product of expressions with smaller degrees. We have two
main tools for this: grouping and using identities. Applying the former tool
includes dividing the expression into groups with a common factors.
For example,
a 2 + ab+ be+ ca = (a 2 + ab) + (ac+ be) = a( a + b) + c(a+ b) = (a + c) (a + b).
Example
1.14.
Factor the following expression :
xyz + 3xy + 2xz - yz + 6x - 3y - 2z - 6.
Solution.
We group the above expression as
(xyz + 3xy) + (2xz + 6x) - (yz + 3y) - (2z + 6).
12
Basic properties of polynomials - Part I
Now we can factor out a z+ 3 from every group,
( z + 3) (xy + 2x - y - 2).
Again, we can group the expression xy + 2x - y - 2 as
x(y + 2) - (y + 2) = (x - l)(y + 2).
Therefore, our expression can be factored as
(x - l)(y + 2)(z + 3).
The latter case refers to identities for factoring. For example, we can factor
the expression
First , note that a 2 - b2 + b2 - c2 + c2 - a 2 = O. Hence by Euler's identity,
Now by the conjugate identity, we find that
Example
equations
1.15.
(Mathematics and Youth Journal 2004) Solve the system of
x3 + y3
2y3 + z3
3z
Solution.
3
+x
3
4y
2
4z
2
4x
2
-
5y + 3x + 4
5z + 6y + 6
-
5x + 9z + 8.
-
Write the system as:
x 3 - 3x - 2
-y 3 + 4y 2 - 5y + 2
2y 3 - 6y - 4
-z 3 + 4z 2 - 5z + 2
3z 3 - 9z - 6
-x 3 + 4x 2 - 5x + 2.
Both sides can now be factored as
(x - 2) (x + 1)2
2 (y - 2) (y + 1) 2
3(z-2)(z+1)2
(2 - y)(y - 1)2
(2 - z)(z - 1)2
(2 - x)(x - 1)2
Now if x = 2, then y = 2 or y = l. If y = 2, then from the second and the
third equation, we get z = 2. If y = l, then comparing the second and the
Basic properties of polynomials - Part I
13
third equation, we have no solutions. This means that if one among x, y, z
is 2, then x = y = z = 2. So, assume that x, y, z -=f.2 and multiply all the
equations. We find that
2
2
(x - 1) (y - 1) (z - 1)2 + 6(x + 1)2 (y + 1)2(z + 1)2 = O
which clearly has no solution .
Example 1.16. (Mathematics and Youth Journal 2005) Let x, y, z, t be real
numbers. Consider the polynomial:
F (x, Y, z, t) = 9 (x 2 y2 + y2 z 2 + z 2 t 2 + t 2 x 2 ) + 6xz (y 2 + t 2 )
- 6yt(x 2 + z 2 ) - 4xyzt.
(i) Prove that the polynomial can be factored into the product of two quadratic
polynomials.
(ii) Find the minimum. value of the polynomial F if xy + zt = l.
Solution. (i) It is obvious that
F (x, y, z, t) = (3x 2 + 3z 2 + 2xz) (3y2 + 3t 2 - 2yt).
(ii) Note that
2
F (x, y, z, t) = 4 ( (x + z)2 + (x-z)
)
((y+t)
2
2
2
+ (y - t) 2 ) ,
then by the Cauchy-Schwarz inequality, we have
F(x ,y,z,t) > 4 ex+
z~t+t)
+ (x- z~y-t)r
= 2(2xy + 2zt) 2 = 8.
Since F(l/ v'2,1/ v'2,1/ v'2,1/ v'2)= 8 this is indeed the minimum.
Note. For the second point, we could have also written F (x, y, z, t) as follows
F (x, y , z, t) = (3yz + xy - zt - 3xt) 2 + 8(xy + zt) 2 > 8.
Example
1.17.
Given the natura! numbers m, n> 2, prove that the number
2n-1
m
-----mn
1
m-1
has a divisor of the form mk + 1, where k is a positive integer.
Solution. Set n+ l = 2r s, where r > O and s is an odd number. Let
m2n-1
d n=
-
-----mn.
m-1
1
Basic properties of polynomials - Part I
14
Now,
mdn = m
2n
- m - mn+ 1 = m
m-1
2n
- 1 - (m n+1 + 1) .
m-1
Note that
ffi2 n -
1
2
--=(m+l)(m
m-1
Since r < n - 1, then m 2r + 1 divides
then
+1)· .. . •(m
m2 n -
1
m-
1
2n - l
+l
)
.
2
and mn+l + 1 = (m r)
s
+ 1,
(m 2 r + 1) I mdnSince gcd (m, m 2r + 1) = 1, we find that (m 2r + 1) I dn and we are done .
Note . We employ a similar approach as in the above problem to solve one that
appeared on a Chinese TST.
Find all m , n 2:'.:2 such that
(i) m + 1 is a prime of the form 4k - 1;
{ii) there exist a prime pand a nonnegative integer a such that
We say that a polynomial is irreducible if it cannot be factored into products of
polynomials with smaller degrees. Otherwis e, we call a polynomial reducible .
These two concepts are context-dependent, that is, polynomial 1 + x 2 is irreducible over JR[x], Q [x], Z [x], but is reducible over C [x] because
1 + x 2 = (x + i) (x - i),
where, i 2 = -1.
Example 1.18. Let Fn be the n-th term of the Fibonacci sequence. Prove
that the polynomial
is reducible over Z[x].
Solution. We can easily find that the polynomial is divisible by (x 2 + x - l).
Indeed, since Fk+I = Fk + Fk - I, then
n
Fnxn+I + Fn+IXn - 1 =
L Fk(xk+I + xk - xk-I)
k=l
= (x 2 + X - l) (Fnxn - I + Fn-IXn- 2 + · · · + F2x + F1),
15
Bas ic prop erties of polynomia ls - P art I
Example
1.19.
Let
P (x) = (x 2 - 12x + 11)4 + 23.
Prov e t hat P( x) cannot be represent ed as t he prod uct of th ree non-const ant
polynom iab with integer coefficients .
Solution. Assume that
P (x) = (x 2 - 12x + 11)4 + 23 = Q(x)H(x)R(x) ,
where Q(x) , H( x) , R (x) have integ er coefficients. Since th e product of the
lead ing coefficients of Q, H and R must equal t he leading coefficient of P ,
which is 1, each must have leading coefficient ±1. If one of the coefficients is
-1 , t hen a second must also be - 1 and we can mul tip ly both by - 1. Thu s we
may assume Q, H , and R are monie . Since P( x ) does not have real root s, then
the degrees of th e po lynom ials Q (x), H (x), R(x) must be even (if for exam ple
Q has odd degre e, th en Q has a real root , hence P( x ) must have a rea l root ).
Th en two of t he po lynomia ls Q (x), H (x) , R (x) must be quadratic. W ith out
loss of genera lity , assume deg Q(x) = deg H( x ) = 2. Since P (l) = P (11) = 23
we can find that Q (1) , Q (11) E {±1 , ±23}. Now
(11 - 1) I Q (11) - Q (1).
Th e above statement is true if and only if Q ( 1) = Q ( 11). Analogously,
H (1) = H (11). Now since one of Q(l) or H (l ) must be equa l to ±1 , without
loss of genera lity assume t hat Q (1) = ±1 , then Q (11) = ± 1. Now ,
Q (x) = (x - 1) (x - 11) ± 1,
but t hen Q(x) has rea l roots, a contra dict ion .
Example 1.20. (Po lish Mathematica l Olympiad 2014) For any integer n> 1,
determin e the sma llest value of the polynomial
Pn(x)
= x 2n + 2x 2n-l + 3x 2n- 2 + ... + (2n - l )x 2 + 2n x
on the set of real numb ers.
Solution. Note that:
x2n + 2x2n- l + x2n - 2
(x + l )2x2n - 2
2x 2n-2 + 4X2n-3 + 2x2n -4
3x2n-4 + 6x2n-5 + 3x2n -6
2(x + 1 )2x2n - 4
kx2n - 2k+2 + 2kx2n-2k+l
3(x + 1) 2x2n-4
+ kx2n-2k = k(x + l )2x2n - 2k_
Basic properties of polynomials - Part I
16
Finall y, write nx 2 + 2nx = n(x + 1)2 - n. Then we can deduce the following
identity:
Pn (x) = (x + 1)2 (x 2n - 2 + 2x 2n- 4 + .. . +(n - l)x 2 + n) - n.
So, Pn (x) > Pn (-1) = -n.
In order to solve the next problem , we need an useful irr educibility criterion.
Eiscust('ill ·s Criterion.
Suppose P(x) = adxd + ad_1xd-I + ... + a0 is a polynomial with integer
coefficients and there is a prime p such that p does not divide ad, but
does divide all of ad-I, ... , a0 , and p 2 does not divide ao. Then P(x) is
irreducible over the integers.
Proof. Suppose on the contrary P(x) = f(x)g( x ) where f( x) =bo+ b1x + · · ·
and g =co+ c 1 x + • • • are non-constant polynomials with integer coefficients.
Since the product of the lead ing coefficients of f and g is ad which is not a
multipl e of p , the leading coefficients of f and g are not multiples of p. Hence
ther e is some least ind ex k such that bk is not a multiple of pand there is some
least index m such that Cm is not a multiple of p. Then from the coefficient
of xk+m we get
Every term on the right hand side of this equation except the middle one is a
multipl e of p, becaus e bo, . .. , bk- 1 and Cm-1, .. . , co are multiples of p. Since
th e middle term is not a multipl e of p, it follows that am+k is not a multiple of
p. Thus m + k = d. But this means that bkxk and CmXm mu st be the leading
terms off and g. In partic ular , k, m > O so bo and co are both multiples of p.
But this would make ao a multiple of p 2 , contrary to the hypothesis. This is
a contradiction and hence P(x) is irr educible over the integers.
Example 1.21. (Titu Andr eescu - Mathematical
least positive integer n for which the polynomial
Reflections S18) Find the
P (x) = xn - 4 + 4n
can be written as a product of four non-constant polynomials with integer
coefficients .
Solution.
We will show that the least numb er is 16. Let us prove that
numbers from 1 through 15 do not work. For 10 < n < 15 t he polynomial
4
P (x) = xn- + 4n is _irr educible over th e ra t ionals , as can be checked using
Eisenstein 's criterion (see below) for the prim es 5, 11, 3, 13, 7, 5, respectively.
For n = 9, we get P (x) = x 5 + 36. If we could factor it in the desired way,
Basic properties of polynomials - Part I
17
t.hcn one of the factors would be lin ear. Hen ce p (x) = x 5 + 36 would hav e an
integer root , which is impossible. Similarly, when n= 8, p (x ) = x 4 + 32 has
an integer root and the same reasoning app lies. For n = 4, 5, 6, 7 at leas t one
of the factors must be constant . For n= 16 we have
x
12
+ 64
'
x 12 + 16x 6 + 64 - 16x 6
(x6 + 8)2 - (4x3)2
(x 6 - 4x 3 + 8)(x 6 + 4x 3 + 8) .
On the other hand,
x 12 + 64
(x4 + 4) (x8 - 4x4 + 16)
((x 2 + 2) 2 - 4x 2 ) (x 8 - 4x 4 + 16)
(x 2 + 2x + 2) (x 2 - 2x + 2)(x 8 - 4x 4 + 16).
Because x 2 + 2x + 2 and x 2 - 2x + 2 have no int eger roots , they are irr educib le.
Hence they must divide x 6 - 4x 3 + 8 and x 6 + 4x 3 + 8. In fact , we have
x 6 - 4x 3 + 8 = (x 2 + 2x + 2)(x 4 - 2x 3 + 2x 2 - 4x + 4)
and
x 6 + 4x 3 + 8 = (x 2 - 2x + 2)( x 4 + 2x 3 + 2x 2 + 4x + 4).
Hence for n = 16 our polynomial is th e product of four non-constant polynomials with integer coefficients.
Example 1.22.
the polynomial
Let n be even and p > nn be a prime number. Prove that
Q (x) = (x - 1) • ... • (x - n)+ p
is irr educible.
Solution. Assume that Q (x) = f (x) g (x) , where f (x ) and g (x ) are nonconstant polynomials with int eger coefficients . Note that whenever x > n or
x <Owe have Q (x) > O. Moreover, for all x E [O, n] we have
(x - 1) · ... · (x - n)+ p > -nn + p > O.
Thus for all real numbers x, we have f (x ) g (x) > O. Without loss of generality,
assume that f (x) > O and g (x) > O. Then for all k = 1, 2, ... , n we have
f (k) g (k) = p implies that f (k) and g (k) E {1,p} . Define the polynomial
f (x) + g (x) - p - 1, which is zero at k = 1, 2, . .. , n, but the degr ee of this
polynomi al is at most n - 1. Thus it must be the zero polynomial and we
have f (x ) + g (x) = p - 1 for all x. But since f and g ar e positiv e this forces
O < f(x) , g(x) < p - 1 for all x . Since f (x ) and g (x) are non-constant , this is
impo ssible.
Basic properties of polynomials - Part I
18
1.4
Values of polynomials
As we have seen, every polynomial P( x) of nonnegative
representation as the following
degree has a generic
where ad, ad-l, ... , a 1 , a 0 are complex numbers. Th e term adxd is called the
leading term and ad is called the leading coefficien t. Furthermore, ao is called
the constant term. The value of the polynomial at x = c, which is denoted by
P( c), is
P(c) = adcd + ad-1cd-l + ... + a1c + ao.
Of special interest are the cases of P(l) and P(-1) , i.e.,
P(l) =ad+ ad- 1 + .. . + ao
which is called th e sum of coefficients and
Example
1.23.
Let
2017
( V2011x -
.J2021
)
= a2017x2011 + a2015x2016 + ... + a1x + ao.
Find the val ue of
(a1 + a3 + · · · + a2011)
Solution.
2
- (ao + a2 + · · · + a20 1s)2.
If we set x = 1, we have
(✓2017 - .J2027)2017 = a2017 + a2016 + ... + a1 + ao
and if we set x = - I, we hav e
+ a2016 + .. . - a1 + ao
(ao + a2 + ... + a201s) - (a1 + a3 + · ••+ a 2017).
-a2017
By multiplying both equalities we find that
( ao
+ a2 + ... + a201s) 2
~
~)2017
( V 2017 - V 2027
-102017.
·
(
2017
v201'7
+ V2027)
Basic properties of polynomials
Example
nomial
1.24.
- Part I
19
(Med it err anea n Competition 2015) Pro ve that for the poły-
P (x) = x 4 - x 3 - 3x 2 - x + 1
there exist infinit ely many positive integers n for which P (3n) is composite .
Solution. Set x = 32n-l _ Th en
p (32n- 1) = 812n-1 _ 27 2n-l _ 3 (g2n-1) _ 3 2n-l + 1.
Taking t he equat ion modulo 5, we find that
p (32n-1) = 1 _ 22n-1 _ 3(-l)2n-1
Since 22n-l
_ 32n-l + l
= (22n-1 + 32n-1)
(mod 5).
+ 32n-l is divisibl e by 5, we are done.
Example 1.25. Let P (x) = a0 + ... + anxn be a polynomial with intege r
coefficients such t ha t P ( -1 ) = O and P ( v'2) E Z. Prov e that th ere exists an
int eger O < k < n such t hat P (k) + ak is even.
Solution. We h ave
P (1) = I:a2
Since P ( -1)
i + I:
a2Hl·
= O, we find that
Th en a 1 must be even. So, P (1) + a1 = 2
L a2i + a1 is even .
Example 1.26. (Zhautykov Mathematical Olympi ad 2014) Find all pol yno mials P (x) wi th int eger coefficients such t hat
Solution.
Let Q (x) = P (x) - x. Clearly, Q (3 + v'5) = O. Furth er mor e,
since Q (x) has integer coefficient s, we have that if D is not a perfect square,
a, b are rat iona l numb ers and Q
numbers c, e, then
Q (a -
(a+bv'JJ)= c +ev'J] for some rational
bVD)
= c - ev'IJ.
Thi s impli es that Q (3 - v'5) = O. Then Q(x) is divisible by
(X -
(3 -
vs))(
X -
(3
+ vs))= x 2 - 6x + 4 .
20
Bas ic prop erti es of polynom ials - Part I
Now,
Q (x ) = P (x ) - x = (x 2 - 6x + 4)R( x )
for some polynomial R(x) with int eger coefficients. Th en
P (x) = x + (x 2 - 6x + 4)R( x ).
Now, take x = l + \1'3.We have
1 = (2 - 4v'3)R(l + v'3).
Analogously , 1 = (2 + 4\1'3) R(l - \1'3). Then
J3)R ( 1 - J3)= 4~.
R (1+
But R( x) has integer coefficients, thus
R (1+
J3)= a + b\1'3and R ( 1 - J3)= a - b\1'3
for some integers a, b. Then R (l + \1'3) R (l - \1'3) = a 2 - 3b2 which is an
integer, contradiction.
An interesting property of polynomials with integer coefficients is that for all
arbitrary integers r, s we have
P(r)-P(s)
=
adrd + ad-1rd-l
+ .. . + a1r + ao
-ads d - ad-1s d-1 - ... - a1s - ao
ad (rd - sd) + ad-1 (rd-l - sd-l)
+ ... + a1(r - s).
Since rk - sk is divisible by r - s, we find that
Since a1, ... , ad all are integers, we find that
8d)
rd
a1 + a2 (r + s) + ... +ad ( --
r-s
= Q(r, s)
is an integer. Thus , we find that
,
Thcormn
Let P(x) be a polynomial
P(r)-P(s).
-----
r-s
.
with integer coefficients.
. .
.
1s an mteger for all d1stmct mtegers, r, s.
Then , the ratio
21
Basic properties of polynomials - Part I
According to above result , there is no polynomial with integer coefficients such
that
P (10) = 2017, P (12) = 2018
.
P(12) - P(l0)
1 .
.
smce
_
=
1s not an mteger.
12 10
2
Example 1.27. Does there exist a polynomial P(x) such that none of its
nonzero coefficients is an integer , P(O) = O and for any distinct integers a, b
P(a) - P(b) .
.
we have
b
1s an mteger?
axm+xn
Solution. The answer is yes. Set P (x) = --where m > n > l are
2
integers. Now
P(a) - P(b) _!(am
- bm
-------+---.
a-b
2
a-b
an - bn)
a-b
Note that
ak - bk
.
.
= ak-I + ak- 2 b + ... + abk- 2 + bk-I - a+ b + (k - 2) ab (mod 2) .
a- b
Then for all k > 2
2
. P(a) - P(b)
a-b
(a + b + (m - 2) ab) + (a + b + (n - 2) ab)
(m + n) ab (mod 2).
.
P(a) - P(b) .
.
Let m, n be of the same panty. Then 2 ·
a_ b
1s an even mteger and
we are done .
If P( x) = xn + an-I xn-l + ... + ao has roots X1, . .. , Xn and
Q(x) = xn+l + bnxn + ... +bo has roots Yl, ... , Yn+l, prove that
Example
1.28.
Solution.
Write
P(x) = (x - x1) · . .. · (x - Xn) and Q(x) = (x - Y1) · ... · (x - Yn+1).
Then
P(y1)· . . . ·P(yn+1) = (y1-x1) · ... ·(y1-Xn)·
... ·(Yn+1-x1)·
... ·(yn+l-Xn)-
This product has n(n+ 1) factors , which is always an even number . Hence we
can rearrange the above product as
(x1 -y1) · · · · · (x1 -Yn+1) · • • • · (xn -y1) ·. • • · (xn - Yn+1) = Q(x1) · ... · Q(xn)
Basic properties of polynomials - Part I
22
and we have the conclusion.
Not e. The previous example can be used to solve the following problem by
Titu Andreescu (Mathematical Reflections U451):
2
3
Let x1, x 2, x3, x 4 be the roots of the polynomial 2018x 4 + x + 2018x - 1.
Evaluate
(xr - X1 + l)(x~ - x2 + l)(x~ - X3 + l)(x~ - X4 + 1).
(Hint: take P(x) = x 3 + 1 and Q(x) = 2018x 4 + x 3 + 2018x
2
-
1.)
Example 1.29. Let P be polynomial of degree at most 10 with integer coefficients such that for all k E {1, 2, ... , 10} there is an integer m such that
P(m) =kand
IP(lO) - P(O)I < 1000. Prove that for all integer k there is an
integer m such that P(m) = k.
Solution. Assume that for all i = 1, 2, ... , 10 there are integers Ci such that
P(Ci) =i.For all i= 1,2, ... ,9 we have
Thus c 1, ... , c10 are consecutive. Since we can always replace P(x) by P(-x)
we may assume without loss of generality that c1, ... , c10 are in increasing
order, therefore Ci= c1 - 1 + i. Set Q (x) = 1 + x - c1. Then,
p (X) - Q (X) = R (X) (X - C1) · ... · (X - ClQ).
for some polynomial R(x) with integer coefficients. Hence,
p (X) = 1 + X - C1+ R (X) (X - C1) · ... · (X - ClQ).
Checking the degree condition we find that R(x) is a constant polynomial, say
R (x) = C. If C =JO,we have
P (10)-P
(O)= 10 + C((lO - c1) · ... · (10 - c10) - (O - c 1 ) · .. . ·(O - c 10 ))
= 1O+ (N+ 20) (N+ 19) · ... · (N+ 11) - (N+ 1O) .....
(N+ 1),
where we have written N= c1 - 1. lt is easy to check that
(N + 20) ( N + 19) · ... · (N + 11) and (N + 1O) .....
(N + 1)
are not equal (the first is larger for N > -10 and the second is larger for
N~ -11) and both are divisible by 10!. Thus
l(N + 20) (N+ 19) · ... (N+ 11) - (N+ 10) · .... (N+ 1)1 > 10!.
Hence, IP(10) - P(O)I > 10! - 10 > 1000, contradiction.
p (X) = 1 + X - C1.
Thus C = O and
23
Basic properties of polynomials - Part I
So, for all integer k take m = k + c 1 - 1 and we get the conclusion.
Another interesting property of polynomials with integer coefficients is the
similarity between P(c) and the c-ary expansion of an integ er number. If we
start with the c-ary expansion of an integer (adad- l .. . aa)c (where as usual
O :S a i :S c - 1 for all O :S i :S d) we can define a polynomial
P(x) = adxd + ad-1Xd-l + .. . + a1x + aa
Th en, we can easily find that
Conversely , if we have a polynomial P and we know that its coefficients ai all
satisfy O :S a i :S c - 1, then we can read off the coefficients of P from the c-ary
expansion of P( c).
Example 1.30. All the coefficients of the polynomial P(x) are either -1 or
1. If P (3) = 130, find this polynomial.
Solution. Write P (x ) = adxd + . . . +aa. If ad= -1, then ·
130 = p (3) :S _3d + 3d-l + .. . + 3 + 1 < 0.
Thus we must have ad= l. Now
130 = p (3) 2 3d - ( 3d-l + ... + 3 + 1) =
1 + 3d
2
implies that d :S 5. If d < 5 then 130 = P (3) :S 34 + 33 + ... + 1 = 121. Thus
d = 5. Now a4 = -1 certainly , then
130 = P (3) = 35 - 34 + 27a3 + 9a2 + 3a1 + aa.
Hence
27a3 + 9a2 + 3a1 +aa= -32
Then a3 = -1, a2 = -1 , a 1 = aa = l and P (x) = x 5 - x 4 - x 3 - x 2 + x + l.
Example 1.31. (Tournament of Towns 2012) Vlad claims that a nonconstant polynomial P(x) with nonn egative integer coefficients is uniquely
det ermined by th e values of P(2) and P(P(2)). Is Vlad right?
Solution.
Vlad is right. Let P (x) = adxd + ... + aa be a non-constant
polynomial with nonnegative integer coefficients. Set
P(2) = b = ad2d + . .. +aa> ad+ ... + aa.
It follows that all the coefficients of P( x) are nonn egative integers in the
range from O to b - l. Thus the coefficients of P( x ) can be read off from the
Basic properti es of polynomials - Part I
2-1
b-aJ·y expam;ion of P(P(2)) = P(b) = (adad-I ···ao)b- Thus P(x) is uniquely
determin ed.
Example 1.32. (Korea.n Mathematical Olympiad 2001) Let n and N be
positiv e integers and define Pn as the set of all polynomials
J( x ) = ao + ... + anxn
such that
(i) For all j = 0.1 , ... , n we have lail< N.
(ii) Th e set {j I 1 < j ~ n , aj = N} has at most two elements. Find the number of elements in the set {f(2N) I J( x ) E Pn}Solution.
Set h(x ) = N(l + x + ... + xn). We must count the number of
different va.lues of J(2N) + h(2N) for J(x) E Pn, If none of the coefficients
of f a.re equal to N, then (J + h) (2N) is a k-digit number in base 2N for
some k < n+ 1. Thus (J + h) (2N) can take on all integer values from O to
(2N)n +i - 1. If only a 11 = N , then (J + h) (2N) = (2Nr+ 1 + M, where M is
a k-digit number in base 2N for some k < n. Thus (J + h) (2N) can take on
all integer values from (2Nyn.+1 to (2Nr+ 1 + (2Nr - 1. If an = an-I = N,
then (f + h) (2N) = (2N)n+i + (2Nr + M, where M is a k-digit number in
base 2N for some k ~ n - 1. Thus (f + h) (2N) can take on all integer values
from (2N)7H 1 + (2Nr to (2Nyn.+1 + (2N)n + (2Nr - 1 - 1. Combining these
three cases, we see that (J + h) (2N) can take on all integer va.lues from O to
(2N)1'+ 1 + (2Nr + (2Nr- 1 - 1. However it is also elear that since at most
two coefficients off+ h are 2N , the largest possible value of (J + h) (2N) is
(2N - 1) (1 + 2N + ... + (2Nr-
2
)
+ (2N)(2Nr-
1
+ (2N)(2Nt
which is equal to
(2Nr+
1
+ (2Nt + (2Nr-
1 -
1.
Thus these are the only va.lues of (f + h) (2N) that occur and hence the number of elements in the set is (2N)n - 1 ((2N) 2 + 2N + 1).
We provide a wonderful problem from Russian Olympia.ds 2003 based on the
c-ary interpretation of the polynomial value P(c).
Example 1.33. (Alexander Khrabrov - Russian Mathematical Olympiad
2003) Let P(x) and Q(x ) be polynomials with nonnegative integer coefficients
such that all of the coefficients of P( x) are less or equal than m. If there exist
integers a < b such that P (a) = Q (a), P (b) = Q (b) and b > m, prove that
P(x) = Q(x ).
Solution. Assume that
P (x) =
n
k
i=O
i= l
L c;xi and Q (x ) = L dixi
Basic properties of polynomials - Part J
25
and O::::;Ci ::::;m < b. Then we see that P(b) written in base bis (CnCn-1 ... CO)b.
If all the coefficients of Q are less than b our work is clone. If not, then there
exists a smallest index i such that di 2: b. Th en we can write di = bq + r with
O = r < b. Now construct the polynomial Q 1 from Q by changing di+l to
di+I + q and di tor. Then it 's obvious that Q 1 (b) = Q(b). Now, what about
Q1(a) and Q(a)? We can see that
(bq + r) ai + di+ 1ai+ 1
dia i + di+1ai+I
> (aq + r) ai + di+ 1ai+ 1
> rai+ (di+I + r) ai+ 1
This implies that Q1 (a) < Q(a). If we iterate this process , then the index
where di first eąuals or exceeds b can only increase. Thus in at most n steps,
we get a polynomial Q8 all of whose coefficients are less than b and with
P(b) = Q(b) = Q 5 (b). The eąuality P(b) = Q8 (b) implies that Pand Q8 must
be eąual, but the ineąuality Q 8 (a) < Q(a) contradicts our assumption. Then
our first assumption was wrong and we have P(x) = Q(x) .
Example 1.34.
of degree 100
(Aleksander Golovanov - Tuymada 2007) Two polynomials
f (x ) = a10ox100+ a99X99+ ... + a1x + ao
and
100+ b99x99+ ... + b1x + bo
g (x) = b10ox
are given and it is known that the coefficients of the polynomials are permutation of each other and for all i we have ai =I bi. Do there exist two such
polynomials such that for all real x we have f(x) 2: g(x)?
Solution. Let h (x) = (x - 1)100= c10ox100+ ... +co. We know that
h (l) = c100+ c99+ . . . +co= O.
Set
bo= 2,
b1 = 2 + co, b2 = 2 +co+ ci,
where bk = 2 +co+ c1 +·· · +Ck-I and
b100= 2 + co+ c1 + ... + c99= 2 - c100= l.
Now set ak = bk + ck. Then for all O < k ::; 99 we have ak
a100=bo= 2. Then
J(x)-g(x)
= h(x) 2: O.
Basi c prop erties of polynomials - Pa rt I
26
1.5
Division , GCD of polynomials
We know t hat for any posit ive int eger b and any int egers a dividing a by b
with a remainder shows t hat t here are unique int egers ą, r such that
a = bq + r,
O ś r ś b - l.
For polynom ials we get a very similar result using polyno mial division. First
consider polynomial s with rat ional coefficients. Exa ct ly th e sa me argument s
would work for eit her real or compl ex coefficient s, tho.ugh as we will see below
integer coefficients require a litt le extra care . Suppo se B( x) is any nonzero
polynomial. Th en for any polynomi als A (x), there are uniqu e polynomials
Q(x ), R(x ) such that
A(x ) = B( x) Q(x ) + R(x) ,
deg R(x) < deg B( x).
So, instead of a remainder with sma ller magnitude, i.e., satisfying the inequality O ~ r ~ b- 1, we have the ana logous condition of a remaind er with smaller
degree. Th e above statement, is called the Polynomials Division Th eorem. It
can be proved by using simple indu ction on th e degre e of A( x) or equivalentl y
by discussing long division of polynomials . When R (x) = O, we say that the
polynomial A(x ) is divisible by the polynomial B( x ).
For examp le, let A (x) = 3x4 - x 3 + x 2 - x + 1 and B (x) = x 2 + x + 2. Then
long division gives
4
3x - x
3
+ x 2 - x + 1 = (x 2 + x + 2) (3x2 - 4x - 1) + 8x + 3.
Hence Q (x) = 3x2 - 4x - 1 and R (x) = 8x + 3. Furthermore , if we want to
divide an arbitra ry polynomial P( x) by x 2 + 2, by the Polynomials Di vision
Theorem, we find that
P (x) = (x 2 + 2) Q(x) + R(x).
Since <leg R(x) < deg(x 2 + 2) = 2, we can write that R(x) = ax + b, whence,
P (x) = (x 2 + 2) Q (x) + ax + b.
If A( x ) and B( x) have integer coefficients , th en we can not always find polynomials Q(x) and R(x) with int eger coefficients such that
A(x) = B(x)Q( x ) + R(x) and deg R(x) < deg B( x).
For example, if B(x) = 2x 2 + 1 and Q # O has integ er coefficients then the
leading coefficient of B( x) Q(x)+R( x) will always be a multiple of 2. 'He nce we
cannot write A(x) = x 3 in this form. However if we assume B( x ) is a monie
polynomial, then this problem disappears and we have the following result.
Ba sic prop erti es of polynom ials - Part I
27
Corollary
If A(x) is a pol ynomial with int eger coefficient s and B( x) is a monie
pol ynomial with intege r coefficients, then there are uniqu e polynom ials
Q( x ), R( x) with int eger coefficients such that
A( x ) = B( x )Q( x) + R(x) ,
<leg R(x) < <legB( x).
Thi s can b e proved eith er by imit ating the proof above an d seeing that as long
as B is monie we can still do long division or by derivin g it as a corollary of the
result for rat ional coefficient s. By that result, we find A( x) = B( x )Q(x)+R(x)
wher e Q and R have rational coefficients. Now supp ose qkxk is th e highest
degr ee monomial in Q for which th e coe fficient is not an integer. Th en looking
at th e coefficient of xk+<leg B, we get a contradiction . Thu s Q and henc e
R(x) = A( x ) - B( x )Q( x) ha ve integer coefficients.
Example 1.35. Let P( x ) and Q(x) be polynomi als of second degr ee with
int eger coefficients. Prove t hat th ere is a polynomial R(x) with integ er coefficients and degree at mo st two such that
R(8)R(l2)R(2017)
= P(8)P(l2)P(2017)Q(8)Q(12)Q(2017).
Solution.
Define , T(x) = P( x )Q(x). Th en , degT(x) = 4 and T(x) has
integer coefficients. Now we must prov e that there is a polynomial R( x) with
integ er coefficients and degree at most two such that
R(8)R(l2)R(2017)
= T(8)T(l2)T(2017)
Now, we divide the polynomial T(x) by the monie polynomial
(x - 8)(x - 12)(x - 2017) ,
i.e. , we write
T (x) = (x - 8) (x - 12) (x - 2017) H (x) + R (x),
where deg R(x) < deg (x - 8) (x - 12) (x - 2017) = 3, i.e., deg R( x)
Now, take x = 8, 12, 2017 in the above identity. We obtain
T (8) = R (8) ,
T (12) = R (12),
T (2017) = R (2017),
i.e. ,
R (8) R (12) R (2017) = T (8) T (12) T (2017).
and we are done.
< 2.
Basic properties of polynomials - Part I
28
. .
(GCD) of two polynomials A(x), B(x) with
By the Greates_t Common Divisor .
I nomial D(x) of greatest degree that
rational coeffic1entswe mean a momc po Y
divides both polynomials.
.
LCM) of two polynomials A(x), B(x) with
ial L(x) of least degree divisi ble
By the Least Common Multiple ( .
rational coefficients we mean a momc polynom
by both polynomials.
1.36. (Italian Mathematical_ Ol~mpiad 2015, _Distri~c:~~:t~t~!
P(x) , Q(x) be two monie polynomials w1th mteger c~efficients 8 is
GCD is (x - l)(x - 2) and their least common multiple (LCM)
Example
(x - 1)2(x - 2)3 (x - 3)(x + 1).
Ifwe further assume degP(x) < degQ( x ), how many polynomials P(x) exiSt
for which there is a Q( x) satisfying these conditions?
Solution. Let D(x) = gcd(P(x), Q(x)) and L(x) = lcm(P(x), Q(x)) .
Analogously to the case of integers , one can find that
_ P(x)Q(x)
(
)
L x D(x) '
i.e.,
L(x)D(x) = P(x)Q(x).
Hence,
P(x)Q(x) = (x - 1)3 (x - 2)4 (x - 3)(x + 1).
The polynomial (x - 1)(x - 2) is common to both polynomials P (x) and Q(x).
After cancelling off these factors, we see that
P(x)
(x-l)(x-2)
(
x-1
Qit)
)= (x - l)(x - 2) (x - 3)(x + 1).
x-2
2
The factors on the right must be split between P(x) and Q(x), and since the
factors on the left are relatively prime we must give both factors of (x - 2) 2 to
the same polynomial. Thus this is possibl e in 24 = 16 ways. Since the product
P(x)Q(x) is of degree 9, we find that degP(x) # degQ(x), thus in exactly
16
2 = 8 cases we have <legP(x) < <legQ(x).
To find the GCD of two polynomials with rational coefficients, assume that
degA( x) > deg B(x). Then , by the Polynomials Division Theorem, we find
that th ere are polynomials Qi (x), R 1(x) such that
A(x) = B(x)Q1(x) + R1(x) ,
degR 1(x) < degB(x).
If R1(x) = O, then B(x) divides A(x) and hence B(x) (divided by its leading
coefficient) is the greatest common divisor. If R 1 (x) # O, then any common
Basic properties of polynomials - Part J
29
divisor of A(x) a~d -!3(x) also divides R 1 (x). Thus B(x) and R 1 (x) have the
same GCD. Contmumg , we divide B(x) by Ri(x) , so
B( x ) = R1(x)Q2(x) + R2(x),
deg R2(x) < deg R 1(x) < deg B(x).
By iterating this procedure, since the degree always decreases, we must eventually getto a case where Rn+1 (x) = O. In this case, Rn(x) (divided by its
leading coefficient) is the greatest common divisor of A(x) , B(x). If Rn(x) is a
constant polynomial , then the polynomials A(x), B(x) have no non-constant
common divisors and they are called relatively prime or coprime. Moreover ,
since
R1(x) = A(x) - B(x)Q 1 (x)
we have
Continuing in this way, we eventually find that there are polynomials
R(x), S(x) , such that
Rn(x) = R(x)A(x) + S(x)B(x).
The interesting case is when A(x), B(x) are coprime. In this case dividing
through by the constant polynomial Rn shows that there are polynomials
R(x) , S(x) such that
1 = R(x)A(x) + S(x)B(x).
Thus we have just proved the following corollary.
Bezout·s
identity
for polynomials
For any two coprime polynomials P(x) , Q(x) with rational coefficients,
there are polynomials R(x), S(x) with rational coefficients such that
1 = R(x)P(x) + S(x)Q(x).
A complex number a is called algebraic if there is a nonzero polynomial P(x)
with integer coefficients such that P (a) = O. For example , v'2+ v13is an
algebraic number since P ( v'2+ v13)
= O, where
P (x) = x4 - 10x2 + 1.
A nonzero polynomial P(x) with integer coefficients of minimal degree for
which P (a) = O is called a minimal polynomial of an algebraic number a. By
the minimality of degree condition, we have that this polynomial is irreducible
over Q [x] and hence over Z [x]. Moreover, let R (a) = O for some arbitrary
Basic properties of polynomials - Part I
30
polynomial R(x) with int eger coefficients . By th e minimal degree assumption,
<legP( x) ::; <leg R(x). Th en by th e Polynomials Divi sion Theor em, one can
write
R(x) = P(x)Q( x) + S(x)
for some polynomials Q(x), S(x) with rational coefficients, where <legS(x) <
<legP(x). Multiplying both sides by th e leas t common denominator of the
coefficients of the polynomials Q(x), S(x), we can assume that Q(x), S(x) have
integer coefficients. Now take x = a and we find that S(a) = O. Hence, the
minimal degree condition implies that S(x) = O, i.e., R(x) is divisible by P(x).
Thus all polynomials with a as a root are multiples of the minimal polynomial.
Example
1.37.
Determine all positive integers n such that the polynomial
p (x) = x4n + x 4(n-1)
+ ... + x4 + 1
Q (x) = x2n + x2(n-1)
+ ... + x2 + 1.
is divisible by
Solution.
If x =I=±1, then
p (x)
Q (x)
(x 4 - l)P (x)
(x 4 - l)Q (x)
x4(n+l)
(x2
_
1
+ 1) (x2(n+l)
(x2(n+l)
1)
+ l)
_ l)(x2(n+l)
+ 1) (x2(n+l)
x2(n+l) + 1
x2 + 1
(x2
-
-
1)
If Q(x) divides P(x), then the last quotient must be a polynomial. If n+ 1
is odd , then x 2(n+l) + 1 is divisible by x 2 + 1 and we are done. If n + 1 is
even, then x 2 + 1 divides x 2(n+l) - 1. Thus the remainder of x 2 (n+l) + 1 when
divided by x 2 + 1 is 2 and we see that Q(x) does not divide P(x). Thus n
must be even and all even numbers have the desired property.
Example 1.38. Let f (x) = ao + a1x + .. . + a4x4 where a4 -f:. O. The
remainders of the polynomial f when divided by x - 2003, x - 2004, x - 2005,
x - 2006, x - 2007 are 24, -6, 4, -6, 24. Find the value off (2008).
Solution. The remainder when a polynomial f(x) is divided by x- a is f(a).
(This is a restatement of the fact that x - a divides f (x) - f (a).) Thus the
hypotheses can be restated as
f (2003) = f (2007) = 24, f (2004) = f (2006) = -6,
and f (2005) = 4.
Basic properties of polynomials - Part I
31
Letting f (x) = g(x - 2005) , we need to find a quartic polynom ial g with
g(O) = 4, g(l) = g(-1) = -6, and g(2) = g(-2) = 24. Since the values of
4
g a.re symmetric, we look for a polynomial of the form g(x) = 4 + ax 2 + bx .
Then we get -6 = g(l) = 4 +a+ b and 24 = g(2) = 4 + 4a + 16b. Solving
4
these two equations we find a= -15 and b = 5. Thus g(x) = 4 - 15x 2 + 5x
is one possibility. This is the only solution since if h( x) is any other quartic
polynomial with the same values at O, ±1 , ±2, th en g(x) - h(x) has degree
at most 4, but has th ese five points as roots, and hence must be the zero
polynomial. Thus
f (x) = 4 - 15(x - 2005) 2 + 5(x - 2005) 4
and
f (2008) = 4 - 15 · 9 + 5 · 81 = 274.
In the solution above we found a fourth degree polynomial f (x) with prescribed values at five points by ad hoc means . One could wonder whether
there is a more systematic way to do this. One such method is the Lagrange
Interpolation Formula. Suppose we want to find a polynomial of degree at
most n for which we have pr escribed values at n+ 1 points, say f (x i ) = ai for
i = O, ... , n. As above there is only one such polynomial since the difference
of two such polynomials would be a polynomial of degree at most n with n+ l
roots.
The polynomial (x - xo)(x - x 1) · · · (x - Xk-1)(x - Xk+1) · · · (x - xn) vanishes
at xo, x 1, ... , Xk-1, Xk+l, . . . , Xn, that is, at every one of the Xi except Xk.
Therefore the polynomial
(x - xo)(x - x1) · · · (x - Xk-1)(x - Xk+I) · · · (x - Xn)
vanishes at every Xi at Xk and takes the value 1 at Xk- Hence
is a polynomial of degree at most n (sin ce each summand has degree at most
n) that takes the value ak at x = Xk- This is the extreme ly powerful Lagran ge
Interpolation Formula. As an examp le of its utility note that it proves that
a polynomial f(x) of degree n that takes on rational values at n+ l rational
values of x has ratio nal coefficients. (If all the Xi and ai are rational, then
the formula above is a polynomial with rational coefficients, but it is the on ly
polynomial of degree at most n with f (xi) = ai.)
Basi c properties of polynomials - Part I
32
For the problem above the Lagrange Int erpo lation Formula reads
(x - 2004) (x - 2005) (x - 2006) (x - 2007) . 24
f(x)=
(-1)(-2)(-3)(-4)
(x - 2003) (x - 2005) (x - 2006) (x - 2007) . (- 6)
+
(1)(-1)(-2)(-3)
(x - 2003) (x - 2004) (x - 2006) (x - 2007) . 4
+
(2)(1)(-1)(-2)
(x - 2003) (x - 2004) (x - 2005) (x - 2007) . (- 6 )
+
(3)(2)(1)(-1)
(x - 2003) (x - 2004) (x - 2005) (x - 2006) . 24 _
+
(4) (3) (2) (1)
Then we again compute f (2008) = 274.
Example 1.39. Let P( x) be a monie fourth degree polynomial such that
f (l + 2n) = 1 + 8n for all n= l , 2, 3, 4. Find the value of P (1).
Solution. It is elear that P(2) = 9, P(5) = 65, P(9) = 513, P(l 7) = 4097.
Assume that P(l) = a. Then ,
p (X) = (X - 2)( X - 5) (X - 9)( X - 17) . a
512
l) (X - 5) (X - 9) (X - 17)
9
+
-315
.
( X - 1) (X - 2) (X - 9) (X - l 7) .
65
+
576
( X - l) (X - 2) (X - 5) (X - 17)
513
+
-1792
.
(x - 1) (x - 2) (x - 5) (x - 9)
4097
+
23040
.
·
(X -
Since P(x) is monie, examining the coefficient of x 4 in both sides, we get
l _ ~_ ~
65 _ 513
4097
1792 + 23040"
- 512 315 + 576
Then, a= 513.
Solution 2. The statement of the problem tells us that the polynomial
P( x) - (x - 1)3 - 1 = P(x) - x 3 + 3x 2 - 3x
has four real roots, i.e. x = 2, 5, 9, 17. Since the polynomial is monie we can
find that P(x) - x 3 + 3x 2 - 3x is also a monie polynomial. Therefore ,
P(x) - x 3 + 3x2 - 3x = (x - 2)(x - 5)(x - 9)(x - 17).
Setting x = l, we find that P(l)
= 1 + 512 = 513.
Basic properties of polynomials - Part J
33
(Dusan Djukic - Serbian TST 2016) Define a family of poly-
Example 1.40.
nomials by
Po (x) = x
3
-
4x and Pn+l (x) = Pn (1 + x) Pn (1 - x) - l.
Prove that x 2016divides P2016 (x) .
Solution. Clearly, Pn(x) = Pn(- x). We have
Pn+2(x)
Pn+l (1 + x) Pn+l (1 - x) - 1
(Pn (2 + x) Pn (-x) - 1) (Pn (2 - x) Pn (x) - 1) - 1
Pn (2 + x) Pn (2 - x) Pn(x) 2 - (Pn (2 + x) + Pn (2 - x)) Pn (x) •
We find that Pn+2(x) is divisible by Pn (x). Then, since Po (2) = O, we find
that Pn (2) = Ofor all even natural numbers n. Setting x = Oin the polynomial
Pn (2 + x) + Pn (2 - x) , we get Pn(2) + Pn(2) = O for all even natural number s
n. Thus Pn (2 + x) + Pn (2 - x) is divisi ble by x and since the x 1 terms cancel
it is divisible by x 2. Now, if there exists some k ~ 2 such that xk(x-2) divides
Pn (x), we have that xk+2(x - 2) divides Pn+2 (x). Since P2(x) is divisible by
x 2(x - 2), we obtain by induction that x 2n(x - 2) divides P2n(x), which gives
the conclusion.
Example
1.41.
Define a family of polynomials by
P1 (x) = 2x, P2 (x) = 2(x 2 + 1)
and for all n > 3,
Pn (x) = 2xPn-1 (x) - (x 2 - l)Pn-2 (x).
Prove that Pn (x) is divisible by x 2 + 1 if and only if n - 2 (mod 4).
Solution.
Since the characteristic polynomial of this constant coefficient
recursion is
T 2 - 2xT + (x 2 - 1) = (T - x - l)(T - x + 1),
1r
we see that the generał solution is Pn(x) = c1(x +
+ c2(x - l)n for some
constants c1 , c2 . Plugging in the given initial values, we find that
Pn (x) = (x + lt
+ (x - lt.
Let n= r + 4k, where O < r < 3. Then
Pn (x) = (x + lY((x + 1)4 l + (x - lY((x - 1)4 l.
Write the above expression as
Basic properties of polynomials - Part I
34
.
h d • ·d d by 1 + x 2 is the same
1
Then the remainder of such polynomia w en ivi e.
fi d
0 1 2 3
as the rema ind er of (-4t((x + lf + (x - lf). Checkmg r = 2 ' ' ' 'we n
that only for r = 2 the above expression is divisible by 1 + x ·
th
Example 1.42. Let p be a prime number greater than 2. Find
e least
positive integer a that can be written as
a= (x - l)f(x)
+ (xp- l + ... + x + l)g( x),
where f(x) and g(x) are polynomials with integer coefficients.
Solution. Let <I>p
(x) = xP- 1 + ... + x + 1. We find that
<I>p(x)= (x - 1)
(I:jx"
-H)
+ P·
J=O
Then ,
p
= <I>p(x)+ (x - 1) (-
I:jxp-l-i)
.
J=O
Thus a::; p. Put x = 1 in the problem statement. Then we see that
a= pg(l) "?_p
We conclude that a = p.
Example 1.43. Let P(x) be a polynomial of degree d > 1 with integer
coefficients. What is the largest number of consecutive integers that can be
contained in the set
{P(x) I x EZ}?
Solution. The answer is d. Let a1 < ... < am be m consecutive integers such
that there are distinct integ ers x1, . .. , Xm such that ·
P (xk) = ak
Vk = 1, 2, . .. , m.
Note that
Then Xk+ l - Xk = ±1. So, the numbers X1 < ... < Xm form an arithmetic
progression of common difference r = ±1. Replacing P(x) by P(-x) if necessary, we may assume the common difference is 1. Take Q (x) = (x - x 1 ) + a 1 .
It is easy to check that Q (xk) = ak for all 1 < k < m. Thus the polynomial
R(x) = P(x) - Q(x) has m distinct real roots x1 < ... < Xm. Now if m > d,
then R (x) = O and P (x) = (x - x 1) + a1, but d > 1, contradiction. Then,
35
Basic properties of polynomials - Part I
m ~ d. Now, we give an examp le of a polynomial of degree d which takes on
d consecut ive int eger values. The polynomial
P(x) = (x - l)(x - 2) • .... (x - d) + x
has P(k) = k for k = 1, .. . , d.
1.6
The composition
of polynomials
Let
and
Q(x) = bcx c + bc-I Xc- l + ... + b1x + bo
with ad and be non zero. Th en,
Th e latter expres sion is equal to adb~xcd+ ... , which is a polynomial of degree
cd. For examp le, let P (x) = x 3 + 1 and Q (x) = x 2 - 1. Then
P(Q( x)) = (Q(x)) 3 + 1 = (x 2 -1 ) 3 + 1 = x 6 -3 x 4 + 3x 2
and
Q (P (x)) = (P(x)) 2 - 1 = (x 3 + 1)2 - 1 = x 6 + 2x 3 .
Th e above example per se shows that the equality P (Q (x)) = Q(P (x)) is not
necessarily true.
Example
1.44.
The polynomials
fi (x) = 2x + 2, h (x) = x 2 + 2x - 2, h (x) = x 3 - 2x + 2
are written on a blackboard. At any time Bob writes on the blackboard a
new polynomial accord ing the following rule: if at some moment the functions
f (x) , g( x), h(x) (not necessa rily distinct) are on the blackboard, then Bob can
write a funct ion of the form f (g(x)), f(x)
+ g(x) - h(x) or f (~~;)(x) by his
own choice. Can Bob get the polynomial
p (x) = xd ± 2xd-1 ± 22xd-2 ± . .. ± 2d-lx + 2m,
where d, m are positive int egers?
Solution. Th e answer is no. Note that fi (-2) = h (-2) = h (-2) = -2 .
Th e prescribed operations don't change the value of functions on x = -2 . Bu t
no polynomial of those kind sat isfies the condition P (-2) = -2.
Basic properties of polynomials - Part I
36
Example 1.45. The polynomials x, x3, ... , x 2d+l, ... are written on a blackboard. One can write polynomials on the blackboard in this way: if at some
moment the polynomials P(x), Q(x) (not necessarily distinct) were on the
blackboard, he can write down any of the polynomials
aP(x) + b,
P(x) + Q(x),
P(Q(x)),
where a > O, b E llł. Can you get the polynomial x 2 k+l - 20x + 17 for some
positive integer k?
Solution. The answer is no. Note that all the polynomials written on the
blackboard are nondecreasing and this property is invariant under the above
operations. Thus any resulting polynomial must be nondecreasing. Let
R(x) = x 2k+l
-
20x + 17.
Then, R(O) = 17 > R(l) = -2, contradiction.
Example 1.46. The arbitrary polynomial P(x) is given. At each step we
can change the polynomial P(x) with any of the polynomials F (P (x) ), xP (x)
or P (x) + x - 1. If we start with the polynomial xk(x - 2) 2n and repeat the
process, can we get the polynomial xl(x - 2) 2m+l7
Solution. Let P(x) = xk(x - 2) 2n. We find that P (1) = 1. The operations
preserve the value of P(l), but then if the polynomial Q(x) = xl(x - 2) 2m+l
is achievable, then we must have Q(l) = 1. But Q(l) = -1, contradiction!
Example 1.4 7. Prove that for every polynomial P there exist polynomials
F and G such that
F (G (x)) - G (F (x)) = P(x).
Solution.
Set G(x) = x + 1. Then the desired equation becomes
F(x + 1) - F(x) = 1 + P(x).
Thus it suffices to show that for any polynomial Q(x) there is a polynomial
F(x) such that F(x + 1) - F(x) = Q(x). We will prove this by induction on
the degree of Q. If Q(x) =cis constant, then we see that F(x) = ex works.
Now, suppose
and let
a xn+l
F(x) = n
+ Fo(x).
n+ 1
We obtain
F (x + 1) - F (x)
=n:
1
((x + 1t+ 1 - xn) + Fo(x + 1) - Fo(x).
Basic properties of polynomials - Part I
37
Note that
(x + 1r+1 - xn+l =(n+ l)xn + n(n + 1)xn-1 + ...
2
is a polynomial of degree n with leading coefficient n+ 1. Thus
an ((x + 1r+ 1 - xn)
n+l
is a polynomial of degree at most n - 1. By the induction hypothesis we can
choose Foso that Fo(x + 1) - F0 (x) = Q 1 (x) , which gives
Q1(x) = Q(x) -
F(x + 1) - F(x) = Q(x),
as desired.
1. 7
Odd and even polynomials
We call a polynomial P(x) an even polynomial if we have P(x) = P(-x) for
all x. Moreover, we call a polynomial P(x) an odd polynomial if we have
P(x) = -P( - x) for all x. If a polynomial P(x) is an even polynomial, then
it cannot have an odd monomial. Indeed, if P is an even polynomial, then
P(x) = -P(-x) which gives
adxd +ad-1xd-l + ... + a1x +ao = ad(-x)d +ad-I (-x)d-l + ... +a1 (-x) + ao.
It follows that a1 = a3 = ... = O. Hence, P (x) = ao + a2x2 + .. . is a
polynomial in x 2 , thus P(x) = Q(x 2 ) for some polynomial Q(x). Furthermore ,
an odd polynomial has no monomial with even degree. Thus, if P( x) is an
odd polynomial, then P(x) = x R(x 2 ) for some polynomial R(x).
Since
= P(x) + P(-x)
P(x) - P(-x)
P( X )
2
+
2
and the farmer t erm is an even polynomial and the latter term is ·an odd
polynomial, we :find that every polynomial can be written as sum of an even
and an odd polynomial.
Example 1.48 . (Polish Mathematical Olympiad 2002) Prove that the graph
of the polynomial P( x) of degree greater than 1 has an axis of symmetry if
and only if there are polynomials F (x) and G (x) with deg G (x) = 2 such that
P(x) = F(G(x)).
Solution. Suppos e P(x) = F(G(x)) for some quadratic polynomial G(x) .
By completing the square we can always write G(x) = c(x - a) 2 + b for some
constants a, b, c. Then we see that G(x) = G(2a - x). Thus,
P(x) = F(G(x)) = F(G(2a - x)) = P(2a - x)
3
Basic prop erties of polynomials - Part J
and th en x = a is the axis of sym1net ry. For the sufficiency part, assume
tha t x = a is the axis of symn1etry of the graph of th e polynomial. Then
P(2a - x) = P( x). Set P1 (x) = P( x + a). Th en,
P1 (x) = P (x + a) = P (a - x) = P1 ( - x) .
Tlm ·. P1 (x) is a.n even polynomial and then P 1 (x) = F( x 2 ) for some polynomial F( x ). Th en, P (x) = F(( x - a)2), and we set G (x) = (x - a)2.
Ba sic properties of polynomials - Part II
2
Basic properties
2.1
Polynomial
of polynomials
39
- Part II
roots
By a root of a polynomi al P( x) we mean a complex numb er r such t hat that
P (r) = O. Th en , the polynom ial P( x ) is divisible by x - r. For examp le,
x = -1 , 2 are roots of t he pol ynomia l Q(x) = x 3 - 3x - 2 because
Q(-1)
Example 2.1.
the polyno mial
= Q(2) = O.
Let a, b, c be integers such that lal , Ibi, lei < 10. Consider
P (x) = x 3 + ax 2 + bx + c
such that IP(2 + v'3)1 < 10- 4 . Prove that P( x) is divisible by x 2 - 4x + 1.
Solution. Since x 2 - 4x + 1 . (x - 2 + v'3)(x - 2 - v'3) , we must prov e that
P (2 ± v'3) = O. Assume the contrary. Then
P (2 +
J3)= (26 + 7a + 2b + c) + (15 + 4a + b) V3 = m + nv'3,
where m = 26 + 7a + 2b + c and n= 15 + 4a + b are int egers. Moreover, by use
of triangl e inequalit y, one can find th at 1ml < 130 and lnl < 65. Therefore,
jm + nv'3j < 130 + 65V3 < 260.
Likewise, Im - nv'31 < 130 + 65\/'3 < 260. Moreover,
IP(2+ v'3)1
= lm+
nv'3
1= :2
__n~
Since, IP(2 + v'3)1 =I=O, we find th at (m ,n) =I=(0,0). Hence, lm 2 - 3n 2 1 > 1,
thus
1
1
jP(2 + v'3)j > Im - nv'31 > 260 > 10-4,
contr adict ion . Hence, (m, n) = (O,O) and then P (2 + v'3) = O. Fin ally, since
P (2 - v'3) = m - n\1'3
, we get P (2 - v'3) = O and we are don e.
Th ere is a theorem called the Fundamental Theorem of Algebra which states
that for every polynomial P(x) of degree d with complex coefficients , the re
are d complex numbers r 1 , ... , rd (not necessarily distinct) such that
P (x) = C (x - r1) · . . . · (x - rd)
40
Basic properties of polynomials - Part II
for some complex number C. This is often phrased as saying that any polynomial of degree d with complex coefficients has d complex roots counted with
multiplicity.
The inquisitive reader may think that the proof of this theorem is straightforward, but it is not easy. Its simplest proof is out of the scope of this
book, but you can use it as criterion to find the number of roots of an arbitrary polynomial. For example in the polynomial P (x) = x 3 - 7x - 6 we
have P (-1) = P (-2) = P (3) = O. Since the polynomial is of degree 3, we
find that x = -1, -2, 3 are the only roots of the polynomial P(x) and then
P (x) = C(x + l)(x + 2)(x - 3) for some complex number C. Comparing the
coefficient of x 3 on both sides of above identity , we find that C = 1.
Example 2.2.
(Italian Mathematical Olympiad 2012, District Round) Let
p(x) and q(x) be two distinct polynomials of degree less than or equal to 3
with integer coefficients such that
p(l) = q(l),
p(-1) = -q(-1),
p(2) = q(2),
p(-2) = -q(-2),
p(3) = q(3),
p( - 3) = -q(-3).
What is the minimum value of (p(0))2 + (ą(0)) 2 ?
Solution. Since
p(-1) + q(-1) = p(-2) + q(-2) = p(-3) + q(-3) = o,
we find that the polynomial p(x) + q(x) has x = -l, -2, -3 as roots, i.e.
p(x) + q(x) = a(x + l)(x + 2)(x + 3)
for some integer a. Analogously, we find that x = l, 2, 3 are roots of the
polynomial p(x) - q(x) . Hence,
p(x) - q(x) = b(x - l)(x - 2)(x - 3)
for some integer b =f:.O. Thus, we find that
p(x) =
q (x) =
1
2 (b(x - l)(x - 2)(x - 3) + a(x + l)(x + 2)(x + 3))
and,
1
2 (- b (x - 1) (x - 2) (x - 3) + a (x + 1) (x + 2) (x + 3)).
Now, take x = O and we get p(O) = 3(a - b) and q(O)= 3(a + b). Ther efore,
(p(0))2 + (ą(0)) 2 = 18 (a2 + b2 ) .
Since a, b are integers with the same parity and b =f:.O, we find that a 2 + b2 > 2,
hence p(0)2 + q(0)2 > 36. The equality case occurs when a, b E {-1 , 1}.
Ba sic propert ies of polynomials - Part II
41
Example 2.3.
Suppose that f (x) is a polynomial of degree 3 with leading
coefficient equal to 2 and
f (2014) = 2015, f (2015) = 2016.
Find the value of J (2016) - f (2013).
Solution. Set g (x) = f (x) - x - 1. Then, J (x) = 1 + x + g(x). Now, we
find that g (2014) = g (2015) = O, hence
g (x) = C (x - 2014) (x - 2105) (x - a)
for some C, a E IR..Comparing the leading coefficient of f(x) with the leading
coefficient of g(x) , we find that C = 2, so
g (x) = 2 (x - 2014) (x - 2105) (x -
a).
Hence,
f (2016) - f (2013)
g (2016) - g (2013) + 3
4 (2016 - a) - 4 (2013 - a) + 3
15.
Example 2.4.
Let P (x) = x 2016 + a201s x 2015 + · · · + a1x + aa such that for
all i= 1, 2, ... , 2015 we have P (i) = 2i - 1. Find the value of
P (O) + P (2016) - 2016!.
Solution. Note that the polynomial Q(x) = P(x)-2x+ l has roots in x = l,
2, ... , 2015. Since degQ(x) = 2016 and Q(x) is monie, we find that there is
only one other real root, say r. Hence,
Q (X) = (X - l) (X - 2) · . .. · (X - 2015) (X - r) .
This implies that
P (x) = 2x - 1 + (x - 1) (x - 2) · ... · (x - 2015) (x - r).
Now, set x = O, 2016 into the above identity. We get
P (O) = -1 + r · 2015!
and
P (2016) = 4031 + (2016 - r) · 2015!.
Hence, P(0) + P(2016) = 4030 + 2016!. Thus, P(0) + P(2016) - 2016! = 4030.
42
Basic properties of polynomials - Part II
By a double root of a polynomial P(x) we mean a complex number r such
that P(x) is divisible by (x - r) 2 . For example x = I is a double root of the
polynomial Q (x) = x 3 - 3x + 2 sin ce
x 3 - 3x + 2 = (x - 1)2 (x + 2).
Example
2.5.
(Croatia Mathematical
P(x)
Olympiad 2008) Let
= ax 2 + bx + c,
where a, b, c are real numbers and a =I=O. The polynomial P(x 2 + 4x - 7) has
x = I as a root and at least one double root. Find the other roots .
Solution. Define Q(x) = P(x 2 + 4x - 7). Then Q(l) = P(-2) = O. Thus,
P(x)
= a (x + 2) (x - r)
for some real number r, so
Q(x) = P (x2 + 4x - 7) = a (x 2 + 4x - 5) (x2 + 4x - 7 - r).
This implies that
Q (x) = a (x - I) (x + 5) (x 2 + 4x - ( 7 + r)) .
As the polynomial Q (x) has at least one double root, we have that this root
must be either x = I or x = -5 or the polynomial x 2 +4x-(7 + r) has a double
root. In the former case , we have r = - 2 and Q (x) = a ((x - I)(x + 5)) 2 • In
the latter case , the discriminant of the polynomial x 2 + 4x - (7 + r) must be
zero, hence 44 + 4r = O, i.e., r = -11. Therefore,
Q (x) = a(x - l)(x + 5)(x + 2) 2 .
Example
2.6.
(Australian Mathematical Olympiad 2003) Let
P(x) = x2003 + a2002x2002
+ ... + ao
be a polynomial with integer coefficients. If Q(x) = (P(x)) 2 - 25, prove that
the number of integer roots of Q(x) is at most 2003.
Solution. We prove the statement by contradiction. Assume Q(x) has more
that 2003 integer roots . Then, one of the polynomials P (x) + 5 or P (x) - 5 has
at least 1002 integer roots. Assume without loss of generality that P(x) - 5
has 1002 integer roots. Then
P(x) - 5 = S(x)(x - x1) · ...
· (x - x1002)
Basic properties of polynomials - Part II
43
for some polynomial S(x) with integer coefficients and distinct integers
x 1 , . . . , x 1002- Let r be a root of the polynomial P( x) + 5. Then,
-10 = (P(r) + 5) - 10 = S(r)(r - x 1) • ... • (r - x1002)It is elear that S (r) is a nonzero integer. Hence
l(r - x1) · ... · (r - x1002)I~ 10.
But (r - x1) · ... · (r - x1002) is a product of 1002 distinct integers. Thus,
l(r - x1) · ... · (r - x1002)I~ 12 · 22 · ... · 5012.
Hence there is no integer r such that P( r) + 5 = O. This implies that all the
integers roots of Q(x) are roots of P(x) - 5. Thus, the total number of such
roots is at most 2003.
Example 2. 7.
(M. Medinkov- Kvant M2369) We have 100 numbers. When
we add 1 to each of them , their product doesn 't change, when we add 1 to each
of them again the product doesn't change and so on. We repeat this procedure
k times and the product doesn't change. What is the maximal possible value
of k?
Solution. Let a1, ... , a100be the list of numbers and set bi = ai - a1. Define
the polynomial
P (x) = (x + b1) · ... · (x + b100).
By the Fundamental Theorem of Algebra, this polynomial cannot assume the
same value more than 100 times. But the hypothesized construction says that
P(a1) = P(l +a1) = · ·· = P(k+a1),
and we find that k ~ 99. To see that 99 can be achieved , we provide an
example. Consider the numbers
-99, -98 , ... , -1,
o.
Then for 99 steps of incrementing the numbers the product is always O.
Example 2.8.
(Mediterranean Mathematical Olympiad 2013) Prove that
th ere are third-degree monie polynomials P(x) and Q(x) such that 9 nonzero
integers whose sum is 72 are the roots of the polynomial P (Q(x)).
Solution. Let P(x) = (x - z1 )(x - z2)(x - z3) and Q(x) = x 3 - ax2 + bx - c.
Then,
P(Q(x)) = (Q(x) - z1)(Q(x) - z2)(Q(x) - z3).
44
Basic properties of polynomials - Part Il
Assume that ai, bi, ci, where i = 1, 2, 3, are the roots of Q(x) - Zi. Then,
comparing the coefficients of x 2 and x , we get
Then, 72 = 3a and a = 24. Likewise, we can easily find that a? + b/ + c? are
the same. If we choose b = 143, we get
a1
= O, b1 = 11, c1 = 13, a2 = 1, b2 = 8, c2 = 15, a3 = 3, b3 = 5, c3 = 16.
Thus, Q(x) = x 3 - 24x 2 + 143x and P(x) = x(x - 120)(x - 240).
Example 2.9.
Does there exist a polynomial P(x) = x 3 + ax 2 + bx + c
such that lei< 2017 and P(x) has three integer roots and IP(34)1 is a prime
number?
Solution. The answer is no. Let P( x) = (x - r )( x - s) (x - t) be a polynomial
satisfying the problems conditions, where r, s, t are integers. Then,
IP (34) I = I(34 - r) (34 - s) (34 - t) I ,
which must be a prime number. Then two of the three factors must be 1.
Without loss of generality assume that
134 - rl = 134- s I = 1.
and 134 - tł is a prime number. It follows that r, s > 33. Since the neare st
prime numbers to 34 are 31 and 37, then ltl > 3.
Therefore, lei= jrstl > 33 x 33 x 3 = 2067 > 2017, a contrad ictio n!
2.2
Integer and rational roots of polynomials
Example 2.10. Let P(x) be a non-constant polynomial with integer coefficients such that P (O) = 2018. What is the maximum number of distinct
integer roots of this polynomial?
Solution.
Let P (x) = adxd + .. . + ao, where d > 1 and ad =/-O. Since
P (O) = 2018, we find that ao = 2018, i.e.
P (x) = adxd + .. . + 2018.
Now assume that x = r is an arbitrary integer root. Then
'
Therefore, r must divide 2018. Since 2018 = 2 · 1009, we get
r E {±1, ±2, ±1009, ±2018}.
Basic properties
of polynomials
45
- Part II
Thus , the extreme case occurs whenever P(x) is divisible by a polynomial of
the form (x - 1) (x + 1) (x ± 2) (x± 1009). In this case, the maximum number
of integer roots are 4.
Theorein
Let P (x) = adxd + · · · + ao be a polynomial with integer coefficients. If
P( x) has an integer root r, then a 0 must be divisible by r.
Similarly , let P (x) = adxd + • • •+ a 0 be a polynom ial with integer coefficients
a
and let r = b be a rational root of P(x), where a and b are integers with b =I=O
and gcd(a , b) = 1. Then ,
ada d + ad- 1a d-lb + · · · + aobd = o.
It follows that adad must be divisible by b and aobd must be divisible by a.
Since gcd(a , b) = 1, we get
a I ao,
b I ad.
The above statement is called the Rational Root Theorem.
Rational Root Theore1n
Let P ( x) = adxd + • • • + ao be a polynomial with integer coefficients.
If P(x) has a rational root r = ~' where a, b are coprime int egers (i.e.,
gcd(a, b) = 1) and b > O. Then,
a I ao,
b I ad.
Corollary 1. ff a monie polynomial P (x) = xd + · · · + ao with integer coefficients has a rational root, then this root must be an integer.
Example 2.11. Let P (x) = adxd + · · · + ao be a polynomial with integ er
coefficients such that all of its roots are rational. Prove that there is a rational
number c such that all the roots of P ( ~) are integers.
Solution.
Consider the polynomial
X )
ad d - Ip ( ad
= x d + ad- 1 x d-1 + · · · + add-1 a 0 .
This is a monie polynomial with int eger coefficients.
Then, by the above
corollary we find that all the rat iona l roots of the polynomial
must be integ ers.
ai- 1P ( :)
Since all the roots of the polynomial P(x) are rational
46
Basic properties of polynomials - Part II
numbers, all the roots of P ( :)
must be rational, hence integers.
Take
c = ad and we are clone.
Example 2.12. Let P(x) and Q(x) be polynomials with integer coefficients
such that a and a+ 2015 are roots of P(x), where a is an integer. Moreover,
let Q(2014) = 2016. Prove that there areno integers b such that Q(P(b)) = 1.
Solution. Set P(x) = (x - a)(x - a - 2015)R( x ) for some polynomial R(x)
with integer coefficients. Assume by contradiction that there is an integer b
such that Q (P(b)) = 1. Then,
Q(x) = (x-P(b))•T(x)+l,
for some polynomial T(x) with integer coefficients. Then,
Q (2014) = (2014 - P (b)) · T (2014) + 1.
Note that P (b) = (b - a) (b - a - 2015) R(b). Since b - a, b - a - 2015 have
different parity , we find that P(b) must be even. Therefore, Q (2014) must be
odd, contradiction!
Example
2.13.
Let a be an integer. Show that the polynomial
P (x) = x 4 - 2003x 3 + (2004 + a) x 2 - 2005x + a
has at most one integer solution. Furthermore, prove that it has no multiple
integer roots .
Solution. First we prove that the polynomial cannot have any odd integer
root. Assume that r is an odd root of P( x) . Then P( r) = O can be rewritten
as
r 3 (r - 2003) + 2004r 2 + a (1 + r 2 ) = 2005r.
The left-hand side is even, but the right-hand side is odd, a contradiction.
Now suppose P(x) has a (necessarily even) root s. We will show that
P(x) :_ P(x) - P(s)
x-s
x-s
also cannot have an even root r. If it did, then
0
= _P_(r_)_-_P_(_s)
r-s
= (r + r 2 s + s 2r + s3 ) - 2003 (r 2 + rs + s 2 ) + (2004 + a) (r + s) - 2005.
3
But the right hand-side is even and the left hand side is odd, a contradiction.
P(x)
Hence -has no integer roots and thus P( x) has at most one integer root
x-s
which cannot be a double root.
47
B asic prop ert ies of polyno mia ls - Pa rt II
Example 2.14. Let P (x ) = aa.~3 + bx 2 + ex + d be a polynomial wit h intege r
coefficients . Assum e th at ad is odd and abc is even . Pr ove t hat at least one
of t he root s of th e pol ynomi al P( x) is irr at ional.
Sol ution . Assum e by contr adi ction t ha.t all the roo ts of t he polynomial P (x)
Pi c .
.
l
are ra t 10na, say - 1or i = 1, 2, 3. Th en by t he Rat ional Root Th eor em , we
Qi
'
find t hat P.i I d and Qi I a. Th en , all th e Pi and q,i are odd . By Vieta's formul a ,
we ca n wri te
-~ = Pl + P2 + P 3 = P1Q 2Q3 + P2 Q1Q3 + p 3Q1Q2
a
QI
!::_= PI P 2
a
Q2
/3 and
QI Q2Q3
Q3
+ P 3 P2 + PI P 3 = P1P2Q 3 + p3p2ą1 + PIP 3 Q2 = 'Y ,
QI Q2
Q3 Q2
Ql Q3
QIQ2Q3
Q
where gcd( /3, a:) = gcd( "f, a:) = 1. Mor eover , since bo t h t he num era.tors and
t he denomin ator s of t he abo ve rational expr essions are odd we find t hat "f, /3,
a: are all odd and th en a, b, c are odd , cont radict ion.
Let n >"2 be a positiv e int eger. Prov e t hat th e pol ynomial
xn + 2xn-I + 3xn- 2 + •••+ nx + n + l = O has no ration al root s.
Solution. Note th a t
Example
2.15.
xn + 2xn -I + 3xn - 2 + · · · + nx + n + l
(xn + xn -l + •••+ x + 1) + · · · + (x + 1) + 1
xn +l - 1 xn - l
x2 - 1 X - l
---+---+··
·+--+-x -l
x -l
x -l
x -l
xn+ 2_1
x -1
-
n -
2
x -l
(n+2) x +n+ 1
(x - 1)2
xn +2 -
It is obvious th at x = l is not a root of the pol ynomial , hence all the roots of
th e pol ynomial
xn + 2xn -l + 3xn - 2 + · · · + nx +n+ l
ar e ind eed roots of the polynomial
xn +2 - (n+ 2) x +n+ l.
Assume that x = 'E.,where q > O, is a root of the aforementioned
Q
polynomial.
Then by the Ration al Root Theorem we find that Q I 1, so Q = 1 and the
root mu st be an integer. N ow, x = - l , O are not roots of the polynomial
Basic properties of polynomials - Part II
48
xn+ 2 - (n+ 2) x +n+ 1 and we prove that this polynomial has no roots with
absolute value greater or eąual than 2. Assume by contradiction that r is a
root and Ir! > 2. Then,
2n+l Ir! <
lrn+l I Ir!
lrn+21
!(n+2)r+
(n+ 1)1
< (n+2)!r!+(n+l)
< (2n + 3) Ir!
Hence, 2n+l < 2n + 3, a contradiction if n> 2. Our proof is complete.
Note. Let P (x) = xn+I + xn + • • • + x + 1. Then,
1) xn + nxn-I + • • • + 1
P' (x) =(n+
and
xn + 2xn-l + 3xn- 2 + · · · + nx + n + 1 = xn P 1 (
!).
We know that all the roots of the polynomial P(x) have absolute value eąual
to 1, then by the Gauss-Lucas Theorem all the roots of P'(x) lie within the
convex hull of the roots of P( x). Thus, they have absolute value less than 1
and then all the roots of xn + 2xn-I + 3xn- 2 + · · · + nx +n+ 1 have absolute
value greater than 1, but we have proved that all of them have absolute value
less than 2.
Example 2.16. (Japanese Mathematical Olympiad 2009) Consider a polynomial P (x) wi th integer coefficients and let n =/=-O be an integer such that
P (n 2 ) = O. Prove that for all positive rational numbers a, we have P (a 2 ) =/=-1.
Solution. Let deg P(x) = d and set P(x) = (x-n 2 )Q(x) for some polynomial
Q(x) with integer coefficients. Assume on the contrary that there exists a
positive rational number a = <j_'where gcd(p, ą) = 1, such that P (a2 ) = 1.
p
Then
We can rewrite this as
(p2d- 2Q (::))
(ą2 _ p2n2) = p2d ,
Since Q has degree d - 1, then p 2d- 2 Q (;)
coefficients. Thus p 2"- 2 Q (~)
is a polynomial with integer
is an integer and so ą2 - p 2 n 2 is a divisor of
Basic properties of polynomials - Pa rt II
49
p d . However, gcd(p, ą2 - p 2n 2) = 1 so we find that
2
ą2 _ p2n2 = (q _ pn) (q +pn)=
±l.
But P, q, n are all positive integers, so we cannot have q +pn= ±1, hence this
gives a contradiction .
2.3
lntermediate
polynomials
value theorem,
increasing
and decreasing
If a polynomial P(x) assumes negative and positive values , then it must have
at least one real root between these values. That is, if P(a) and P(b) have
different signs, then there is at least one real number c such that P(c) = O.
Moreover , if a< b, then a< c < b.
The above statement is called the Intermediate Value Theorem (IVT) and
it is true for all continuous functions and hence polynomials. For example,
consider the polynomial P (x) = x 2 + x - 2. We have
P (O) = -2,
P (2) = 2.
Hence , there is at least one real number c E (O,2) such that P (c) = O. We
could have simply factored to get P (x) = (x - l)(x + 2) and concluded that
c = 1 is a root, but the IVT tells us that there is a root without the need to
factor .
Example 2.17. (Andryi Gogolev) Prove that for every real number a the
polynomial x 4 + a 2 x 3 + 2ax 2 + 3a 2 x + a - 1 has at least one real root.
Solution. Define, P (x) = x 4 + a 3 x 2 + 2ax 2 + 3a 2 x + a - 1. Hence,
P(O) = a - 1,
P(-1)
= -4a
2
+ 3a.
Thus P (O)+ P (- 1) = - (2a - 1)2 < O. This implies that either P (O) < O or
P (-1) < O. Since P(x) > O for all sufficiently large x > O, then P(x) must
have at least one real root in [- 1, oo).
Example
2.18.
Prove that the polynomial
P( x) = x
4
+ ax 3 + bx 2 + ex - -2b - -4l
has a real root for any rea l numbers a, b, c.
Solution.
P (
l)
Note that P ( ~)
+P (
l) =
O. Then the values of P ( ~),
have different sign or both are zero. Either way the polynomial has
at least one real root.
Basic properties of polynomials - Part II
50
Example
2.19.
Let P (x) beany arbitrary polynomial such that
P (8) + P (11) < 19 < P (12) + P (7).
Prove that there are real num bers x, y such that
X+ y = p (x) + p (y) = 19.
Solution. Define Q (x) = P (x) + P (19 - x) - 19. We have Q (8) < O and
Q (7) > O. Hence, we have at least one real root y E (7, 8), i.e.
P(y) + P(19-y)
= 19.
Hence,
19 = y + 19 - y = P (y) + P (19 - y).
We are done.
Example 2.20. (Sergei Berlov - Saint Petersburg Mathematical Olympiad
2005) The polynomials P1 (x), ... , Pn (x) are all linear polynomials whose
roots a 1 , . . . , an are distinct and in increasing order. Prove that the polynomials
P1 (x) P2 (x) · ... · Pn-2 (x) Pn (x) + Pn-1 (x),
P1 (x) · ... · Pn-1 (x) Pn (x) + P2 (x)
all have at least one real root.
Solution. Since the roots a1, ... , an of P1 (x) , . .. , Pn (x) are in increasing
order, we have
Then the sign of Pi (x) differs after and before x = ai. Set
-
Qi (x) = P1 (x) P2 (x) · ... · Pn- i+l (x)Pn (x) + Pn-i+l (x),
Then
...
)
Moreover,
... '
i= 2, . .. , n - 1
Basi c properlies of polynomials - Part II
51
Then ,
Q2(a1 )Q2(an) = Pn- 1 (a1) Pn-1 (an) < O,
Q3(a1)Q3(an) = Pn-2 (a1) Pn-2 (an) < O,
so all the Qi(x) for i= 2, . . . , n-1
have at least one real root in (a1, an)-
Example 2.21. (N. Aghakhanov) Prove that there exists 10 distinct real
numb ers a1, a2, ... , a10 such that the equation
(x - ai) · ... · (x - a10) = (x + a1) · . .. · (x + a10)
has exactly 5 distinct real roots .
Solution.
We construct an examp le. Set a 1 = -7 , a2 = -6, ... , a10 = 2.
Now as sketched here. Th e graphs of the functions y = (x - a 1 ) • . . . • (x - a10)
and y = (x + a1) · . . . · (x + a10) intersect at exact ly 5 points x = O, ±1 , ±2 .
Simplifying the equation by (x-2)(x-l)x(x+
l) (x+2) , we get the equation
(x + 7) (x + 6) · ... · (x + 3) = (x - 7) (x - 6) · . . . · (x - 3)
which reduces to
5x 4 + 235x 2 + 504 = O.
This equation has no real roots, so we are done.
y
'
I
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I
'
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'
I
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'
'I
'
'
'
, \
'
ł
ł
ł
I
I
' ,
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,
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I
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Example 2.22. Prov e that every polynomial of odd degree with real coefficients has at least one real root.
52
Basic properties of polynomials - Part II
Solution. Let P (x) = a 2d+ 1x 2d+l + ... + ao. Without loss of generality,
assume that a2d+1 > O (otherwise, consider the polynomial -P(x)).
Then
for all sufficiently large positive values of x, we have P (x) > O and for all
sufficiently large negative values of x, we have P (x) < O. Therefore, by use of
IVT for polynomials , there is at least one real root between these two values.
Example 2.23. (Ilya Bogdanov - Russian Mathematical Olympiad 2013)
Let P(x) and Q(x) be monie polynomials with real coefficients and
degP(x)
= degQ(x) = 10.
Prove that if the equation P(x) = Q(x) has no real solutions, then
P(x + 1) = Q(x - 1)
has a real solution .
Solution. Let
P (x) = x 10 + agxg + · · · + ao, Q (x) = x 10 + bgxg +···+bo.
Then
P (x) - Q (x) = (ag - bg) xg + ... + (ao - bo).
If ag =I=bg, then the polynomial has odd degree and thus has a real root,
contradicting the problem assumption. Hence, ag= bg. Now
P (x + 1) = x 10 + (10 + ag) xg + •••
Q (x - 1) = x 10 + (-10 + ag) xg + .. • .
Now,
P (x + 1) - Q (x - 1) = 20xg + ... ,
which has odd degree and has a real root.
Example 2.24. (I. Rybanov - Russian Mathematical Olympiad 2001) Let
P(x) be a polynomial of odd degree. Prove that the equation P (P(x)) = O
has at least many distinct solutions as the equation P (x) = O.
Solution.
Let x1 , . .. , Xn be distinct real roots of P(x). It is elear that
real roots of P ( P (x)) are the roots of the polynomial P (x) - Xi. Since the
polynomial P (x) - Xi is of odd degree, then the polynomial P (x) - Xi has at
least one real root. As the polynomials P (x) - Xi for i = 1, 2 ... , n do not
have common roots, there are at least n distinct real roots.
Example 2.25. (Ukrainian Mathematical Olympiad 2011) Prove that for all
real numbers a 1 , .. . , a20 17, where a2017 =I=O, there is a function f : IR.-+ IR.such
that for any real number x we have
a1f(x) + a2f(f(x)) + ... + a2011f(f(. • • f(x)) • · .) = x.
2017 times
Basi c properties of polynomials - Part II
Solution.
Let f(x)
53
= Cx for some nonzero real numb er C. Then
f(J(· · · J(x)) · · ·) = Ckx.
k times
So, the problem leads us to find a real number C such that
2
(a1C + a2C + · · · + a 2017C 2017)x = x .
2017
Then , a2011C
+· · ·+a1C-l
= O. The polynomial a 2017C 2017+· · •+a1C-l
is of odd degree and so certainly has a real root. Take C as a root and we are
clone.
Example 2.26. Consider two polynomials P(x) and Q(x). If a, b are distinct
real numbers such that P(a) = Q(b) and P(b) = Q(a), prove that the equation
P( x) = Q( x ) has at least one real solution.
Solution. Let R(x) = P( x) - Q(x). Now,
R(a)R(b)
(P(a) - Q(a)) (P(b) - Q(b))
(P(a) - P(b)) (P(b) - P(a))
- (P(a) - P(b)) 2 < O.
If P(a) -=I=P(b), then R(a)R(b) < Oand R(x) has at least one real root on (a, b),
say c. Then, O = R(c) = P(c) - Q(c). Thus , P(c) = Q(c). If P(a) = P(b),
then P(a) = P(b) = Q(a) , and the equation P(x) = Q(x) has at least one real
solution .
Example 2.27. Let ao < bo < a1 < b1 < · · · < an < bn, Show that the
roots of the polynomial
P(x) = (x + ao)(x + a1) · ... · (x +an)+ 2017(x +bo)· .. . · (x + bn)
are real.
Solution.
Since sign P(-ai) = (:-l)i and sign P( -bi) = (-l)i+ 1 for i =
O, ... , n , then P(-ai)P(-bi)
< O. It follows that the polynomial P(x) has a
real root in the interval (-b i, -ai), where i= O,... , n . Since <legP(x) = n+ 1,
all the roots of the polynomial are real.
Example 2.28.
(Razvan Gelca - USA TST 2005) Let
f(x)
n
n
k=l
k=l
t~ xk.
= L akxk and g(x) = L 2
1
Prove that if 1 and 2n+l are roots of g(x), then f (x) has a positive root less
than 2n.
Basic properties of polynomials - Part II
54
Note that
Solution.
The above expression is equal to
n
L 2t~ 1 ( (2n+l / - 1) g (2n+l) - g(l)
=
= O.
k=O
If f (2r) = O for some O< r < n we are done. If not , there are i < j such that
f(2 i)f(2J) < O. Thus , the polynomial f(x) has at least one real root in the
interval (2i, 2J) and hence we are done.
Example 2.29. (Bernd Kreussler - Irish Mathematical Olympiad 2010) Let
m > 3 be an odd number. What is the number of real roots of the following
polynornial:
fm (x) = (x - 1) (x - 2) · ... · (x - m + 1) + 1?
Since fm (1) = fm (2) = · · · = fm (m - 1) = 1, we consider the
m - 1 We prove that fm ( 2k - -1) < O. So,
numbers 2k- -1 for k = l, 2, ... , --.
2
2
2
the polynomial f m(x) has two real roots between (2k - 1, 2k) and then it has
m - 1 real roots. Observe that
Solution.
fm ( 2k _
D 4k; +~ .(-D.(-D.....
(- t=
3 ....
2m -
1) +l
Let m > 7. Then , the produ ct has m - 1 > 6 factors. Oniy two (i.e., ±½)
have absolute value less than 1, and the number of negative factors is odd (i.e.,
m-2k), so the product is negative . The absolute value of the product is always
1 1 3 5 3 5
225
(
1)
more than
·
·
·
·
·
=
> 1. This shows that fm 2k 2 2 2 2 2 2 64
2 <O
for k = l, 2, . .. ,
that
m-1
2
, and we are done. The cases m = 3, 5, remain. Note
h (x) = (x - l) (x - 2) + 1 = x 2 - 3x + 3 > O
which has no real roots. Finally,
2
fs (x) = (x - 1) (x - 2) (x - 3) (x - 4) + 1 = (x2 - 5x + 5)
which has two double roots.
Example 2.30. Let k, n be positive integers. Prove that there are monie
polynomials P1 ( x) , ... , Pk (x) of degree n with n distinct real roots such that
Basic properties of polynomials
- Part II
55
any polynomial which is a sum of a subset of these polynomials also has n
distinct real roots.
Solution.
For all 1 < i < k, define polynomial pi (x) as
Pi(x) = (x - i) (x - (i+ k)) .. . .. (x - (i+ (n - 1) k)).
Then all such polynomials are monie of degree n with exactly n distinct real
roots. Moreover , assume O < j :'.Sn- 1. Then all the roots of the polynomial
Pi (x) lie on the intervals (j k + ½, (j + 1)k + ½).Further, we can easily find
that Pi (jk + ½)> O whenever j = n (mod 2) and Pi (jk + ½)< O otherwise.
Define
Q (x) = Pi 1 (x) + Pi2 (x) +···+Pi
s (x)
where i1, . .. , is E {l , 2, ... , k}. It is elear that Q (jk + ½),Q ((j + l)k + ½)
have different signs, hence the polynom ial Q(x) is of degree n with exactly n
distinct real roots.
We say that a function f (x) is increasing on the interval (a, b) if for all a <
r < s < b, we have f(r) < f(s) . If the inequality is strict (i.e. , f(r) < f(s)
for all a < r < s < b) , we say that the function is strictly increasing. For
examp le, the functi on f (x) = x + x 3 is strictly increasing on the real line.
We say that a function f (x) is decreasing on the interval ( a , b) if for all a <
r < s < b, we hav e f(r) > f(s) . If the inequality is strict (i.e. , f(r) > f( s)
for all a < r < s < b), we say that t he function is strictly increasing. For
example, the function f (x) = -x 3 is strictly decreasing on the real line.
Example 2.31. For an even polynomial P( x) of degree 10 with nonnegative
coefficients, what is the maximum number of real roots of the equation
P (x) = r
for some real number r?
Solution. Since P(x) is even, it is the sum of even degree monomials, i.e.
4
P (x) = a10x10+ asx 8 + a5x6 + a4x + a2x 2 + ao
where a 10 > O and a8 , a6 , a4, a2, ao > O. This impli es that the function P (x) is
monotonically increasing for x > O. Therefore, its graph cuts every horizontal
line at most once with x > O. Moreover, if x = xo is any solution of equation
P(x) = r, then x = -xo is a solution too. Hen ce, the equation P (x) = r has
at most two real roots.
Example 2.32. All the coefficients of the polynomi al P(x) are positive.
Prove that the following system of equations has no distinct positive roots:
P(x)
{ P(y)
P(y)P(y + 1)
= P( x) P( x + 1).
Basic properties of polynomials - Part II
56
Solution. Since all the coefficients of the polynomial P( x) are positive, then
the polynomial is a strictly increasing function for x > O. Let x =I-y. Multiplying both equations, we get
P(x + l)P(y
+ 1) = 1.
lf x < y , then x + l < y + l, so P(x + l) < 1 < P(y + 1). Hence
P(x) = P(y)P(y
contradiction.
trivial.
+ 1) > P(y),
Let x = y. Then P(x + l) = P(y + 1) = 1 and the system is
Example 2.33. (Radu Gologan - Romanian TST 2001) Consider a polynomial P (x) of degree at least 2 such that all of its roots are distinct and real.
Prove that there is a nonzero rational number r such that all the roots of the
polynomial P(x + r) - P(x) are real.
Solution.
Let x1 < · · · < Xn be the roots of P(x). Then on each interval
[xk, Xk+1], we have that P is zero at the endpoints and cannot change sign in
the interior. Hence P has a local maximum in Yk E (xk, Xk+1) if Pis positive
on this interval and a local minimum Yk E (xk, Xk+i) if Pis negative. Set
R=
min {Yk - Xk, Xk+l - Yk}.
0'.Sk'.Sn-1
Then for all O < r < R the polynomial Q(x) = P(x + r) - P(x) has only
real roots. Indeed, assume without loss of generality that Yk E (xk, Xk+i) is a
maximum point . Then,
and so there is Zk E (xk, Yk] such that Q(zk) = O. Thus we have found n - l
real roots of the polynomial Q(x) which has degree n - l. Hence all its roots
are real. Since the rational numbers are dense on JR,choose a rational number
r E (O, R) n Q and we are done .
Example
2.34.
(K. Kokhas) Is it possible that the graph of a polynomial of
degree 20 intersects the graph of the function y =
points? 2
Solution.
: in exactly 30 distinct
X 0
The answer is no . Let the graph of P(x) = a20x20+· · ·+ao, where
a20 I=O, intersect y = : 0 at exactly 30 distinct points. Then the equation
X
1
a20x20 + · · · + ao = ---:w,
X
2
This problem needs the concept of the derivative, if you are not familiar with it, don't
worry , you can skip it!
Basic prop erties of polynomials - Part II
57
i.e.,
Q(x) = a20x60+ ... + aox40 - l
has exact ly 30 real roots. Then , the derivative of the polynomial has at least
29 distinct real roots, but
Q' (x) = 6Oa20x 59 + · · · + 4Oaox39 = x 39 ( 6Oa20x 20+ · ••+ 4Oao)
has at most 21 distinct real roots , a contradiction!
Example 2.35. (Chinese TST 2017) Find all integer pairs (m , n) such that
there exist two monie polynomials P(x), Q(x) such that degP(x) = m,
deg Q(x) = n and for all real num bers t we have
P (Q (t)) =/=Q (P (t)).
Solution. Assume m ::; n. If m = 1, then P (x) = x + a. If a = O, then
P (x) = x and hence
P (Q (x)) = Q (P (x)) = Q(x) .
Therefore , a=/=-O. If n= l , then Q (x) = x + b. In this case,
P(Q( x)) = Q(P(x))
= x +a+b .
So, n> l. Now, Q(P(x)) - P(Q( x)) = Q(x+ a) - Q(x) - a is of degree
n - l . Then if n is an even numb er, the polynomial Q (x + a) - Q (x) - a is
of odd degree and hence it has at least one real root, that is, there is a real
number t such that
P(Q(t)) = Q(P(t))
So, n must be odd and n > 3. Now we provide an example. Let
P (x) = x + l and Q (x) = x + xn .
Then, since n is even xn > O and hence
Q (P (x)) - P (Q (x)) = (x + lt - xn > O.
Thus, (m, n) = (1, 2k + 1), where k is a positive integer , is indeed a solution.
Now, let n > m > 2 and let n be an even number. We provide an example.
Let P (x) = xm and Q (x) = 2 + xn. Then,
P(Q (x)) = (xn + 2)m > Xmn+ 2m > xmn + 2 = Q(P(x)).
Thus (m, n) = (m, 2k) , where k is a positive integ er and m > 2, is indeed a
soluti on . If n is an odd number , we provide an example. Let P (x) = xm and
m
Q (x) = (x + ar, where a> 2m-l . Then,
P (Q (x)) = (x + a)mn > Q (P (x)) = (xm + at.
58
Basic properties of polynomials - Part II
For sake of this, we prove that
Note that if x > O, we have (x + a)m - xm > am > a. If x +a< O, then take
y = -x - a. We have y > O, therefore
Finally, if x < O < x + a, set b = x + a, c = -x.
Then, b, c > O. Moreover,
b + c = a, then
Thus (m, n) = (m, 2k + 1) for all positive integers k and m > 2 is indeed a
solution. Hence the answer is (m, n) = (m, 2k + 1) for all positive integers
k, m and (m, n) = (m, 2k) for all positive integers k and m > 2.
Second degree polynomials
59
3
Second degree polynomials
3.1
The form ax 2 + bx + c
A ąuadratic polynomial is a polynomial of the form of ax 2 + bx + c for some
real numb er s a, b, c with a i=-O. Furthermore , as we have seen before , a is
called t he leading coefficient and c is called the constant term.
Example 3.1.
(Alexander Khrabrov - Saint Petersburg ivlat hematical
Olympiad 2015) Do es t here exist a monie ąuadratic polynomial P( x) with
int eger coefficients such that P (P ( v'2))= O?
Solution. The answer is yes. Consider the polynomial P (x) = x 2 - l.
Example 3.2.
(Alexander Khrabrov) Does there exist a ąuadratic polynomial P (x) such that two of its coefficients are integers and
p
Solution.
(-1)
2017
= _1
2018'
p
(-1) _1
2018
=
?
2017.
If P(x) is sucha polynomial then
1
1
1
1
nd
P( x) = 2017 + 2018 - x for x = 2017 a
x = 2018.
Hence we can write any polynomial with these pr escr ibed values as
p (X) =
2;17
+ 2;18
- X + C ( X - 2;17)
(X - 2;18)
for some C. Taking C = 2017 · 2018 gives a polynomial P whose x 2 and x
coefficients are integers .
Example 3.3.
Let a, b, c be distinct real numbers such that for a quadrati c
polynomi al f (x) we hav e
f (a) = be, f (b) = ac, f (c) = ab.
Prov e that
f (a + b + c) = ab + ac + be.
Solution.
Let f (x) = kx 2 +l x+ m where k , l, mare real numb ers then:
k (a2 - b2 ) + l (a - b) = f (a) - f (b) = c(b- a)
k (a2 - c2 ) + l (a - c) = f (a) - f (c) = b(c - a)
Since a - b, a - c i=-O we have
k (a+ b) + l + c = k (a+ c) + l + b = O
Second degree polynomials
60
Thus k (b - c) = b - c. Since b - c =IO we have k = 1 then we compute that
l = -a - b - c and hence m = ab + ac + be. Thus
f (a + b + c) = (a + b + c)2 - ( a + b + c) (a + b + c) + ab + ac + be
= ab+ ac+ be
Solution 2. Note that af (a)= bf (b) = cf (c) = abc, hence the third degree
equation xf (x ) = abc has three real roots a, b, c thus
x f (x) - abc = D (x - a) (x - b) (x - c) .
Set x = O. We find that -abc = - Dabc and if abc =IOwe conclude that D = 1.
(If without loss of generality a= O, then we have f(x) = D(x - b)(x - c) and
we still conclude that D = 1.) Thus
x f (x ) - abc = (x - a) (x - b) (x - c) .
Set x = a + b + c. We find that
(a + b + c) f (a + b + c) - abc = (a + b) (b + c) (c + a) .
Thus,
(a + b + c) f (a + b + c) = (a + b) (b + c) (c + a) + abc
= (a + b + c) (ab + be + ca)
Then , f (a+ b + c) =ab+ be+ ca.
Example 3.4.
(G. Zhukov - Moscow Mathematical Olympiad 2014) Does
there exist a quadratic polynomial f (x) = ax 2 + bx + c with integer coefficients
and a not divisible by 2014 such that all the numbers f(l), f(2), . .. , !(2014)
have different remainders when divided by 2014?
Solution. Let f (x) = 1007x2 +1008x = 1007x (x + l)+x. Since 1007x(x+l)
is divisible by 2014, then
f (x) = x (mod 2014) .
Hence , f (1), f (2), ... , f (2014) have different remainders when divided by 2014.
For solving problems with unknown polynomials, we must compare the coefficients of x 2 , x 1 , x 0 on both sides of the equation . Then, set them equal to
each other .
Example 3.5.
(Peter Boyvalenkov) Find all polynomials f (x) = x 2 +ax+b
such that for all real numbers x we have
2f (x 2 - 1) = f (x - 1) f (x + 1) + x 4 + 6x 2 - 15.
Second degree polynomials
Solution.
61
Set x = 1, then we get
2f (O)= f (O)f (2) - 8.
Set x = -1 , then we get
2f (O) = f (-2) f (O)- 8.
It is obvious that
f (O) # O. Thus ,
f (2) = f (-2).
Thus a= O, hence the first equality implies that
2b = b (4 + b) - 8 or ( b - 2) (b + 4) = O.
Setting x = O gives 2(b + 1) = (b + 1) 2 - 15, which shows that only b = -4 is
possible, sof (x) = x 2 - 4 is the only possibility and it is easy to check that
this is a solution .
2. Comparing the coefficients of x 3 in both sides, we find a = O.
Now, from the coefficients of x 2 we find b = -4 and then f (x) = x 2 - 4 is the
only possibility and it is easy to check that this is a solution .
Solution
3.2
The discriminant
An interesting issue for quadratic polynomials is the existence of a criterion
that gives the number of roots of any arbitrary quadratic polynomial. This
universal criterion is based on the discriminant. Consider the polynomial
P (x ) = ax 2 + bx + c,
where a# O. Then ,
4aP (x) = 4a2 x 2 + 4abx + 4ac = (2ax + b)2 - (b2 - 4ac).
Put D = b2 - 4ac. Then , 4aP (x) = (2ax + b)2 - D. Clearly , if D < O, th en
4aP (x) > O for all real numbers x, so it does not have a real root. If D = O,
th en 4aP (x ) = (2ax + b)2 and the equation P (x) = O reduces to the equation
2ax + b = O. Thus we have only one root at x = - ba. Finally, if D > O, th en
2
the equation P (x) = O becomes the equation (2ax + b)2 = D and we hav e
. ·
± y'D . Th"1s 1mportant
.
. 1s
. ca 11
two d 1stmct
rea 1 roots x = -- b--quantity
ed t h e
2a
discriminant. Th erefore, th e discriminant provid es the following classification.
62
Second degree polynomials
of o.r 2 --1 lu f c
Discrin1iuaut
For a polynomial P (x) = ax 2 + bx + c, the quanti ty D = b2 - 4ac, is
called the discriminant and its sign leads to three cases .
Case
Example
Description
D > O
The polynomial P (x) has P (x) = x'2 - 5x + 6 with
roots x = 2, 3 and D = 1
two distinct real roots .
D=O
The polynomial P (x ) has P (x) = x'2 - 6x + 9 with
one real root , which is a the double root x = 3 and
double root.
The polynomial P (x) has
no real root.
D<O
Corollary
2. Let P(x)
D=O
P (x) = x'2 -5x+7 with no
real roots and D = -3
= ax 2 + bx + c have two distinct real roots, say r , s.
JI5 . In other words,
The absolute value of their differences is Ir - si = -
D = a 2 (r - s) 2 .
a-
Corollary 3. If D < O and a> O, then P (x) > O for all real numbers x. If
D < O and a< O, then P (x) < O for all real numbers x.
Example 3.6.
Let D > O be the discriminant of a monie quadratic polynomial P(x) . Find the number of roots of the polynomial P (x) + P(x + JD).
Solution. Assume P (x) = x 2 + bx + c. Then, D = b2 - 4c. So,
P(x)+P(x+v'D)
x
2
+ bx + c + (x + v'D) 2 + b(x + v'D) + c
2x 2 + 2 ( b +
JD)x + 2c + D + bv'D .
lts discriminant is
Hence the polynomial has a double root.
Solution 2. Since D > O, we find that P(x) has two distinct real roots, say
r, s. Moreover, assume that r < s. Then, s - r = ../D.Hence,
Q (x) = P (x) + P (x +
JD)= P (x) + P(x + s - r).
We know that Q (r) = P (r) + P (s) = O. Hence, the polynomial Q(x) h~ a
root at x = r. Moreover, we know that P(r + s - x) = P(x). Hence, we claim
Second degree polynomials
63
that x = r is an axis of symmetry of the graph of Q(x). Note that
P(2r-
Q(2r-x)
-
x) +P(r+s-
x)
P (r + s - (x + s - r)) + P (r + s - x)
P( x +s-r)+P(x)
Q (x).
Hence , the point (r, O) is the center of symmetry of graph of Q(x). Therefore,
the graph of Q(x) intersects the x-axis at exactly one point.
Solution 3. Write the equation P( x) +P(x+v'D) = Oas P(x+v'D) = -P( x).
Meanwhile, we must find the points of intersection of the graphs of P( x + ..JJ5)
and -P( x). By what we have found from the second solution, one can find
that the parabolas P(x+ v'D)and P(x) intersect at only one point, i.e. , x = r.
Since the graph obtained from these two graphs is symmetric with respect to
x = r, we can find that the graph of -P( x) intersects the graph of P( x + ..JJ5)
at exactly one point, which is x = r.
Example 3. 7.
(N. Aghakhanov - Russian Mathematical Olympiad 2013)
Let a, b, c be distinct real numbers. Prove that at least two of the following
equations have a real root:
(x - a) (x - b) = x - c,
Solution.
(x - c) (x - a) = x - b,
(x - c) (x - b) = x - a.
Define
f (x)
g (x)
h (x)
(x - a) (x - b) - (x - c)
(x - c) (x - a) - (x - b)
(x - c) (x - b) - (x - a).
Assume on the contrary that at least two of the three polynomials above have
no real roots. Since their leading coefficients are positive, then they must be
positive everywhere. Assum e without loss that f (x) > O and g (x) > O. Then
f (x) + g ( x) = (x - a) (2x - c - b) - (2x - b - c) = (2x - b - c) (x - a - 1) ,
must be positive, but clearly has at least one real root, contradiction!
Solution 2. Without loss of generality assume a < b < c and let
h (x) = (x - c)(x - b) - (x - a)
as above. Then h (a) = (a - c) (a - b) > O and h (b) = - (b - a) < O, thus
there exists a real root in the interval (a, b) and then another real root must
exist outside such interval. Analogously for
g (x) = (x - c) (x - a) - (x - b),
Second degree polynomials
64
note that g (a) = - (a - b) > O and g (b) = (b - c) (b - a) < O, so there exists
a real root on the interval (a, b) and then another real root must exist outside
such interval.
Solution 3. As in the first solution, assume f (x) > O and g (x) > O. Then
their discriminants must be negative, so
(a + b + 1)2 < 4 (c + ab) ,
(a + c + 1)2 < 4 (b + ac) .
We rewrite the above inequalities as
(a-b+1)
2
<4(c-b),
(c-a-1)
2
<4(b-c),
contradiction, because c - b and b - c cannot both be positive .
Example 3.8.
real roots:
Prove that at least one of the following equations has no
ax 2 + 2bx + 2c = O, bx2 + 2cx + 2a = O, cx 2 + 2ax + 2b = O.
Solution. Assume by contradiction that all the above equations have real
roots . Then, their discriminants must be nonnegative which says that
b2 > 2ac > O, c2 > 2ab > O, a 2 > 2bc > O.
Multiplying all the above inequalities, we get
a 2b2 c2 >
8a 2 b2c2 > O'
which gives a contradiction, and we are done.
Example 3.9.
(K. Sikhov - Saint Petersburg Mathematical Olympiad
2013) The numbers a1, ... , a10 are given. None of the following equations
has more than one real root:
.. . '
Prove that ai < 4 for all i.
Solution. By the problem assumption, all the quadratics have non-positive
discriminants. Thus we have
.. . '
Assume on the contrary that some ai exceeds 4, without loss of generality,
a1 > 4. Then, a2 > a 1 > 4 and then a3 > a2 > 4. Continuing in this way, we
eventually get a 1 > a10 > .. . > a1 > 4, contradiction!
Second degree polynomials
65
Solution 2. Take the largest number among a 1 , . .. , a 10 , say a 1 . It suffices to
prove that a1 < 4. Indeed, as above one can find that
a1
2
< 4a2 < 4a1.
Then , O< a1 < 4 and we are done.
Example 3.10. (F. Petrov- Saint Petersburg Mathematical Olympiad 2009)
Let f, g, h be quadratic polynomials such that the discriminants of the polynomials
f(x),
g(x),
h(x),
f(x)+g(x),
g(x)+h(x)
,
f(x)+h(x)
are all equal to 1. Find the discriminant of the polynomial f(x) + g(x) + h(x) .
Solution. We provide the following nice lemma.
Lemma. Denote the discriminant of the polynomial P(x) by Dp. Then,
Df+g+h = Df+g + Dg+h + Dt+h - Dt - Dg - Dh.
Proof. Let f (x) = a1x 2 + b1x + c1, g (x) = a2x 2 + b2x + c2 and
h (x) = a3x 2 + b3x + c3.
Then, we m':lst prove the following identity
2
(b1 + b2 + b3) - 4(a1 + a2 + a3)(c1 + c2 + c3)
= L(b1 + b2) 2 - 4 (a1 + a2) (c1 + c2) bi+ 4
L
L
L a1c1,
cyc
cyc
cyc
cyc
which is an easily checked algebraic identity.
Back to our problem, now we find that D f +g+h = O.
Remark. If we keep going, we will actually reach the conclusion that there
cannot be three such polynomials f, g, h. Indeed, since we have shown that
the discriminant is zero we have f + g + h = k(x - /3)2 for some k =I=-O and /3.
Then,
f(x + /3) + g(x + /3) + h(x + /3)= kx 2.
Now write
f (x + /3) = ax 2 + bx + c and g (x + /3) + h (x + /3) = Ax 2 + Bx + C.
Hence,
a + A = k,
B + b = O,
c + C = O.
Then 1 = B 2 - 4AC = b2 + 4 (k - a) c = b2 - 4ac +kc= 1 + kc. Since k =f-O,
we must have c = O. Analogously, the constant terms of g(x) and h(x) must
be zero. Then, the absolute value of the coefficient of x in every polynomial
f (x + /3),g(x + /3) and h(x + /3) must be 1 and then their sum cannot be zero.
But,
f(x + /3) + g(x + /3) + h(x + /3)= kx 2,
contradiction!
66
Second degree polynomials
3.3
Roots
In t his sectiou we use what we have learned before. That is, the role of
th e discriminaut , th e int erm ediate value theorem and th e decomposition of a
polynomial into a product of linear factors. We start with the discriminant.
Example 3.11. (F. Pet rov- Saint Pet ersburg Math ematical Olympiad 2016)
Th e graph of the quadratic polynomial 2ax 2 + bx + c, with positive leading
coefficient , is sketched on the piane. Ea ch of following lines intersects that
graph at most once:
y = ax +b ,
y = bx +c ,
y = ax+c,
y = bx+a,
y = cx+b,
y = cx+a
What is the maxima! value of ~?
a
Solution.
Consider the equation 2ax 2 + bx + c = ax + c. It is elear that
2ax 2 + (b - a) x = O.
This equation has roots at x = O, a - b, which are distinct unless a = b.
2a
Now, consider the equation 2ax 2 + bx + c = ax + b, which since a= b reduces
to
x2 = a 2a c =
~ ( 1 - ~) .
Since it must have at most one solution, we find that ~ > 1. Finally, consider
a
the equation, 2ax 2 + bx + c =ex+ a, i.e., 2ax 2 + (a - c) x + (c - a) = O. This
is equivalent to
2x
2+(1 - ~)X+(~-1) = 0.
Its discriminant is
The problem assumption entails that
.
C
1.e., - < 9. The equality occurs when c = 9a and a= b.
a
Example 3.12. (Saint Petersburg Mathematical Olympiad 2011) The monie
quadratic polynomial f (x) has exactly one real root and the equation
J (2x - 3) + J (3x + 1) = O
also has only one real root. Find the polynomial f (x).
Second degree polynomials
67
Solution. Let x = r be the roo t o f t h e polynom1al
. f (x). Since fis monie with
1
on 1y one
. h equahty
. if and only if x = r . Since we
b .rea h root ' we have f (x) >
- 0 wit
mu st em t e equality case of the resulting inequ ality f (2x-3)+ f (3x + 1) > O.
We must have
-
2x - 3 = 3x + 1 = r.
Now, the only solution of the equat 1·0 n 2x - 3 = 3X + 1 lS
· X = - 4 , W h.lC h g1ves
·
2
2
r = -11. Thus f (x) = (x + 11) = x + 22x + 121.
Solution 2. We know that the polynomial ax 2 + bx + c has only one real root
2
whenever b = 4ac . Assume that J (x) = (x - r) 2 . Then ,
f (2x - 3) + f (3x + 1)
(2x - 3 - r) 2 + (3x + 1 - r) 2
13x2 - (l0r + 6) x + 2r 2 + 4r + 10.
The above polynomial has only one real root , thus
(l0r + 6) 2 = 52(2r 2 + 4r + 10).
It follows that r = -11 and we are done.
An interesting issue for any quadratic polynomial P (x) = ax 2 + bx + c is that
whenever P(x) = P(y) for sorne x, y, then
O= P (x) - P (y) = (x - y) (a(x + y) + b)
b
im plies that x = y or x + y = - -a . The vertical line x = - {a is called the
axis of symmetry of the graph of the quad rati c polynomial (which is called a
parabola). This observation references the prop ert y that points (x , P(x)) an d
(y, P(y)) in the plane are symrnetric with respect to the axis of symm etry of
t he grap h of th e pol ynomial. Moreover, t he poin t ( - {a, 4~), which is on the
gra ph, is called the vertex of the parabola. In the following example, we use
the syrnrnetry technique.
Example 3.13. (Fedor Petrov - Saint Pet ersburg Mat hernati cal Olympiad
2016) Doe s there exist a quadr at ic polynomial P( x) with real coefficients such
that each of the following equations have at least one integer root?
P( x) = P(6 x - 1),
P( t) = P(3 - 15t).
Let P(x) be an arb itr ary second degree polynornial such that
P(a) = P(b) for sorne a, b. We find that either a = b or a, b are syrnmetric
with respect to th e axis of syrnmetry of th e graph of P(x). Keepin g this fact in
rnind, since neit her t he equati on x = 6x - 1 nor the equation t = 3 - 15t have
integ er solut ions, we find that x, 6x-1 and t , 3-15t are syrnmet ric with resp ect
Solution.
Second degree polynomials
68
to the axis of symmetry of the graph of P( x). Hence, x + 6x - 1 = t + 3 - 15t,
i.e. , 7x + 14t = 4, which has obviously no integer solution (x, t).
Example 3.14. (All Russian l\!Iathematical Olympiad 2016) Given the
quadratic polynomials fi(x), h(x) , ... , J100(x) with the same coefficients of
x 2 and x but with different constant terms. Each of the polynomials has two
real roots. For each polynomial fi(x) one root is denoted by Xi, What values
can the sum below have:
fi (x100) + h (x1) + · · · + f10o(x99)?
Solution.
Let fi (x) = ax 2 + bx + Ci. We have
h (x1) = ax1 2 + bx1 + c2 = ax1 2 + bx1 + c1 + c2 - c1 = c2 - c1.
Thus,
. fi (xwo) + h (x1) + · · · + !100(x99)
c2 - c1 + c3 - c2 + · · · + c1 - ewo
o.
Example 3.15. (Saint Petersburg Mathematical Olympiad 2011) The
teacher gave Dima and Serezha two quadratic polynomials each, wrote four
numbers a, b, c, don the blackboard and suggested to the students to plug them
into their quadratic polynomials. Serezha obtained the values 1, 3, 5, 7. Dima
managed to substitute only the first three numbers and obtained 17, 15, 13.
At the same time Dima was going to substitute the fourth number (i.e., d), it
turned out that the teacher had already erased the numbers from the board.
What would be the value of Dima's polynomial at this point?
Solution. Note that
f (a)+ g (a)
1 + 17
f(b)+g(b)
J(c)+g(c)
3 + 15
18
18
5 + 13
18.
Hence the second degree polynomial f (x)+g (x) = 18 has three distinct roots,
a, b, c. Then for all real numbers x we have
f(x)+g(x)=18.
Hence g ( d) = 18 - f (d) = 18 - 7 = 11.
Example 3.16. A quadratic polynomial J(x) has two distinct roots. For all
real numbers a and b the inequality f (a2 + b2 ) > f(2ab) holds. Prove that at
least one of the roots of this polynomial is negative.
Second degree polynomials
69
Solution. Set b = O. Then f (a2 ) > f(O). It follows that for all positive real
numbers z we have f (z) > f(O), thus we conclude that the parabola y = J(x)
is concave upward . Now, assume the contrary. Then, the minimal value of
the polynomial is attained at some point t > O. Thus for all x =/-t, we have
f(x) > f(t), in particular f(O) > f(t). Set z= t. Then,
f (t) > f (O) > f (t),
contradiction.
Solution
2. Set a= -b. Then, f (2a 2 ) > f (-2a 2 ). For all z > O, we have
f (z) > J(-z).
Assume that the parabola y = f(x) is concave upward. If both roots are
positive, then the smallest value of the polynomial occurs at some t > O.
Thus , as above, for all x =/-t we have f (x) > J(t). Now, set z = t. Then,
J(t) > J(-t) > f (t) (since t =/--t) , a contradiction. Suppose the parabola is
concave downward. Set b = 2a then
Then for all z> O we have f (¾z)> f(z). Now the polynomial has a maxima!
value , the point t such that f (x) < f(t) for all x =I-t, again assume contrary
and set z = t then
J(t) > f ( ~t) > f (t),
contradiction!
Example 3.17. (Tuymada 2012) A quadratic polynomial P(x) having two
real roots, for all x satisfies the inequality
P (x3 + x) > P (x 2 + 1).
Find the sum of the roots of the polynomial P( x).
Solution. Since x 3 +x-x 2 - l = (x-1) (1+ x 2 ), we find that x 3 +x > x 2 + l
whenever x >land x 3 +x < x 2 + l whenever x < 1 and x 3 +x = x 2 + l only
when x = l. Now we prove that the leading coefficient of the polynomial P(x)
is positive . If not, for sufficiently large x, the function P(x) is increasing, then
x 3 + x > x2 + l for large enough x implies that P (x 3 + x ) < P (x2 + 1), a
contradiction. Thus, the polynomial has minimum point , say t. For all x =f.t
we have P(x) > P(t) . Also there exists only one real number z such that
z 3 + z = t (since the polynomial x 3 + x is an increasing function!). Take x = z
in the original inequality
P(t) = P (z 3 + z ) > P (1+ z 2 ) > P(t).
70
Second degree polynomials
If 1 + z 2 -/- t, then we have P(t) = P (z3 + z) > P (1 + z 2 ) > P(t), a contradiction . Thus , t = z 3 + z = 1 + z 2 , and z = 1, t = 2. It follows that the
minimal value of the polynomial occurs at x = 2, making its sum of roots
equal to 4.
Example 3.18. (A.Golovanov) Let a, b,c be nonzero real numbers and suppose that the equation ax + 2:.= b has a real root. Prove that at least one of
X
the following equations has a real root.
C
ax + - = b - 1,
X
C
ax + - = b + 1.
X
C
Solution. Assume by contradiction that the equation ax + - = b has a real
X
root but neither of the equations
C
C
X
X
ax + - = b - 1 ax + - = b + 1
have one . We rewrite the above as
ax 2 -bx+c=O,
ax 2 -(b-l)x+c=O,
ax 2 -(b+l)x+c=0.
Since a-/- O, these are quadratics and the discriminants give
b2 - 4ac > O,
(b - 1)2 - 4ac < O,
(b + 1)2 - 4ac < O.
Adding up the last two inequalities, we get 2b2 + 2 - 8ac < O, so b2 - 4ac < -1,
which contradicts the fact that b2 - 4ac > O.
Solution
2. If a, c > O, the graph of the function ax + 2:.is similar to the
X
figure below.
y
y=b
X
71
Second degree polynomials
If b > O, the above figure shows that the equation ax + c = b + I has exactly
X
two real roots. Furthermore, if b < O, according to the above figure, we can
deduce that the equation ax + c = b - I has exactly two real roots. Finally,
X
if a > O and c < O, the graph of the function ax + ~ is similar to the figure
X
below.
y
y=b
X
According the figure above, both of the equations ax + ~ = b - I and ax + c =
X
X
b + I have two real roots.
Example 3.19. (Alexander Khrabrov) Let f(x), g(x), h(x) be three
quadratic polynomials with the same leading coefficients and distinct coefficients of x such that none of them has a real root. Prove that there is a real
number D such that the polynomials f (x) + Dg (x) and f (x) + Dh (x) have
a common root.
Solution. Write
g (x) = ax 2 + bx + c,
h (x) = ax 2 + dx + e, where b =/d.
Then the equation g(x) = h(x) reduces to bx+c = dx+e, which has a unique
e-c
root r = -. Since g(x) has no real roots, we know that g(r) =/O. Therefore
b-d
we can take D
=-:i;!, +
then f (r)
Dg (r) = f (r) + Dh (r) = O and we are
done.
As before, if P (t) P (y) < O for some real numbers t, y, then P(x) has a
real root r E (t, y) . Moreover, if a quadratic polynomial P(x) with positive
Second degree polynomials
72
leading coefficient is such that P (z) < O for some real number z, we can find
that P( x) has two distinct real roots. In particular, we must have one real
root in (-oo, z) and another in (z, +oo).
Example
3.20 .
(Saint Petersburg Mathematical Olympiad) The polynomial
f (x) = x 2 + ax + b
has roots in (O,1). Suppose there is a p E (O, 1) with p =/
-a
2
such that
f (b - f(p)) > f(p). Prove that there exist distinct real numbers p, q E (O, 1)
such that f (p) = f (q).
Solution. Let h (x) = x 2 +cx+d be an arbitrary monie quadratic polynomial.
Then
h (h(O)) = h(d) = d(d + c + 1) = h(O)h(l).
Applying this to g(x) = f(x) - f(p) = x 2 + ax + (b - f(p)), we get
O< f (b - f(p)) - f(p) = g (b - f(p))
= g(O)g(l).
Since g (p) = O, we have that g(x) has a real root in (O, 1). From g (O)g (l) > O,
we find that either g( x) has two real roots in that interval or it has a double
-a
root. But since p =I
, we see that g cannot have a double root at p. Thus ,
2
there exists a real number q =Ipin this interval such that g (q) = O. Then ,
o= g(q) = f (q) - f (p)
and we are done.
Example
3.21.
(Mathematics in School) If a, b, c are integers and
f (x) = ax 2 + bx + c
is a polynomial such that f (f(l)) = 1 and such that the equation f(x) = x
has an integer root, prove that f (l) = 1.
Solution.
Let f(l) = m . Then, f (f (1)) = f (m) = 1. Now we have the
system
a+b+c=m
2
am +bm+ c = l.
Hence
(m - 1) (am+ a+ b + l) = O.
Thus m = 1 or b =-(am+
a+ 1). Assume that b = -(am+ a+ 1). Then ,
c = 1 + m + am The discriminant of the polynomial f (x) - x is
(b - 1) 2 - 4ac = (am - a) 2 + 4.
Second degree polynomials
73
Since the polynomial has an integer root, then the above discriminant must
be perfect square, i.e.,
(am - a ) 2 + 4 = d 2
for some integer d. Hence,
(ldl- lam - al) (ldl+ lam - al) = 4.
Since both of the above factors have the same parity, both must be even and
thus
ldl - lam - al > 2.
Since lam - al = lal Im- li > 1, we must have ldl> 3. But then
4 = (ldl- lam - al) (ldl+ lam - al) > 2 (ldl+ lam - al) > 4,
Therefore, m = 1.
contradiction.
3.4
Vieta's formulas
Let r , s be the roots of ax 2 + bx + c. It is elear that one can write
ax 2 + bx + c = a( x - r) (x - s).
Since a(x - r)(x - s) = ax 2 - a(r + s)x + ars , examining the coefficient of x
and the constant term , we obtain
r+s=-
b
- ,
a
C
rs = -.
a
Moreover,
a 2 (r - s) 2 = a 2 ((r + s) 2 - 4rs) = b2 - 4ac = D.
Combining these two results we can write any symmetric polynomial in r and
s in terms of a, b, and c. For example , if r, s are roots of the polynomial
x 2 + 4x - 1 = O, then
r 3 + s 3 = (r + s) 3 - 3rs (r + s) = - 43 - 12 = -76.
Example 3.22.
th e valu e of
If z, t are roots of the polynomial x 2 + 2x + 4 = O, compute
(z +t) 7 - z 7 -t 7 .
Solution. It is elear that z 2 +2 z +4 = O. Multiply this by z -2 . We have that
2
z 3 = 8. Likewise , t 3 = 8. Then, t7 = (t 3 ) t = 64t and similarily z 7 = 64z,
that is,
(z + t) 7 - z 7 - t 7 = (z + t) 7 - 64 (z + t) .
Second degree polynomials
74
Since z+ t = -2 we find that (z+ t) 7 - 64 (z+ t) = -128 + 128 = O.
Example 3.23. For each positive integer n, let the roots of the quadratic
equation x 2 + (2n + 1) x + n 2 = O be an, bn. Determine the value of
1
1
1
(1 + a3) (1 + b3) + (1 + a4) (1 + b4) + · · · + (1 + a20) (1 + b20)·
Solution. Let x 2 + (2n + 1) x + n 2 = (x - an) (x - bn)- Substitute x = -1
in the preceding identity. We get (1 + an) (1 + bn) = n 2 - 2n . Therefore , the
20
1
, which is equal to
problem is reduced to finding the sum ~ 2
L.t n - 2n
n=3
20
1
2~
Example
3.24.
(
1
1)
1(
1
1
1)
531
n - 2 - n = 2 1 + 2 - 19 - 20 = 760 '
(Russian Mathematical Olympiad 1994) Let
P (x) = ax 2 + bx + c with a< b.
For all real numbers x, we have P(x) > O. Find the minimal value of the ratio
a+b+c
b-a
.
Solution.
b2
Since P(x) > O, we find that c > a. Then,
4
a + b + c > 4a 2 + 4ab + b2
b-a
4a(b-a)
Set t = b - a> O. Hence,
4a 2 +4ab+b
2
_
4a 2 +4a(a+t)+(a+t)
2
9a 2 +6at+t
4a(b - a)
2
4at
4at
3 9a 2 + t 2
3 3
= 2 + 4at
> 2 + 2 = 3,
where the inequality follows from the AM-GM inequality . Equality occurs
when 3a = t = b - a, i.e., b = 4a. Then, c = 4a and
P (x) = a(x + 2)2 .
Solution
2. Note that
P (-2) = 4a - 2b + c =(a+ b + c) - 3(b - a) > O
Since b - a > O, we find that
a+ b+ c > 3 _
b-a
Equality again occurs for P(x) = a(x + 2)2 .
Second degree polynornials
3.5
75
Solving inequalities
For a polyuomi al P (x ) = ax 2 + bx + c, where a > o the maximum value on the
· . 1(s, t ] occurs at eith
. er x = s or x = t. Moreover if s > -- b ort< --, b
mterva
· ·
'
2a
2a
the mm1mum value on th e int erval [s, t] occurs at the other endpoint. Finally,
if - 2ba E [s, t], th en th e minimum value on the interval [s, t] occurs at -:a·
All th e aforemention ed facts can be proven by straightforward computations
or by insp ecting th e graph of th e parabola.
Note. If a < O, consider th e polynomial -P(x), and all the minimum values
become maximum values and vice versa in the above lines.
(Moldova TST 2004) Let a, b, c be side-lengths of a triangle.
Example 3.25.
Prove that
a
2
(~
-
1)+ b2 (~ - 1)+ c G-1)> O.
2
Solution. Assume without loss of generality that bis the shortest side. Since
-C > 2 - -a., 1t su ffices to prove that
a
C
Rewriting this as
f(a)=
2
b
) a+
( ~-1
(c
,;-~
2
2
b ) a+b 2 -c, 2
we see that f (a) is a quadratic polynomial with negative leading coefficient.
Therefore to prove the inequality f (a) > O for a E [b,b + c), it suffices to prove
it for a at the ends of the intervals. For these we have
f (b) =O> O
and
and we are done.
Example 3.26. (Sergei Berlov - Russian Mathematical Olympiad 2002) Let
J,g be two monie quadratic polynomials such that they take negative values
on disjoint intervals. Prove that there exist a, /3 E IR such that for any real
number x we have af (x) + {3g(x) > O.
Solution. Assume f (x) < O on (x1, x2) and g (x) < O on (x3, x4). Without
loss of generality, assume that x 2 < x3. We draw the tangent lines to the
76
Second degree polynomials
graph of o:f (x) at x = x 2 and to the graph of (3g(x) at x = x3. Choose o:,(3
such that the absolute value of the slope of tangent lines at these points is
equal but with different signs. So, the equation of the tangent line at x = x2
is y = ax + b1 and the equation of tangent line at x = x3 be y = -ax + b2,
.
.
b2 - b1 b2 + b1)
where a > O. These two lines intersect at ( --, --. It 1s obv1ous
2a
2
b2 + b1
that
> O (see below figure).
2
Note that at x = x3, x2, we have
hence b1 + b2 = ax3 - ax2 = a (x3 - x2) > O. Since the equations
o:f (x) = ax + b1, (3g(x) = -ax
+ b2
have double roots, we find that for all x we have
o:f (x) > ax + b1 , (3g (x) > - ax + b2
Indeed, the graph of the functions o:f (x), (3g (x) are always above than ax +
b1 , -ax + b2. Adding these two inequalities we find that
o:f (x) + (3g (x) > b1 + b2 > O.
Solution 2. If the two parabolas y = f(x), y = g(x) intersect at (xo, Yo),
where Yo > O, then
f (x) =Yo+ (x - xo) (x - x1),
g (x) =Yo+ (x - xo) (x - x2) •
S econd degree polynomials
77
First suppose that x1 < x 2 < x 0. Then J(x) , g(x) > o for x > xo so the
intervals where they take negative values are both in x < x 0 . We compute
f (x) - g (x) = (x - xo)(x2 - x1).
Thus for all x < xo, we have f (x) < g(x). Thus, whenever g(x) < O, we
have that also f(x) < O, contradiction. We get a similar contradiction if
x2 < X1 < Xo so we may suppose without loss of generality that x1 < xo < x2.
Now,
h (x) = aj (x) + {3g (x) =(a+
/3)Yo+ (x - xo) ((a+ /3)x - (ax1 + /3x2)).
We choose a , /3such that ax 1 + {3x2 = (a + {3) x 0. Then ,
aj (x) + {3g (x) =(a+ /3)Yo+ (a+ /3)(x - xo)2 2: (a+ /3)Yo> O.
Another important technique is using the discriminant of a polynomial. We
can either use the positivity or negativity of the discriminant to prove inequalities on the values of the polynomial or use th e number of roots to give an
inequalit y on the discriminant . In thi s procedur e we follow one of these paths:
Strategy
(i) Find the number of roots of a polynomial.
(ii) Determine the sign of the discriminant.
Or
(i) Determine the sign of the discriminant (we hope that it will be
nonpositive).
(ii) Determine the sign of the polynomial from the leading coefficient.
Example 3.27. (Saint Petersburg Mathematical Olympiad 1998)
Let x , y , z , t satisfy the inequality
(x + y + z + t) 2 > 4(x 2 + y 2 + z 2 + t 2 ) .
Prove that there exists a real number a such that
(x - a) (y - a)+ (z - a) (t - a) = O.
Solution.
Write the equat ion (x - a) (y - a)+ (z - a) (t - a) as
2a2 - ( x + y + z + t) a + xy + zt = O
Such an a exists if
2
(x + y + z + t) > 8(xy + zt ).
But this holds since the hypothesis followed by the AM-GM inequality gives
(x + y + z+ t) 2 > 4(x 2 + y
2
+ z 2 + t 2 ) > 8(xy + zt).
Second degree polynomials
78
Solution
2. We can rewrite the original inequality as
2
2
(x - y) 2 + (x - z) 2 + (x - t)2 + (y - z) 2 + (y - t) + (z - t) < O.
Then, x = y =z=
3.6
t. Now set a= x and we are done.
Miscellaneous
problems
In this section , we provide some challenging and interesting problems. Most
of them, at first sight, have little resemblance to a quadratic polynomial. The
method behind the solution is defining a second degree polynomial, then using
its properties.
Example 3.28. (Singaporean Mathematical Olympiad 2012) Let a, b, c, d be
distinct positive real numbers such that
(b2012_ c2012) (b2012_ d2012) = 201 1.
(a2012 _ c2012) (a2012_ d2012) = 2011 ,
Find the value of (cd) 2012- (ab) 2012.
Solution. Set a 2012= A, b2012= B, c2012= C, d2012 = D. One can easily
find that x = A , B are roots of the polynomial
(x - C) (x - D) = 2011,
i.e., x 2 - (C + D) x + CD - 2011 = O. Then,
A+B=C+D,
AB=CD-2011
Thus (cd) 2012- (ab) 2012= CD - AB= 2011.
Example
that
3.29.
(Michał Rolinek)
Let a, b, c be positive real numbers such
(a + c) (b2 + ac) = 4a.
Find the maximal value of b + c and find all triples (a, b, c) for which the value
is attained.
Solution. Write the equality as (c + t) (b2 + et) = 4t, i.e.
ct 2 + (b2 + c2 - 4) t + cb2 = O.
The above quadratic equation hast= a as a positive root. Then b2+c2-4 < O.
Otherwise, for all positive real numbers t the left-hand side is positive too.
Moreover, the discriminant must be nonnegativ e, hence
(b2 + c2 - 4)
2
-
4c2 b2 > O.
Second degree polynomials
2
Now, jb + c
2
-
79
4j > l2bcl, i.e., 4 - c2 - b2 > 2bc. Then , (b + c) 2 < 4 and then
b + c < 2.
For b + c = 2, the equation has t = a as a double root.
Since the product of the roots is b2 and the root is double, then a 2 = b2 , so
a= band (a,b,c) = (s,s,2- s) for some O< s < 2.
Example 3.30. (J. Simsa - Czech-Slovak Mathematical
Suppose that the real numbers x, y, z satisfy the equalities
Olympiad 2015)
15 (x + y + z ) = 12 (xy + yz + zx) = 10(x 2 + y 2 + z 2 )
and that at least one of them is different from zero.
(i) Prove that x + y + z = 4.
(ii) Find the smallest interval [a, b] containing all numbers x, y, z satisfying
the above equalities.
Solution. (i) From the problem condition we find that
Xy
+ XZ + ZX =
a
,
12
x2 + y2 + z2 = .!!:._
10
We have (x + y + z) 2 = x 2 + y2 + z 2 + 2(xy + yz + zx ), then
a2
4
225 = 15a.
a
= 4.
15
a
(ii) Now, xy + yz + zx =
= 5. Since x + y = 4 - z, we have
12
Hence, a = 60 and x + y + z =
xy = 5 - z (x + y) = 5 - z (4 - z)= z 2 - 4z + 5.
Consider the quadratic equation t 2 + (z - 4) t + z 2 - 4z + 5 = O which has x, y
as real roots . Thus, the discriminant must be nonnegative, hence
D = (z - 4)2 - 4 (z2 - 4z + 5) = -(3z - 2)(z - 2).
Then z E
[~,2]and we are done. Similarly
x,y E
rn,2]-
Since (x, y , z) = (1, 1, 2) and (x , y, z) = (5/3, 5/3, 2/3) satisfy the equalities ,
this is the smallest possible interval.
Example 3.31. (Saint Petersburg Mathematical Olympiad 2007) Let a, b, c
be distinct natural numbers such that (a+ b) (a+ c) = (b + c) 2 . Prove that
(b - c)2 > 8(b + c).
Second degree polynomials
80
Solution.
Clearly b c must be of the same parity. Define
' '
2
f (x) = (x + b)(x + c) - (b + c)2 = x 2 + (b + c)x - (b2 +be+ c ).
Then, f (a) = O. It follows that the polynomial f(x) has an integer root and
so its discriminant must be a square
2
D = (b + c)2 + 4 (b2 +be+ c2 ) = (b - c)2 + (2b + 2c) .
Thus, D > (2b+2c) 2 . Since Dis an even perfect square, then D > (2b+2c+2) 2 .
Hence,
(b - c)2 > (2b + 2c + 2)2 - (2b + 2c)2 = 8 (b + c) + 4
Solution 2. As above, b, c are of the same parity, so b + c is even. Then both
a+ c and a+ b must be even . Hence, we can write
a + c = kn 2 ,
a + b = km 2
for same k > 2, where b + c = kmn and m =I-n. Hence
(b-c) 2 = k2 (m 2 -n 2 ) 2 = k2 (m-n)2(m+n)
2
> 2k· 1 ·4mn = 8kmn = 8(b+c).
Note that m =I-n implies that (m + n) 2 > 4mn.
3. 7
More advanced problems
In the last section of this chapter, we provide same challenging problems which
are a synthesis of the approaches we proposed here. It is possible that the
reader may face such problems in a mathematics competition. But trying
one's hand at the problems and expanding the arsenał of strategies will lead
to success .
Example
3.32.
Let a, b, c be real numbers. Define
Ll (a, b,c) = max (la - bi, lb - cl, le - al).
lax
2
Assume that for all x E [O,1] we have
+ bx +
value of k such that Ll (a, b,c) < k.
Solution. Let f (x) = ax 2 + bx + c. Since
f (O)= c, f
(21) 4 + 2 +
= a
b
ci< 1. Find the minimal
c and f (1) =a+ b + c,
we can say that
a= 2f (O)+ 2f (1) - 4f
(½), =
b
- 3f (O)- f (1) +4!
G)·
Second degree polynomials
81
Th erefore,
Ja-bi= 5/(0)+3/(1)-8/G)
:o;5J/(0) J+31f(!)J+8
Analogously,
Jb - ej = -4/ (O) - f (l) + 4/
t(D
< 16.
G):,;
g
and
Jc-aJ=
-/(0)-2/(1)+4/G)
<7 .
Hence,
~(a,b ,c):s;l6.
The equali ty case occurs if f (x) = 8x 2 - 8x + 1.
Example
3.33.
(Russian Mathematical Olympiad 2011) Let
P(x)
= x 4 + ax 3 + bx 2 + ex+ d, Q(x) = x 2 + px + ą.
If both P and Q take negative values in a common inte rval with length greater
than 2, prove that there exists x 0 such that P(x 0 ) < Q(x 0 ).
Solution. By assumption P and Q have two common real roots at the end points of the interva l where th ey take negative values. By a trans lation we
may assume these are at x = O and at x = r > 2. It follows that
Q(x) = x(x - r) and R(x) =
~i:!
is a monie quadratic polynomial. Assume on the cont rary that P(x) > Q(x)
for all x . Th en for x < O or for x > r, we have Q(x) > O and hence R(x) > 1.
For O< x < r, we have Q(x) < O and hence R(x) < l. It follows (by the IVT
applied to R(x) - 1), that we must have R(O) = R(r) = 1. Thus
R(x) = 1 + x(x - r).
But since Pand Q are both negative on (O,r), we must have R(x) > Oon (O,r)
and from this formula we com pute that R(r/2) = 1-r 2 /4 < O, a contradict ion.
Solution
2. By problem assumption, we have
Q (x) = (x - x1) (x - x2),
where x 2 - x 1 > 2. It is elear that x2,x1 are also roots of P(x). Hence,
P (x) = (x - xi)(x - x2) (x 2 + Ax + B).
82
Second degree polynomials
Assume the contrary. Then for all real numbers x we have P(x) > Q(x). Now
P (x) - Q (x) = (x - x 1) (x - x2) (x 2 + Ax + B - l) > O.
Hence, x1, x 2 must be the roots of x 2 + Ax + B - l. Otherwise, the polynomial
changes its signs in the neighborhoods of x 1 and x2. So,
+ X2 = -A, X1X2 = B - l.
The polynomial x 2 + Ax + B must be nonnegative, otherwise it has two roots
X1
say x3, X4 and then the polynomial P (x) takes negative values outside (x1, x2),
a contradiction. Then the discriminant of the polynomial x 2 + Ax + B must
be negative. But
contradiction. Thus, our assumption is wrong.
Example 3.34. (A. Golovanov) Does there exist a quadratic polynomial
P(x) that on natural numbers assumes only values that are powers of two?
Solution. Let P (x) = ax 2 + bx + c. If a < O, then for all sufficiently large
values of x the polynomial is negative which contradicts the statement of the
problem. Thus, a > O. Then for all sufficiently large positive integers n we
have f (n+ 1) > f(n). Since both of them must be powers of two, then
f(n + 1) > 2f(n),
i.e.
(a(n + 1))
Hence ,
2
+ b(n + 1) + c > 2 (an 2 + bn + c).
an 2 + (b - 2a) n+ c - a - b < O.
Since a > O, for all sufficiently large positive integers n the above inequality is
false, contradiction!
Example
mials
3.35.
(Cristinel Mortici - Gazeta Matematica) Find all polyno-
P (x) = ax 2 + bx + c,
where a is a real number and b,c are integers such that for all positive integers
n we have
n - 2,/n < P ( 1+ ~ + ... +
Solution.
}n)< n - ,In.
Using the inequality
1
----;==
2Jn+l
--
1
< Jn + l - ./n< 2 t;;;'
yn
83
Second degree polynomials
we get
I
I
2Jn + I - 2 < I + v'2+ ... + Jn < 2vfn,- 1.
1
. ~ + .. . + - - can be made arbitrarily large by taking n
v2
../n
large enough. Therefore if a < O we get
lt is elear that I+
P ( 1+
~ + ... + ),, ) < O < n - 2\/r!
for large n, which contradicts the assumed lower bound . Thus, a > O. Then
for all sufficiently large C, the polynomial P( x) must be increasing for x > C.
Hence
~ + ... + ;,, ) < P(2vln, - I).
P(2Jn + 1 - 2) < P (I+
So, we have following inequalities
P (2vn - 1) > n - 2vfn,, P (2Jn + 1 - 2) < n - Jn < n +
I
2 - Jn + l.
Setting 2.,/n - 1 = x in the first of these and 2Jn + 1 - 2 = x in the second,
we get the following inequalities
x2
3
X
p (x) > 4 - 2 - 4'
x2
x
I
P(x) < - +- - 4
2
2'
for infinitely many x. Then,
x2
x2
3
x
l
3
x2
4
4
x
2
2
1
1
x
4 - 2 - 4 < p (x) < 4 + 2 - 2'
Le. ,
x2
x
2
l
- - - - - < ax 2 + bx + c < - + - - 4
Dividing by x 2 ,
1
1
b
3
c
l
4 - 2x - 4x 2 < a + ; + x 2 < 4 + 2x - 2x 2 •
Taking x sufficiently large, we find that
1
1
-4 <a<-.
- - 4
Hence, a =
1·Now we get th e following inequality
x
2
3
4
x
2
I
2
- - - - < bx + c < - - - .
Second degree polynomials
84
Dividing by x and taking x sufficiently large, we obtain
1
1
-- 2 -< b -< -2
Since bis an integer then b = O. Now, set n= 1. We get
1
-1 < p (O)= 4 + C < o.
Hence --
5
1
< c < --4· Since cis an integer, we find that c = -1. Hence,
4
P(x)=--1
x2
4
is the only possibility.
To see that this polynomial satisfies the desired inequalities we have to show
that
2(vn-1)
< 1 +~+
... +)n<
2/n-
vn+ 1.
These are easy to derive from the inequalities above using
2Jn
+ 1 - 2 > 2(Jn - 1) and 2vfn - 1 < 2/ n - Jn + 1.
(To prove the second of these square both sides and cancel.)
85
Third degree polynomials
4
Third degree polynomials
4.1
Roots and graph
The graph of a third degree polynomial P(x) = ax 3 + bx 2 +ex+ d with a> O
is similar to one of the below graphs. The right figure is a graph of a strictly
increasing function which has only one real root. The left graph is a function
which is strictly increasing on (-oo, a) U (b,+oo) and it is decreasing on (a, b).
In this case we can have one, two (in this case we have a double root) or
three real roots. This depends on the relative location of the coordinates of
points (a, P(a)) and (b, P(b)) (see the figure and consider the intersections of
horizontal lines and the graph). For example, the graphs of the polynomials
x 3 and x 3 - 3x are respectively instances of the former case and the latter
case.
I
GJ [O
I
Corollary 4. The polynomial ax 3 + bx 2 + ex + d = O has either one or three
simple real roots or one simple real root and one double root.
Example 4.1.
(Alessandro Ventullo - Mathematical Reflections U403)
Find all polynomials P(x) E IR[x]of degree at most 3 such that
p ( 1 _ x(3x/
1))-
(P(x)) 2 + p ( x(3x -1) _ 1) = 1
2
for all x E IR.
Solution. Let P(x) = ax 3 + bx 2 +ex+ d be a third-degree polynomial with
a, b, e, d E IR, a -=/-O, such that
Third degree polynomials
86
Setting x = -1 , x = O, x = 1 into the given relation, we obtain
P(O) - (P(-1)) 2 + P(l)
P(l) - (P(0)) 2 + P(-1)
=
2
P(-1) - (P(1)) + P(O) =
1
1
1.
Adding these three equations side by side, we get
(P(-1)
so P(-1)
-1) 2 + (P(O) -1) 2 + (P(l) -1)
2
= O,
= P(O) = P(l) = 1. lt follows that
d
=
=
a+b+c+d
-a+b-c+d
1
1
1.
We obtain d = 1, b = O and c = -a. So, P(x) = ax 3 - ax + 1.
Setting x = 2 into the given relation, we have
P(-6)
- (P(2) )2 + P( 4) = 1,
which gives
-36a 2 - 162a + 1 = 1.
Likewise, setting x = -2 into the given relation, we have
P(-4)
- (P(-2))
2
+ P(6) = 1,
which gives
-36a 2 + 162a + 1 = 1.
Subtracting, we get 264a = O, i.e., a= O. So, we obtain P(x)
= 1.
Solution 2. As before, we get P(-1) = P(O) = P(l) = 1.
Consider Q(x) = P(x)-1. Since Q(-1) = Q(O) = Q(l) = Oand degQ(x) = 3,
then Q(x) = a(x - l)x(x + 1) for some a E JR. Clearly, Q(x) is an odd
polynomial. Setting x = 2 and x = -2 into the given relation and using the
fact that P(x) = Q(x) + 1, we get the equations
Q(-6)+1-(Q(2)+1)2
+Q(4)+1
Q( - 4) + 1 - (Q( - 2) + 1)2 + Q(6) + 1
1
1.
Adding both equations side by side and using the fact that Q(x) is odd, we
get
2 - (Q(2) + 1)2 - (-Q(2) + 1)2 = o.
An easy computation gives Q(2) = O. So, Q(x) has four roots and it follows
that it must be Q(x) = O for all x E JR,i.e., P(x) = 1 for all x E JR.
Third degree polynomials
87
Example 4.2.
(Michel Bataille - Crux Mathematicorum
complex roots of the polynomial
4224) Find the
6
16x - 24x5 + 12x4 + 8x3 - 12x2 + 6x - 1.
Solution.
Observe that
6
16x - 24x5 +
12x4 + 8x3 - 12x2 + 6x - 1
(2x2 ) 3 + (2x2 ) 3 + (2x - 1)3 - 3(2x2 )(2x 2 )(2x-:--1).
Using the identity
3
3
3
a + b + c
-
3abc = (a + b + c) (a 2 + b2 + c2 - ab - be - ca),
we get
(2x2 ) 3 + (2x2 ) 3 + (2x-1) 3 -3(2x 2 )(2x 2 )(2x-1) = (4x2 +2x-1)(2x
2 -2x+
1)2 .
.
. l ic1ty
. ) are x = -1 ± v'5, -1 ±-.i
T hust he roots (1gnoring
mu l tip
4
2
Example 4.3.
(Russian Mathematical Olympiad) The following figure is
the graph of the polynomial f(x) = ax 3 + bx 2 +ex+ d. The projections of
AB, CD , EF on the line Ox are A' B', C' D', E' F', respectively.
y
o
X
Prove that C' D' = A' B' + E' F'.
Solution. Assume that the points A, D, E lie on y =pand the points B, C, F
lie on y = q. Then the equations f(x) = q and f(x) = p have three real roots
X1 < x2 < x3 and YI < Y2 < y3, respectively. Note also that
X1 + X2 + X3
b
= Yl + Y2 + Y3 = --.a
Third degree polynomials
88
As shown in the figure we obtain that
A'= y 1 ,
B' = x 1 ,
C' = x2,
D' = Y2,
E' = Y3,
Now, A' B' = x 1 - y 1, E' F' = x 3 - y 3 and C' D' = y2 prove that
Y2 - x2 = x3 - Y3 + X1 - Y1
x2.
F' = X3.
We now need to
or that x 1 + x 2 + x 3 = y 1 + y2 + y 3 which is obviously true.
It is important to notice that when we find that x = r is a root of the polynomial ax 3 + bx 2 + ex + d, then ar 3 + br 2 + er + d = O. We will use this simple
fact in different situations, like the following.
Example 4.4.
Let p be a root of the polynomial x 3 + bx + e = O. Prove
that b2 > 4pe.
Solution. Note that p 3 = -bp- e. If b2 < 4pe, then
b2 < 4pe = 4p(-p 3 - bp).
Now, 4p4 + 4bp2 + b2 = (b + 2p2 ) 2 < O, a contradiction.
Example
4.5.
(Russian Mathematical Olympiad 2014) The equation
2
3
x + *X + *X + * = 0
is written on a blackboard. Pete and Bazil take turn to replace the asterisks
with rational numbers. At first, Pete replaces any asterisk, then Bazil replaces
one of the remaining ones and finally Pete replaces the last one. Is it true that
Pete has a strategy so that the difference of two real roots of the obtained
equation is 2014?
Solution. Pete can play so that O and 2014 are the roots of the equation.
Indeed, on his first move Pete makes the constant term zero. After Bazil's
move, we can assume that the polynomial written on the blackboard is one of
3
2
x + ax + *X = O, x 3 + *X2 + ax = O.
In the former case Pete replaces the* with -2014(a+2014), while in the latter
.
20142 + a
case he replaces the * w1th 2014
Solution 2. We prove that Pete can play so that the polynomial
3
2
x + *X + *X + *
is divisible by x 2 - s 2 for any s, which yields s, -s as roots of the polynomial.
Taking s = 1007 solves the problem asked. On his first move, Pete sets the
coefficient of x to -s 2 . After Bazil's move, the blackboard will have either
x 3 + ax 2 - s 2x +*or x 3 + *X2 - s 2 x + e. In the former case, Pete replaces the
e
* with -as 2 and in the latter case he sets - 2 instead of*·
s
Third degreepolynomials
Example 4.6.
such that
89
Let P(x) be a third degree polynomial with real coefficients
IP(l)I = IP(2)I = IP(3)I = IP(5)I = IP(6)I = IP(7)I = 12.
Find all the possible values of IP(O)I.
Solution. Without loss of generality we can assume that the leading coefficient of the polynomial P( x) is positive. Since the two equations P( x )-12 = O
and P( x ) + 12 = O have at most three real roots, we obtain that each of them
has exactly three real roots which are in the set {l , 2, 3, 5, 6, 7}. Then, the
graph of the polynomial P(x) must be S-shaped as you can see below.
Now, we can easily prove that x = 2, 3, 7 are the roots of P(x) = -12 while
x = 1,5, 6 are the roots of P(x) = 12, hence
P(x) = C(x - 2)(x - 3)(x - 7) + 12 = C(x - l)(x - 5)(x - 6) - 12.
Setting x = O in the above equality, we get that
P(O) = -42C + 12 = -30C - 12.
Then C = 2 and P(O) = -72. Since we assumed that the leading coefficient
of the polynomial is positive we have also to consider the case in which is
negative. In this case we have C = -2 and P(O) = 72. So, IP(O)I= 72.
Solution 2. Let P(x) = ax 3 + bx 2 +ex+ d. Then, the sum of the roots and
the sum of the pairwise products of the roots of the equations P(x) = 12 and
P(x) = -12 are equal. Hence, the sum of their squares must be equal. Now
Third degree polynomials
90
we have to partition the numbers 1, 2, 3, 5, 6, 7 into two sets with equa l sums
and equal sums of squares. The only possibility is {2, 3, 7} and {1, 5, 6} and ,
now we can continue as above (or we can consider the product of the roots) .
4.2
Vieta's formulas
lf r, s, t are roots of the polynomial ax 3 + bx2 + ex + d, we can write
ax 3 + bx2 +ex+ d = a(x - r)(x - s)(x - t).
Comparing the coefficients of x 2 , x 1 , x 0 on both sides, we have
r+s+t
=--
b
a
e
a
rs + st+ tr = d
rst = -- .
a
Example 4. 7.
Let r, s, t be the roots of the equation
x 3 + 9x 2 - 9x - 8 = O.
Find the value of
(r + s)(s + t)(t + r).
Solution.
Note that
P(x) = x 3 + 9x 2 - 9x - 8 = (x - r)(x - s)(x - t).
Since r + s + t = -9, we obtain
(r + s)(s + t)(t + r) = (-9 - r)(-9 - s)(-9 - t) = P(-9) = 73.
Example
4.8.
Let a < b < e be the three real roots of the equation
x 3 - 3x + 1 = O.
Then,
1 - a + -1 - -b + -1- c.
(1.) F"m d t he va 1ue of A = -l+a
l+b
l+c
(ii) Write an equation with roots a2 - 2, b2 - 2, c2 - 2.
(iii) Find the value of (a2 - c)(b2 - a)(c 2 - b).
Solution. (i) Note that
A+ 3 =1-a+l+l-b+l+l
l+a
l+b
-c +l=
l+c
2(
1
l+a
+-1-+_1_)·
l+b
l+ c
Third degree polynomials
Moreover , let x =
(:
91
1
1
+a. Then , a= x - 1. Since a3 _ 3a + 1 = O, we obtain
1
r-
- 1
3 (:
-1)
+ 1 = o,
i.e.
3x 3 - 3x + 1 = o.
1
1
1
Then, 1 + a, 1 + b, + c are the roots of the previous equation. Hence ,
1
1
1
--+--+--=O
l+a
l+b
1
l+c
'
and A= -3.
(ii) Note that a+ b + c = O, ab+ be+ ca = -3 and abc= -1. Moreover ,
P(x) = x
3
-
3x + 1 = (x - a) (x - b) (x - c).
Now, we can find that
Also,
( a 2 - 2) (b - 2) + (b - 2) (c - 2) + (c - 2) (a - 2)
2
2
2
2
2
= a 2 b2 + b2 c2 + c2 a 2 - 4 (a 2 + b2 + c2 ) + 12
=(ab+ ac+ bc)2- 2abc( ~)
-4(6)
+ 12 = -3 ,
o
and finally
(a 2 - 2) (b2 - 2) (c 2 - 2)
= ( v'2- a)(v'2- b)(v'2- c)(-v'2 - a)(- v'2- b)(-v'2 - c)
=P(h)P(-h)
= (v'2+1) (-v'2+1) =-1.
Then a 2 - 2, b2 - 2, c2 - 2 are the roots of the polynomial x 3 - 3x + 1 = O.
This conclusion implies that {a 2 - 2,b 2 - 2,c 2 - 2} = {a,b,c}.
(iii) Note that
71> O,
( 9)= 125
37
> O, P(l) = -1 < O, (8)= 125
> 0.
5
P(-2)
P(O)
= -1 < O, p -5
P
Third degree polynomials
92
Th en , we get that
-2 <a< -
9
5
<O<
b< 1< c<
8
5,
which means that b2 < c2 < a 2 and that b2 - 2 < c2 - 2 < a 2 - 2. We obtain
that b2 - 2 = a, c2 - 2 = b and a 2 - 2 = c, thus
a 2 - c = b2 - a = c2 - b = 2,
which leads to (a 2 - c)(b2 - a)(c 2 - b) = 8.
Example
4.9.
(Singaporean Mathematical Olympiad 2013) Let
3
a+ b + c = 14, a 2 + b2 + c2 = 84 and a3 + b3 + c = 584.
Find max(a , b, c).
Solution. It is elear that ab + ac+ be = 56 and using the identity
a3 + b3 + c3 - 3abc = (a + b + c)( a 2 + b2 + c2 - ab - ac - be)
we find that abc = 64. Hence a, b, c are roots of the polynomial
x 3 - 14x 2 + 56x - 64.
By factoring we find the roots of the polynomial
x 3 - 14x2 + 56x - 64 = x 3 - 64 - 14(x 2 - 4x) = (x - 4)(x 2 - lOx + 16)
= (x - 4)(x - 2)(x - 8).
Hence {a , b,c} = {2, 4,8} which gives max(a , b,c ) = 8.
Example 4.10. (Saint Petersburg Mathematical Olympiad 2008) Let P(x)
be a cubic polynomial with integer coefficients and three positive irrational
roots . Prove that the roots can't form a geometrie progression.
Solution. Let P( x) = a3x3 +a 2x2 +a1x+ao , where a3, a2, a1, ao are integers.
Assume that the roots of the polynomial form a geometrie progression with
ratio q, ląl =/ 1, O and that the roots of the polynomial are a, aq, aq2 . By
Vieta 's formulas we find that
a(l + q + ą2) = - a 2 and a2 q (1 + q + ą 2 ) = ai .
a3
Hence,
a3
a2 ą ( 1 + ą + ą2 )
a1
aq = a (1 + ą + ą2 ) = - a2'
which means that one of the roots of the polynomial must be rational, contradiction.
Third degree polynomials
93
Example 4.11. The polynomial x 3 + ax 2 + bx + c, with c =I O, has three
distinct real roots. Prove that the polynomial x3 _ bx 2 + acx - c2 also has
three real roots .
Solution. Assume that r, s, t are the roots of x3 + ax 2 + bx + c. Then ,
r + s + t = -a ,
rs +st+ tr = -b ,
rst = - c,
which means that
rs ·st+ rt ·st+ rt • rs = rst(r + s + t) = ac
and
rs ·st · tr = (rst) 2 = c2 .
Hence , rs, rt , ts are roots of the polynomial x 3 - bx 2 + acx - c2 .
Example 4.12. (Titu Andreescu - Mathematical Reflections S189) Let a , b, c
be real numbers such that a < 3. All the zeros of the polynomial
P( x) = x 3 + ax 2 + bx + c
are negative real numbers . Prove that b + c =I4.
Solution. Let
P(x) = x 3 + ax 2 + bx + c = (x + r)(x + s)(x + t)
for some positive real numbers r, s, t. Then, by Vieta's formula , we obtain
that
a= r + s + t, b = rs + tr + sr, c = rst.
By the AM-GM Inequality , we get
~>
3 -
~
( )
1/ 2
>cl / 3_
3
-
since a < 3, this means that c < 1, b < 3, sob+ c =I-4.
4.3
More advanced
problems
Suppose we have a cubic polynomial with real coefficients and we want to
study how many real roots it has. In generał such a polynomial has the form
T( x ) = mx3 + nx 2 + kx + u where m, n, k, u are real and m =I-O. If we replace
T(x) by T(x), its roots don't change. Thus we can restrict our study to monie
m
polynomials which are of the form T(x)
if we consider the polynomial T (x -
= x 3 + ax 2 + bx + c. It 's elear that
i) instead of
T(x), then all the roots
Third degree polynomials
94
get a shift of ~- In particular, the reality or complexity of the roots doesn't
3
change . Note that
T ( x - -a) = x 3 + (b - a 2 )x + c - -ab - -a3 = x 3 + px + q.
3
3
9
Hence , if we are interested in the roots of cubic polynomials with real coeffi.cients, then we can restrict to the polynomials of the form x 3 + px + q.
Now, consider the expression
It 's elear that if r, s, t are real, then D > O and that if we have a double root,
then D = O. Consider now the case in which we have a real root, let's say r,
and two complex roots which are conjugate, say s = x + iy, t = x - iy, where
x, y E JRand i 2 = -1. Now,
(r - s) 2(s - t)2(t - r) 2 = (r - x - iy) 2(r - x + iy)2(2iy) 2
-4y2((r - x)2 + y2)2 < O.
D
Hence, the above expression has the same use as the discriminant
quadratic polynomial.
of the
Example 4.13. For a polynomial x 3 + px + q compute the value of D as a
function of p and q.
Solution. Assume that r, s , t are the three roots of the polynomial x 3 +px+q.
Then
r + s + t = O, rs +st+ tr = p, rst = -q.
We also have that r 2 + s 2 + t 2 = -2p. We also have
r 2 + s 2 + rs = (r + s )2 - rs = t 2 - rs = -rs - st - tr = -p,
which means that
(r - s) 2 = -p - 3rs.
It follows that:
2
D = (r - s)2(s - t) (t - r) 2 = -(p + 3rs)(p + 3st)(p + 3tr)
3
3
D = -(p + 3p + 9p · rst (r + s + t) +27r 2s 2t 2) = -(4p 3 + 27q2).
'-..,-,'
o
Now we have the following important criterion:
Th ird degree polynom ials
Diseriminant
95
of ., : . /V - ,1
For the polynomial x 3 +p:r+ą. the quantity D = -(4p3 +27ą2 ) , characterizes the situation of its roots.
If
4p + 27ą 2 = 108 (ą + P )
4
27
2
3
3
<O
'
then P(x) has three distinct rea1 roots.
If
2
3
4
27
2
4p + 27ą = 108 (ą + P ) > O
3
'
then P(x) has one real root and two complex roots.
If
3
4p
+ 27 q2 = 108 ( ~ + ;; ) = o,
then P(x) has one double root and the other root is either a simple root
or equal to the double root.
Remark. We can find the coordinates of local maxima and minima in the graph
of the polynomial x 3 + px + q. It 's obvious that we can consider the function
x 3 + px , since an horizontal translation doesn t change the y-coordin ate. At
these points the equation x 3 + px = m has a doubl e root , hence we can name
the roots as r , r , t. Now ,
x 3 + px - m = (x - r) 2 ( x - t).
By Vieta 's formulas we obtain that 2r + t = O, r 2 + 2rt = p , m = tr 2 . Then ,
2
p = -3r 2 and m = -2r 3 • Hence, r = ±Ma:nd
m = :i: :
which
M,
means that we have a local minimum and maximum if p < O.
Example 4.14. Let a, b, c be non-zero real numbers. Find the maxima! number of real roots of the polynomial
P(x) = (ax 3 +bx+c) (bx 3 +cx+a) (cx 3 +ax+b).
Solution. Let
P 1 (x) = ax 3 + bx + c,
3
P2(x) = bx +ex+ a,
P3(x ) = cx 3 + ax + b.
Assume that a1, a 2 , a3 are the roots of P1 ( x) , b1, ~ , b3 are the roots of P2 ( x)
and c1 , c2 , c3 are the roots of P3(x). If all of them are real , then
+ a~ + a~,
bi+ b~ + b~ and cy+ c~ +
must be positive. Note that
ci
ay
Third degree polynomials
96
However
(-2D(-2D(-2~)= -8,
which is negative . This proves that the polynomial P(x) has at least two nonreal roots and at most 7 real roots. We now prove that there are a, b, c for
which P(x) has 7 real roots. Take (a, b, c) = (1, -6, 4). Then the first factor
ax 3 + bx + c = x 3 - 6x + 4 = (x - 2) (x 2 + 2x - 2)
has three real roots at x = 2 and x = - l ± v'3.The second factor
bx3 +ex+ a= -6x 3 + 4x + 1 = f(x)
has three roots since f(-1) = 3 > O, f(-1/2) = -1/4 < O, f(O) = 1 > O,
/(3/4) = 47/32 > O, and /(1) = -1 < O. The third factor
cx 3 + ax + b = 4x 3 + x - 6 = g(x)
has only one real root which is between 1 and 2 since g(l) = -1 < O and
g(2) = 28 > O. The roots off and g lie in the intervals (-1, -1/2), (-1/2 , O),
(3/4 , 1), and (1, 2) hence they are distinct and not equal to 2 or -1 ± v'3.
Thus we get 7 distinct real roots.
ą2
p3
Solution 2. We know that whenever D = - + - < O, the polynomial
4
27 -
Q(x) = x 3 + px + q
has three real roots. Assume now that P1(x), P2(x) and P3 (x) have all three
real roots . Then,
c2
-+-<O
2
4a
b3
27a 3 -
'
a2
c3
4b
27b3 -
-+-<O
2
'
b2
a3
4c2 + 27c3 < O,
be a
which means that ;, b' ~ < O and so are their products. However, we know
that their products is equal to 1, contradiction. Now take the example provided in the first solution.
Example 4.15. The polynomial ax 3 - x 2 + bx - l = O has three positive
real roots. Prove that
(i) O < 3ab < 1;
(ii) b > 9a;
(iii) b > v'3.
97
Third degree polynomials
Solution.
(i) Let r, s , t be the roots of the given polynomial. We have
r+s+t=-,
1
a
b
r s + st + tr = - ,
a
1
rst = -.
a
Hence, a > O and then b > O, which implies that ab > O. Now,
(r+s+t)
2
> 3(rs+st+tr),
. 1
b
1.e - 2 > 3-, then O< 3ab < 1.
a
a
-
(ii) Since (r + s + t)(rs +st+
tr) > 9rst, we obtain b2 > ~, hence b > 9a.
a
a
3
> 2.
(iii) By the inequality (rs +st+ tr) 2 > 3rst(r + s + t) we get
a
a
Hence, b2 > 3 and since b > O, we obtain b > v'3.
b:
Example 4.16. The polynomial x 3 + v'3(a - l)x 2 - 6ax + b = O has three
real roots. Prove that
Solution.
Let r, s, t be the roots of the given polynomial. We get that
r + s + t = -v'3 (a - 1),
r s + st + tr = -6a,
r st = -b.
By the AM-GM Inequality,
{/jbf =Viri . lsl. ltl < ✓r2 + s32+ t2
J(r + s + t)2 -;(rs + st +tr)
✓ 3 ( 1 - a)2 + 12a -_ Ia+ 1.I
3
3
.
Hence Ibi< la+ 11
Example 4.1 7. The polynomial x 3 + qx + r, with r, q =/=-O, has u, v, w as real
roots. Prove that the polynomial r 2 x 3 + q2 x + q3 has three real roots outside
the interval (-1, 3).
Solution.
We have to find the relation between the roots x1, x2, x 3 of the
polynomial r 2 x 3 + q2 x + q3 and the roots u, v, w of the polynomial x 3 + qx + r.
Set
q
q
q
X3 = -C .
X2 = -b,
X1 = -a,
r
r
r
Then,
Third degree polynomials
98
and a + b + c = O. Similarly, we obtain
ab + ac + be = q,
abc = r .
Hence, {a , b, c} = {u,v,w} . Now without loss of generality
q
q
X1 = -U,
X2 = -V,
r
r
q
X3
= -W,
r
hence
X1 = 1u = - UV+
T
2
VW+
UW.
u= -l - u(v + w) = (v + w) - 1.
UVW
VW
VW
If X1 E (-1, 3), then lx1 - li < 2. But
(v+w)
lx1-ll=
2
-2
VW
v2 +w2
VW
> 2,
contradiction.
Example
4.18.
Let a, b, c, d be positive real numbers such that
a < b < c < d and abcd = l.
Prove that if r is a root of the polynomial
x
3
-
(a
1 + -;;
1+d
1) ,
+ b + c + d)x 2 + (ab + ac + be + cd + db + da )x - ( ~1 + b
then r > b.
Solution. Let
P(x)
= x 3 -(a+b+c+d)x
2
+(ab+ac+bc+cd+db+da)x-
1 1 1 1)
( ~ + b +-;;+ d .
Since abcd = 1, we can write P(x) as
P(x)
= x 3 -(a+b+c+d)x
2
+(ab+ac+bc+cd+db+da)x-(abc+bcd+cda+dab).
Then,
Q(x) = I+ xP(x) = (x - a)(x - b)(x - c)(x - d).
If P(r) = O, then Q(r) = 1 and vice versa (unless r = O). Suppose that r < O.
Then O < a < a - r, O < b < b - r, O < c < c - r and O < d < d - r, which
implies that
Q(r) = (r - a)(r - b)(r - c)(r - d) > abcd = I.
Third d_egree polynomials
99
Let O < r < a. Th en O < a _ r < a, o < b _ r < b, o < c - r < c and
O < d - r < d, which implies Q(r) < abcd = 1. If a:::;r:::; b, then
Q(r) = (r - a)(r - b)(r - c)(r - d) < O.
Hence, if r is a real root of P(x), then r > b.
Example 4.19. (Belarusan Mathematical Olympiad 2000) 3 The real numbers a, b, c satisfy the equality
2a
3
-
3
3
b + 2c - 6a 2 b + 3ab2 - 3c2 a - 3c2 b + 6abc = O.
If a< b, determine max(b , c).
Solution. Let
f(x)
= 2x 3 - 3(a + b)x 2 + (6ab)x + 2a 3 - b3 - 6a2 b + 3ab2 .
Then f(c) = O. Note that f '(x) = 6[x2 - (a+b)x+ab] = 6(x-a)(x-b),
hence
3
3
f '(a) = f '(b) = O. Moreover f (a) = (a - b) < O and f(b) = 2(a - b) < O,
which means that the y coordinates of the local minima and maxima of the
function are negative. The graph of the function is similar to the one below
and f (x) has only one real root, leading to the conclusion that c > b > a.
------------:-t
3 This problem needs the concep o f the derivative , if you are not familiar with it, don't
worry, you can skip it.
Fourth degree polynomials
101
5
Fourth degree polynomials
5.1
Solving equations
In this section, we provide two examples for solving fourth degree polynomial
equations that require some elementary algebra .
Example 5.1.
Solve the following equations:
2
(i) (x + x + 3) (x 2 + 3x + 3) = 3x 2 ;
2
(ii) (x - 12x - 64) (x 2 + 30x + 125) + 8000 = O.
Solution. (i) We have
(x
2
+ x + 3) (x 2 + 3x +3) = (x 2 + 2x + 3 - x) (x 2 + 2x + 3 + x)
= (x 2 + 2x + 3) 2 - x 2 .
2
So, we get (x + 2x + 3) 2 = 4x 2 , i.e., x 2 + 2x + 3 = ±2x, which gives x 2 + 3 = O
2
or x + 4x + 3 = O. Thus, the equation has four roots x1 = iv'3, x2 = -iv'3 ,
X3
= -1, X4 = -3.
(ii) Rewrite the equation as follows
(x + 4) (x - 16) (x + 5) (x + 25) + 8000 = O.
Hence , (x 2 + 9x + 20) (x 2 + 9x - 400) + 8000 = O.
Now, (x 2 + 9x) 2 - 380 (x 2 + 9x) = O.
Then, (x 2 + 9x) (x 2 + 9x - 380) = O.
Thus, the equation has four roots
X1 =
5.2
Vieta's
-9 - ✓1601
2
, X2 =
-9 + v'1601
, X3 = 0, X4 = -9.
2
formulas
Let x 1 , x 2, x 3, x4 be the roots of the polynomial a4x4 + a3x 3 + a2x 2 + a 1x + a 0 .
As we have seen in the previous chapters, we can deduce that
4
= - a3'
LXi
i =l
~
a4
x ·x·_a2
~
i
l<i<"<4
J_
L
XiXjXk
J -
a4
= --a1 ,
a4
l ~i<j<k~4
X1X2X3X4
'
ao
= -.
a4
102
Fourth degree polynomials
Example 5.2.
nomial
Find all real numbers a such that all the roots of the poly-
P (x) = x 4 + ax 2 + a 2x - 1
have equal modulus.
Solution. Let x 1, x2, x 3 , x 4 be four real roots of the polynomial P(x) .
Then,
Note that, by Vieta's formulas
4
1 = lx1x2x3x4I = lx11 .
Hence, 1 = lx1I= lx2I= lx3I = lx4I. Now, we prove that the polynomial P(x)
has two real roots. Since P(O) = -1 and lim P(x) = lim P(x) = +oo,
x --+-oo
x--++oo
then the polynomial has two real roots, one in the interval (-oo, O) and another
in the interval (O, +oo). Since its absolute values are equal to 1, we get
P(l)
= P(-1) = O.
Then,
1 + a + a2 - 1 = 1 + a - a2 - 1 = O.
Hence, a= O. Then
P (x) = x 4 - 1 = (x - l)(x + l)(x + i)(x - i)
and clearly satisfies the problem conditions.
Example 5.3.
The polynomial x 4 + ax 3 + 3x 2 + bx + 1 has four non-real
roots with absolute value equal to 1. Prove that 2 < lal < ~ and a = b.
2
Solution. Assume that x1 = p+iq, x2 = p- iq, X 3 = r+ i s, x4 = r-is , where
p, q, r, s are real numbers, are the four real roots of the polynomial. Since the
absolute values of the roots are equal to 1, we find that
p2
+ q2 = r2 + 8 2 = 1.
Moreover, (x - x1) (x - x2) = x 2 -2px+l
imply that
and (x - x3) (x - x 4) = x 2 -2rx+l
(x - 2px + 1) (x2 - 2rx + 1) = x 4 + ax 3 + 3x 2 + bx + 1.
2
Comparing the coefficients of x 3 , x 2 , x on both sides, we find that
a= b = -2 (p + r),
1
pr = 4·
Fourth degree polynomials
103
Now,
lal= 2lp+rl = 2 p+ 4~ .
We know that P, r E (-1, 1), so
1
4 = pr = IPrl< IPIand hence
G,1)u(-1
,-D·
pE
The range of the function
1
4x
f(x)=x+-
5
on the preceding interval is [1 ~) U (-~ -1] thus 1 < p + _!__< - and
'4
4'
'
4p
4
1
5
then 2 < 2 p + P = lal< . The equality case a= ±2 cannot occur , since
4
2
in this case we have
x 4 ± 2x 3 + 3x 2 ± 2x + 1 = (x 2 ± x + 1)2
5
and hence two double roots rather than four roots. Thus we have 2 < lal< - ,
2
as desired.
Example 5.4.
(Great Britain, IMO Longlisted 1987) Prove that if the
4
3
equation x + ax + bx + c = O has all real roots, then ab< O.
Solution. Let x, y, z, t be the real roots of the above equation. By Vieta's
formulas, we have
xy + xz + xt + yz + yt + zt = O.
Without loss of generality, assume that x + y + z =I=O (otherwise any sum of
three of the roots is zero, which implies a = O and we are done). Then,
xy + yz + zx + t (x + y +z)=
xy + yz + zx
O, h ence t = ------.
x+y+z
By Vieta's formulas, we find that
a= -t-(x+y+z)
and
xy + yz + zx
- (x+y+z)
x+y+z
= ----
(xy + y z + zx ) 2
b = -t (xy + yz + zx) - xyz = ..;__------'- - xyz.
x+y+z
Hence,
1
ab= (
x+y+z
)2 [(xy+y z + zx )2 -xyz
(x + y + z )][xy+yz+zx-(x+y+z)2].
Fourth degree polynomials
104
Note that
xy + yz + zx - (x + y + z)
+ y 2 + z 2)
y) 2 + (x + z)2] < O.
- (xy + yz + ZX + X
2
_![(x + y) 2 +(z+
2
2
Furthermore, replacing x, y, z in this inequality with yz, zx , xy we get
(xy + yz + zx) 2 - xyz (x + y + z) > O.
Therefore
ab =
1
(x+y+z)
2
( ( xy
+ y z + zx) 2 - xy z (x + y + z))
2'.0
2
. (xy + yz + zx - (x + y + z) )
~o
and we are done.
5.3
Number of real roots and graph
The graph of a fourth degree polynomial with positive leading coefficient is
one of the following. In the left figure, we have two local minima and one local
maximum. The right figure has one absolute minimum. Depending on the
coordinates of these local maxima and minima, the fourth degree polynomial
can have either 4, 3, 2, 1 real roots (see the figure and note the intersections of
horizontal lines and the graph).
Example 5.5.
Does there exist a positive integer d and real numbers
ao, a1, ... , ad so that the graph of the function
105
Fourth degree polynomials
is as in the figure below?
X
Solution. Let g ( x) = adxd + ... + a 0 and h (x) = a 0xd + ... + ad. If d is even,
we have f(-1) = g(-1) and if dis odd, we have f(-1) = -g(-1). In either
case,
f (-1) = lg(-1)1- lh(-1)1 = o.
Similarly, f (1) = lg(l)I - lh(l)I = O. According to the graph, we find that
f(x) has exactly two roots, which we have found (i.e., -1, 1), hence according
to the graph the function is positive everywhere on the interval (-1, 1). So,
f (O) = laol- ladl> O. Moreover, since
lim
x-+±oo
f (x) = x-+±oo
lim (ladl- laol)xd
= +oo,
then ladl- laol> O, contradiction!
Example 5.6.
(Sergei Berlov - Russian Mathematical Olympiad 1993) The
2
equation x + ax + b = O has two distinct real roots . Prove that the polynomial
x 4 + ax 3 + (b - 2) x 2 - ax + 1 = O has four distinct real roots.
Solution. First we prove following lemma.
Lemma . Let c be a real number . Then the equation x - ~ = c has two real
X
roots.
Proof. Write the equation as x 2 -cx-1 = O. Its discriminant is D = 4+c 2 > O,
so it has two real roots. Our proof is complete.
Let r and s be the roots of x 2 + ax + b = O. By our lemma, each of the
equations x - ~ = r, x - ~ = s has two real roots. Since r 2 + ar + b = O and
X
X
s 2 +as+ b = O, these are the roots of the quartic equation
2
(x - : )
+ a ( X - ~) + b = 0.
Multiplying by x4, we have x 4 + ax 3 + (b - 2) x 2 - ax + 1 = O. Thus, we have
four distinct real numbers satisfying the above equation.
Fourth degree polynomials
106
Example 5. 7.
(Saint Petersburg Mathematical Olympiad 2010) Let f (x)
be an arbitrary polynomial. Is it possible to find a polynomial g(x) of degree
4 such that f(g( x)) has no real roots?
Solution. The answer is yes. If f(g( x)) has ras a root , then g(r) must be a
root of J(x). If f( x) has no real root , we are done. Assume f( x) has r~al roots
x 1 , ... ,x 8 andleta bethelargestoneamongthem.
Now, setg(x) = X +l+a.
Then , for any real number x we have g (x) > a > x1,. • •, Xs- Thus, f(g( x))
has no real roots.
As you have seen, the second and the third degree polynomials with real
coefficients have functions of their coefficients which characterize whether the
polynomial has real roots or not and how many real roots it has. A natural
question is whether a similar polynomial exists for a fourth degree polynomial.
It is straightforward that the polynomial
(xi - Xj ) 2 doesn 't work! For
example, consider the polynomial
II
x 4 + 2x 3 - x - 2 = (x 2 + x + I) (x 2 + x + 2)
II
2
which has four complex roots, but surprisingly the value of
(x i - Xj ) is
positive (why?) 4 • On the other hand , when the polynomial has four real roots ,
the aforementioned value is clearly positive, thus this function does not work.
In the next example we show that sucha function doesn 't exist at all.
Example 5.8.
(Iranian Mathematical Olympiad 2012) Prove that there
doesn 't exist a four variable polynomial P( a, b, c, d) such that the fourth degree
polynomial x 4 + ax 3 + bx 2 + ex + d with real coefficients can be written as a
product of four linear polynomials if and only if P( a, b, c, d) > O.
Solution.
Assume the contrary. If we put a = c = O, the polynomial
4
2
x + bx + d can be written as a product of four linear polynomials if and
only if the polynomial t 2 + bt + d has two nonnegative real roots. Therefore ,
P(0 , b, O, d) > O if and only if b < O, d > O, b2 - 4d > O. Fix b < O. Con2
sider Q(d) = P(0, b, O,d) > O which is nonnegative if and only if O < d < b •
b2
- 4
Hence, Q (
= O for all b < O. Therefore , it must be zero for all b, i.e.,
4)
( ~)
P O,b, O,4
= O, but this means that x 4 + bx2 + ~ must have four real
4
roots for all b, a contradiction since x 4 + bx 2 + ~ = (x2 + ~)
2
. . b, we h ave no real roots .
a pos1t1ve
4
2
and choosing
~ttis a simple complex number exercise and it uses the properties of complex conjugates .
Try l .1
107
Fourth degree polynomials
5.4
Miscellaneous
In the last section of this chapter we provide some challenging problems which
needs the synthesis of approaches we proposed here. It is possible that in any
mathematical competition the reader face such problems. Working on these
problems help the reader to assess his learnt instructions better.
2
Example 5.9.
Let P (x) = x 4 +ax 3 +bx 2 +ex+ 1 such that lal,Ibi, lei< .
Prove that for all real numbers x we have P (x) > O.
Solution. Observe that P (1) = 2+a+b+e > Oand P (-1) = 2-a+b-e
Finally, if lxl =/ 1,we have
IP(x)I
3
> O.
lx4 + ax 3 + bx2 +ex+ li
> jx4 +1j-jax 3 +bx2 +exj
> x 4 + 1- laljx3 1-Ibilx2 1- leilxl
>
x•+ 1- G)(lx31+ lx21+ lxl)
+;lxl+1)
(lxl- 2(1x21
> o.
1)
Example 5.10. (Dorin Andrica - Mathematical Reflections J410)
a, b, e, d be real numbers such that a 2 < 2b and e2 < 2bd. Prove that
Let
x 4 + ax 3 + bx2 + ex + d > O
for all real numbers x.
Solution. Notice that 2b > a 2 > O and 2bd > e2 > O. Sob> O. Let
P( x) = x
2
b
+ ax + 2
b
Q(x) = x 2 +ex+ d.
2
The discriminant of P is non-positive. Hence, P > O =} P • x 2 > O. The
discriminant of Q is negative, so Q > O. Adding up the results above yields to
the desired conclusion.
Solution 2. Notice that 2b > a 2 > O and 2bd > e2 > O and for all real
numbers x we have
2
x +ax +bx +ex+d = x
4
3
2 [ (
x + -a)
2
2
2b - a
+ --4
2
l+- + -e) +--- e
b (x
2
2
b
2bd -
2b
2
>O
.
108
Fourth degree polynomials
Example 5.11. (Vladimir Cerbu - Mathematical Reflections S455) Let a
and b be real numbers such that all the roots of the polynomial
are real numbers. Prove that J ( - ~) < ~ .
1
Solution. Let x 1 , x2, x3, x4 be the roots of the given polynomial. By Vieta's
formulas, we get
+ X2 + X3 + X4
X1X2 + X1X3 + X1X4 + X2X3 + X2X4 + X3X4
+ 2_ + 2_ +
-X1X2X3X4
XI
(2XI
X2
X3
2-)
1
o
a
X4
X1X2X3X4
b.
The first and second equation gives
Using Cauchy-Schwarz's Inequality,
1 = X 21 + ( X22 + X32 + X42) > X 21 + 1 (X2 + X3 + X4 )2 = X12 + 1 ( 1 - X1 )2 .
3
3
Hence we have
1
--2 -< X1 -< 1.
Similarly, we have -
1
2 < x2, X3, X4 < l. Hence we get
f (l) > O# a+ b > O.
It is sufficient to prove that
!)
# a > 2b.
f ( - 2 -< ~
16
-
Let b < O. From a+ b > O, we have a > O. In this case a > 2b is true. Let
b > O. Then, x1x2x3x4 > O and we have
a > 2b #
1
X1
1
1
1
X2
X3
X4 -
+ - + - + - < -2.
(1)
In this case, two roots are positive and two roots are negative. Assume that
x1, x2
> O and X3, X4 < O. Since
-1
< X4 < 1, we get 2x 4 + 1 > O, 1 - x 4 > O
109
Fourth degree polynomials
and x1x2x3
< O. Hence, we get
+ X2 + X3) - X1X2X3 > 2x1x2X3X4
X4(X1 + X2 + X3) _ 2_ >
2
x~(x1
~
X4
X1X2X3
-x1x2
-
X1X3 -
X2X3
_
2_ > 2
X4 -
X1X2X3
~
and (1) is proved.
1
1
1
1
XI
X2
X3
X4
- + - + - + - < -2
111
On roots of polynomials - elementary problems
6
On roots of polynomials
- elementary
6.1
Vieta's formulas in the generał case
problems
Let x 1, x2, ... , xd be roots of the polynomial adxd+ . . . +a 0. Similarly to what
we have already done we can write
d
L
Xi
i=l
z:
XiXj
Xi 1Xh · .•.
· Xik
l,Si<j,Sd
z: Il
ad-l
ad
ad-2
ad
---
(-lt
i1 , ... ,ik
Xl · ... · Xd
ad-k
ak
(-l)dao
ad
Note. The polynomials on the left-hand sides of the above relations are called
elementary symmetric polynomials of degree k.
(Polish Mathematical Olympiad 2004) Let c be a real number such that the polynomial
Example
6.1.
P( x) = x 5 - 5x 3 + 4x - c
has five distinct real roots x1, x2, x3, X4, X5. Determine, depending on c, the
sum of the absolute values of the coefficients of the polynomial
Q(x) = (x - xi)(x - x~)(x - x~)(x - x~)(x - xg).
Solution.
From the equality
(x2 - xI)(x 2 - x~)(x 2 - x~)(x 2 - x~)(x 2 - xg)
(x- x1)(x-x2)(x -x3)(x -x4)(x -xs)
·(x + x1)(x + x2)(x + x3)(x + x4)(x + xs)
P(x) · (-P(-x))
(x 5 - 5x 3 + 4x - c) • (x 5 - 5x 3 + 4x + c)
(x 5 - 5x 3 + 4x) 2 - c2
x 10 - 10x8 + 33x 6 - 40x 4 + 16x2 - c2
we conclude that
Q(x) = x 5 - 10x4 + 33x 3 - 40x 2 + 16x - c2 Vx > O.
On roots of polynomials - elementary problems
112
Expressions on both sides of this equality are polynomials . Hence , the obtained
formula is true for any real number x. Thus , the sum of the absolute values
of the coefficients of Q(x) is
2
1 + 10 + 33 + 40 + 16 + c2 = 100 + c .
Example 6.2.
(Mathemati cs In School Journal) Let a1, ... , a100be the real
roots (with multiplicity) of the polynomial
2
P(x) = (x - a1x + a2)(x 2 - a3x + a4) · .. . · (x 2 - a99X + a100).
Find the 100 numbers .
Solution. By the problem assumption we find that
P(x) = (x - a1) · ... · (x - a100).
By comparing the coefficients of x 99, x 98 in both polynomials assigned as P(x) ,
we get
50
= 0,
L
L
a2k-1a2z-1 =
aiaj,
1$k<l$50
1$i < j$100
If we write the first equality as
I:a2k
k=l
50
100
k=l
j=l
L a2k-1 = L aj
and square it, we obtain
50
100
k=l
j=l
L a~k-1 + 2 1$k<l$50
L a2k-1a2z-1 =La/+
2
L
aiaj,
1$i<j$100
50
Hence,
L a~k = O and a2 = a4 = ... = a100= O, which means that
k=l
50
P(x) = x (x - a1)(x - a3) · . .. · (x - a99).
It follows that a1, a3, ... , agg are roots of the polynomial. So, the answer is
a2 = a4 = ... = a100= O (and chose a1, a3, ... , a99 arbitrary.)
Example 6.3.
Let a1, .. . , an be the roots of the equation
x"-(;)a1xn-l
+ ... + Ht(;)a~xn-k
+ ... + (-lJ"a: = O.
113
On roots of polynomials - elementary problems
Pro ve that a1 = ... = an.
Solution. Let k be an index which maximizes lakI- Since a 1 , ...
of the polynomial , by Vieta 's formulas we have
, an are roots
By triangle inequality, we have
<
<
LIT lai1· ai2 · • • • · aikl
LIT lai1I· lai2I· •••· laiki
(:)
latl-
Thus we must have equality throughout the calculation above and hence
Since the sum of roots is
a1, ... , an must have the same sign.
Example
(Crux Mathematicorum)
6.4.
P(x) = xn - 2nxn-l
Given that the polynomial
+ 2n(n - l)xn- 2 + ... + a0
has only real roots, find all real roots.
Solution. Let r1, ... , rn be the roots of the polynomial taken with multiplicity. By Vieta's formulas, we have
Then,
which means that
n
L(ri- 2) = Lr/ - 4 L ri + 4n = O.
2
i=l
On roots of polynomials - elementary problems
114
Since r1
Example
...
, rn are all real , we obtain
6.5.
that r1 = .. • = rn = 2.
(Nguyen Viet Hung- Mathematical Reflections S369) Given
the polynomial
P(x) = xn + a1xn-l + ... +an-IX+
an
having n real roots (not necessarily distinct) in the interval [O,1], prove that
3ar + 2a1 - 8a2 < 1.
Solution.
Let x 1 , x 2, ... , Xn be the roots of the polynomial P(x). Then,
a1 = -(x1 + x2 + ... + Xn)
and
so we can rewrite our inequality as
3(x1 + x2 + ... + Xn)2 - 2(x1 + x2 + ... + Xn)
- 4((x1 + X2 + ... + Xn)2 - (Xi+ X~+ ... + x~))
< 1,
which is equivalent to
Since x1, x2, ... , Xn E [O, 1) we have that
2
2
2
X1 + X2 + . . . + Xn <XI+ X2 + ... + Xn,
SO
4(xi + x~+ ... + x;J- 2(x1 + x2 + . .. + Xn)
< 4(x1 + X2 + ... + Xn) - 2(x1 + x2 + ... + Xn)
= 2(x1 + x2 + ... + Xn) < (x1 + X2 + ... + Xn) 2 + 1,
where the last inequality follows from (x 1 + x 2 + ... + Xn - 1)2 > O.
Equality holds when x1 = 1 and x2 = X3 = ... = Xn = O.
Example 6.6.
Are there polynomials of the form xn ± xn-l ± ... ± x ± 1
such that all their roots are real?
Solution. The case n = 1 is obvious. Assume that n > 1 and let r 1 , . . . , rn
be the roots of the polynomial. By Vieta's formulas, we have
115
On roots of polynomials - elementary problems
Hence,
Lr/=
3 and
Lrirj = -1.
Moreover, ri2r2 2 · . .. • rn 2 = (±1) 2 = 1. By the AM-GM inequality we obtain
Lr/
3
n
n ~-----
~ z/r12r22 .. ; .rn2
- = ---
n
= 1,
which yields n :; 3. For n = 3 the equality case of AM-GM occurs, hence
r12 = r2 2 = r3 2 = 1. Then , the desired polynomial is (x 2 - 1) (x ± l) • For
n= 2 we get the polynomial x 2 ± x - I, while for n= l we have x ± l.
6.2
Inequalities
between
coefficients
and roots
In this section we combine Vieta's formulas and some elementary facts about
polynomial roots with the well-known inequalities like AM-GM, CauchySchwarz and other elementary inequalities.
Example 6. 7.
Let
P(x) = aox2008+ a1x2001+ ... + a2008
be a polynomial with 2008 distinct positive real roots . Prove that
2007aI > 4016a2ao.
Solution. Let P(x) = ao(x - r1) · ... · (x - r2008) with r1, ...
Then th e desired inequality is reduced to
2
2
1
2001(a ) > 4016a .
ao
ao
Using the Vieta's formulas we obtain the following inequality
2008
2001(
2
L Ti) > 4016 L
i= l
TiTj.
19<j$2008
Now we have to prove that
2008
2007 Lr;
>2
i= l
2008
so we add
L r;to both sides. We get
i= l
L
1$ i<j$2008
rirj,
, r2008
> O.
On roots of polynomials - elementary problems
116
which is obvious by using the Cauchy-Schwarz Inequality (and the observation
that we cannot have equality) .
Example 6.8.
Consider the polynomial
P(x) =aoxd+ .. . +ad
with positive real roots. Prove that
Solution. Assume that r 1 , ... , rn are the positive roots of the polynomial
P(x). By Vieta 's formulas, we have that
By the AM-GM Inequality , we obtain
By adding d to the inequalities we get
d- 1
Lakad - k
aoad
Example 6.9.
_.
-L (d)= (2d)
2
> d-l
k= l
k
k=l
d
2.
Let d be an even number and
a polynomial such that ad, ao > O and
2
a1
+ · · · + ad2
< 4min(a8, a~)
I
-
d- l
.
Prove that P(x) 2: O for all nonnegative real numbers x.
Solution. By Cauchy-Schwarz Inequality, we have
On roots of polynomials
117
- elementary problems
Hence ,
min(ad , ao)(l + rd -
~ · Jr2 + r4 + ... + r2d-2).
d-1
Now we prove that
l+rd
> ~.
-
d-1
Jr2+r4+
... +r2d-2
for all r. Since d is even both sides are positive , hence we can square the above
inequality and we get
(d - 1)(1 + rd)2 ~ 4(r 2 + r 4 + ... + r 2d- 2 ).
Once we set d = 2m and r 2 = s , we have to prove that
(2m - 1)(1 + sm) 2 > 4(s + s 2 + ... + s 2m- 1).
This follows by adding the inequalities
2(1 + 8 m)2 ~ 2 (sk + 8 m-k + 8 m+k + 8 2m-k)
for k = 1, . .. , m - 1
(which hold since they factor as 2(1 + sm)(l - sk)(l - sm-k) > O) and the
inequality (1 + sm) 2 ~ 4sm.
The next two problems need more focus and skills in working with inequalities.
Example 6.10. Let P(x) = xd - a1xd-l + ... + (- ltad be a polynomial
such that its roots lie in the interval [a, ,B] where a ~ -1. Prove that:
(l+a)d
Solution.
'.S(-l)dP(-1)
< (l+,B)d.
Let a1, . . . , rd be the roots of P(x). Then we have
d
P(x)
= II(x - ri)i=l
Hence we have
d
(1 + a)d < (-l)d P(-1)
=
II(1 + ri) < (1 + ,B)d.
i=l
On roots of polynomials - elementary problems
118
Example 6.11. Let P( x ) = xd + ad- Ixd-I
that all its root s !ie in (2, +oo) . Prov e th at
+ ... + ao be a polynomial such
Moreover, if all the roots of th e polynomial lie in (O,2) prove that
2d + 2d- 2 ad -2 + 2d- 4ad-4 + ... < -1.
2d-Iad -I + 2d- 3ad-3 + · · ·
Solution.
Assume that rI, .. . , rd are the roots of the polynomial P(x) . Then,
P(x)
= (x - rI) · ... · (x - rd)
and
( rI - 1) · . .. · (r d - l)
(1 + (rI - 2)] · ... · (1 + (rd - 2)]
> 1 + rI - 2 + ... + r d - 2
d
L ri - 2d + 1 =-ad - I+ 1 - 2d,
i=I
where the last eąuality is deduced from the Vieta's formulas. Hence,
(-l)d P(l)
+ ad-I > 1 - 2d.
For the second part, note that since all the roots lie on (O,2), then P(2) > O
(if you can 't see this, notice that P(2) = (2 - rI) · . . . · (2 - rd) > O, but
P(2) = 2d + ad-I2d-I + ... + ao > 0). Thus,
2d + 2d-2 ad - 2 + 2d-4 ad-4
+ · · · > - (2d-I ad-I + 2d-3 ad-3 + ... ) ,
and we get the desired ineąuality. (Note that ad-I, ad_ 3, ... are negative.)
The next two problems need more elaborate work with the concept of roots
and Vieta's formulas.
Example 6.12. (IMC 2017) Let p(x) be a non-constant
real coefficients. For every positive integer n, let
polynomial with
Prove that there are only finitely many numbers n such that all roots of ąn(x)
are real.
Solution. We will make use of a nice observation.
119
On roots of polynomials - elementary problems
d
Suppose the polynomial r(x)
= Lak
xk has only real roots , say x1, • • • ,Xd-
k=O
Then,
which gives a~_ 1 - 2adad-2 > O.
Assume that
p( x) = axk + bxk-l
+ cxk- 2 + ....
Wlthout loss of generality assume that a > O, otherwise multiply by -1.
It follows that
qn(x)
= 2axn+k + (a(k - n)+ 2b)xn +k-l
+ ( n(n - 1) + k(k - 1) a+ (k - n - l)b + 2c) xn+ k - 2 + ...
2
Axn+k + Bxn+k-1
+ cxn+k-2 + ...
Assume by contradiction that all the roots of the polynomial qn( x) are real
for infinitely many n . Then, from our earlier observation, we get that
B 2 -2AC
> O.
Putting the pieces above together, we compute
B 2 - 2AC = - an 2 + dn + e
for some constants d, e. Now for sufficiently large n we have B 2 - 2AC < O,
which means that the polynomial qn(x) has non-real roots for infinitely many
n.
Example 6.13. (Plamen Penchev - Bulgarian Mathematical Olympiad
2013) Let n > 2 be a positive integer and a1 < a2 < . .. < a2n be real
numbers. Let
2n
S =Lai,
i=l
A1 =
L
i<j
a2ia2j,
A2 = ~ a2 i-ia
2j- l•
i<j
Prove that
(n - l)S 2 > 4n(A1 + A2).
Solution. First we prove the following lemma.
Lemma. Let P(x) = bnxn + bn-1xn - l + .. :+ bo a polynomial with n real roots .
Then,
On roots of polynomials - elementary problems
120
Proof. Dividing both sides of the inequality that we want to prove by b;, we
get
2
bn-1)
(n - 1) ( -
bn-2
2 2n · -b-.
bn
n
Let x 1 , . . . , Xn be the real roots of the polynomial. By Vieta's formulas we
need to prove that
(n - l)(x 1 + ... + Xn)
2
L
> 2n (
XiXj)
1:Si<j:Sn
i.e.
n
(n- 1) Lx; > 2 L
i=l
XiXj-
1'.Si<j'.Sn
We now add the n - l inequalities
n
LXT> X1X2 + X2X3 + · · · + XnXl
i=l
n
LXT
> X1X3 + X2X4 + · · · + XnX2, ...
i=l
and finally
n
LXT> X1Xn
+ X2X1 + •.. + XnXn-1·
i=l
The equality occurs when all the roots are the same .
Back to our problem, let
be a polynomial of degree n with n real roots (check the signs of P( a 2i) and
P(a 2i_ 1 )). The conditions of our lemma are satisfied. Hence, it is easy to
check that the inequality of our lemma leads to the inequality
We removed the equality because the polynomial has n distinct real roots and
the equality is not possible .
121
On roots of polynomials - elementary problems
6.3
Miscellaneous
problems
In this last section we will provide some challenging problems that need the
synthesis of the approaches we have been seen previously. lt is possible that ,
in any mathematical competition, the reader will face this kind of problem.
Working on these problems help the reader to assess his proficiency better.
Example 6.14. (Mathematics In School Journal) Let P(x) be a polynomial
of degree 54. Is there a polynomial Q(x) of degree 37 such that all the numbers
1, 2, .. . , 1998 are the roots of the polynomial P(Q(x))?
Solution.
Assume by contradiction that such polynomials exists. Let
x1, ... , Xm be the real roots of P(x) , where m ~ 54. Since the polynomial
Q(x) is of degree 37, then the numbers 1, ... , 1998 must be the roots of the
equations Q(x) = Xk for 1 ~ k ~ m. We have at most 37m ~ 37 · 54 = 1998
real roots. Since we have exactly 1998 real roots, then m = 54 and each
of the equations of the form Q(x) = Xk has exactly 37 real roots in the set
{1, ... , 1998}. Assume Q(x) = a31x37 + a35x 36 + ... + ao. Then, by Vieta's
36
formulas, the sum of the roots of any equation of the form Q(x) = Xk is - a ,
a37
which is an integer (since the roots of the equation Q(x) = Xk are integers).
Hence the numbers 1, ... , 1998 can be partiotioned in 54 disjoint sets with
equal sums, which means that
1 + ... + 1998
54
1998 · 1999
54 · 2
37 · 1999
2
must be an integer, contradiction. So, sucha polynomial doesn 't exist.
Example 6.15. (Crux Mathematicorum)
n 2: 2, the equation
Prove that for all positive integers
has a root in ( 1, 1 + ~) .
= xn + x-n - x - 1.
First we need to prove that f
+ ~) > O. For n= 2 the claim is obvious.
Solution. Let f(x)
(1
Assume n 2: 3. Then,
l)n
( 1+- n =l+n
It follows that
(1)
n-1
- + n(n-1) - 21 + ... >2+-->2+-.
2
n
n
2n
1
n
On roots of polynomials - elementary problems
122
We now define
= x2n - xn+l - xn + 1 = (x - l)h(x) ,
g(x) = xnf(x)
where
h(x) = x2n-1 + x2n-2 + ... + xn+l _ xn-1 _ xn-2 _ . . . _ x _ 1.
Since
1(1+
:)
> O, we obtain that
g(l+
~) , h(l +:)>O
and
h(l) = -1 < O.
(1,1+ ~) and so there is
Hence, there is a root for h(x) within the interval
a root also for J(x) .
Example 6.16. Let P(x) and Q (x) be polynomials with real coefficients
having at least one real root such that
pc;17
+X+
(Q(x))·)
=
Q(2;17
+X+ (P(x))·).
Prove that P(x) = Q(x).
Solution. Assume that P(r) = O, Q(s) = O and consider the polynomial
R(x) = (P(x)) 4 - (Q(x)) 4 .
Since R(r) :S O and R(s) 2: O, then there is a real number to such that
R(to) = O. Setting x = to into the original equation, we obtain
P(20\7
+to+ (Q(to))4) = Q(20\7
+to+ (P(to))4).
1
+to+ (Q(to)) 4 > to, then R(t 1) = O. Continuing in this way,
2017
we find an increasing sequenc e tn of real numbers such that P(tn) = Q(tn) for
all n. It follows that the equation P(x) = Q(x) has infinitely many real roots
and thus for all x we have P(x) = Q(x).
Set t1 =
Example 6.17. (Nguyen Viet Hung - Mathematieal Refleetions U412) Let
P(x) be a monie polynomial with real eoefficients, of degree n, whieh has n
b2)n
real roots. Prove that if a> O and P(c) :S ( -;
then P(ax 2 + 2bx + c) has
at least one real root.
Solution. If P(x) is an n-th degree monie polynomial with real eoefficients
and with n real roots a1, ... , O'.n, then
P(x) = (x - a1)(x - a2) · · · (x - an).
123
On roots of polynomials - elem entary probl ems
Hence,
and
P(ax 2 + 2bx + c) = (ax2 + 2bx + c - a 1 )
x (ax 2 + 2bx + c - a 2 ) • • • (ax 2 + 2bx + c - Dn)Assum e by contradiction that P(a x 2 + 2bx + c) has no real roots. Then , eac h
factor has negat ive discriminant, i.e.
b2 - a(c - 01) < O
b2 - a( c - 02) < O
b2 < a(c - 01)
b2 < a(c - 02)
b2 < a(c - Dn)
Multiplying side by side all these inequaliti es, we get
i.e.
contradiction.
Example 6.18. (Belarusan Mathematical Olympiad 2016) Let P(x) and
Q(x) be two polynomials such that degP( x ) = degQ(x). We define the polynomial PQ(x) so that the coeffi.cients of x 2 k and x 2 k+l are taken from P(x)
and Q(x ), respectively. For example, let
P(x) = x 3 + 2x 2 + 4x + 1 and Q(x) = 3x3 + x 2 + 2.
Then,
PQ(x) = 3x3 + 2x 2 + 1 and Qp(x) = x 3 + x
2
+ 4x + 2.
Prove that
(i) There are polynomials P(x), Q(x) having no real roots but with
PQ(x ), Qp(x) having at least one real root.
(ii) What is the smallest degree of the polynomials P(x), Q(x) satisfying the
previous point?
Solution. (i) Consider P(x) = 4x4 + 4x3 + 1 and Q(x) = x 4 + 4x + 4. It
is easy to prove that Q(x), P(x) > O (Indeed 4x4 + 1 > 3x4 + 1 > 4lxi3,by
AM-GM . The first inequality is an eąuality only at x = O and the second only
at lxl = 1. Thus the two ineąualities cannot both be equalities, and P(x) > O.
On roots of polynomials - elementary problems
124
Similarly x 4 + 4 > x 4 + 3 > 4lxl implies that Q(x) > O.), which means that
th ey have no real roots. Now, we obtain
PQ(x) = 4x4 + 4x + 1,
Hence Pq(
Qp(x) = x
4
+ 4x 3 + 4.
-D
< O and Pq(O) > O. Moreover, Qp(-2) < O and Qp(O) > O,
thus both PQ(x) and Qp(x) have real roots .
(ii) Since P(x) and Q(x) have no real roots, we get that the degrees of the
polynomials are even. We need to prove that they can't be quadratic. In fact,
if
2
P(x) = a1x 2 + b1x + c1 and Q(x) = a2x + b2x + c2
have negative discriminants , then
PQ(x) = a1x 2 + b2x+ c1 and Qp(x) = a2x 2 + b1x + c2.
Without loss of generality , assume that lb2l > jb1j. It follows that
/::j.pQ
Finally , we obtain
= b22 - 4a1c1 and J::j.Qp = b12 - 4a2c2.
Number theory and polynomials
125
7
Number theory and polynomials
7 .1
Number
theory and low degree polynomials
In this section we will provide some problems that can be solved by using what
we have learnt until now about low degree polynomials.
Example
7.1.
Solution.
Let
Find the number of positive integer pairs (a, b) satisfying
k = b3 - a3 = (~) 2
a2 b
a
a
b'
Then
(~)" +
k(~)'-1
=
O.
Then, the monie polynomial x 3 + kx 2 - 1 = O has a rational root r and this
a
must be an integer dividing -1. It follows that r = 1 (since r = b > O).
Therefore, a = b and a 2 + b2 = 2a 2 < 2018. Hence, a 2 < 1009 and 1 < a :S 31.
We obtain 31 pairs (a, b) satisfying the above relations .
Example 7.2.
(Titu Andreescu - Mathematical Reflections 0433) Let
q, r , s be positive integers such that s 2 - s + 1 = 3qr. Prove that q + r + 1
divides q3 + r 3 - 83 + 3qr8 .
Solution. Since s 2 - 8 + 1 = 3qr, then s 3 = 3qr8 + 3qr - 1. We get
n = q3 + r 3 - 83 + 3qr 8 = q3 + r 3 - 3qr + 1.
Consider the polynomial
J (x) = x 3 - 3xr + (r 3 + 1).
It is elear that J(q) = n and J(-r - 1) = O. This implies that there exists g E
Z[x] such that J(x) = (x+r+ l)g(x). In particular, n= f(q) = (ą+r+ l)g(q)
and the conclusion follows.
Example 7.3.
(Alexander Ivanov - Bulgarian Mathematical Olympiad
2005) Let a, b, c be positive integers such that ab I c(c2 -c+ 1), (c2 + 1) I (a+b).
Prov e that {a, b} = {c, c2 - c + 1}.
Solution. This problem was a nightmare for many students. We will provide
an elegant solution based on deep understanding of second-degree polynomials.
Number theory and polynomials
126
2
Assum e that a ~ b. Then a + b = k( c
)
+ 1 an
d b - k(c2 + 1) - a> a, which
-
k(c2 + 1)
Now,
m eans that a<__.:_--.
2
ab= a[k(c 2 + 1) - a] < c(c
The function
2
-
c + l).
J(x) = x [k(c2 + 1) - x]
.
is increasing on the mterval
(
2
-oo,
l)l
We also get that
·
2
k(c +
2
k(c + 1) > c.
2
-
k(c 2 +
It follows that a, c both belong to ( -oo ,
2
1)]
. a> c,
•
Now, if
a[k(c 2 + 1) - a] = f(a) > f(c) = c[k(c2 + 1) - c] > c(c
2
-
then
c + 1),
contradiction. Hence a < c and
b I [bc-kc( c2 - c+ 1)] = c[k(c2 + 1) -a]But
kc(c
2
-
c+ 1) = kc
2
-
ac.
2
b = k( c2 + 1) - a > kc2 - a > kc2 - ac > kc 2 - c > O.
2
Since kc2 - ac > O and it is divisi ble by b, we obtain that b < kc - ac. However
the above inequality contradicts this, unless kc2 - ac = O and a = kc < c.
Hence, k = 1 and a = c. So,
2
b = k( c2 + 1) - a = c
+1- c
and we are don e.
Th e next example is more generał, it only requires one to work with linear
polynomials.
Example 7.4.
(Alessandro Ventullo - Mathematical Reflections 0422) Let
P( x) be a polynomial with integer coefficients and a nonzero int eger root.
Prov e that if p and ą are distinct odd primes such that P(p) = p < 2q - 1 and
P(q) = q < 2p - 1, then pandą are twin prim es .
Solution. Let r -/=O be a nonzero integ er root of the polynomial P( x) . The
polynomial can be written as
P(x) = (x - r)Q(x),
127
Numb er theory and polynom :ia.ls
where Q(x) is a polynomi al with int egH coefficient :;. Jt follow:; that
P(p) = (p - r) Q(p) = p,
P(ą) = (q -
r)Q(q) = q.
which impli es that p - r E {±1 , ±p} aud ą - r E {± 1, ±ą } , since pandą are
prim e. Th erefore,
r E {p - 1, p + 1, 2p} n {ą - 1, ą + 1, 2ą }.
We conclude that eith er p - 1 = ą + 1 or p + 1 = ą - 1, since p < 2ą - 1 and
q < 2p - 1. So, p - q = ±2 , i.e., p and q are twin prim es.
7.2
P(a) - P (b)
In the first chapter we proved that for a polynomial P( x ) with integer
.
l
.
P(a) - P(b)
coe ffic1ents , t 1e quoti ent ------'---'----'--...:...
is integer for all int egers a # b. This
a-b
fact has an alt ernative formulation , i.e if a b (mod m), then P(a)
P(b)
(mod m) for any polynomial P( x ) with int eger coefficients. For examp le,
since a= b (mod a - b), then P( a) = P(b) (mod a - b).
=
=
Reu1ark
By the above formulation , we can establish two important congruences
for a polynomial P( x) with integ er coefficients and arbitrary integers
n,k .
(i) P(n + kP(n)) = O (mod P(n)) .
(ii) P(nP(n)) = P(0) (mod P(n)).
Example 7.5.
Let P(x) be a polynomial with integ er coefficients such that
P(m) - P(n) divides m 2 - n 2 for all integers m , n. If P(0) = 1 and P(l) = 2,
find the larg est possibl e value of P(l00).
Solution. Since P(0) = 1, then (P(m)-1) I m2 for all int egers m. Therefore,
IP(x) - li < x 2 for all integers x. Hence deg P( x ) < 2 and if we write
P(x) = ax 2 + bx + l , then lal< 1. Since P(l) = 2 gives a+ b = 1, we see that
P( x) must be one of th e thr ee polynomials -x 2 + 2x + 1, x + l , or x 2 + 1. The
first of thes e does not satisfy the hypothes es, but the last two do. It follows
2
that the largest value of P(l00) is P(l00) = 100 + 1 = 10001.
Example 7.6.
(Titu Andre escu - Mathematic al Reflections U421) Find all
pairs a and b of distinct positive integ ers for which there is a polynomial P( x)
with integer coefficients such that
Number theory and polynomials
128
Solution.
Rewrite the equality as
P(a3) _ P(b3) = 7(b + a2 _ a - b2) = 7(a - b)(a + b - l).
We know that (a 3 - b3 ) I (P(a 3) - P(b 3)) , which implies that
(a 3 - b3 ) I 7(a - b)(a + b - l).
Now, for all a -:/=b we get (a2 +ab+ b2) I 7(a + b - l). We can assume that
a> b. Then,
7(a + b - l) > a2 + ab+ b2 = (a + b - l) (a + l) + b2 - b + l > (a + b - l) (a + 1) ·
It follows that 7 > a+ l , i.e., 6 > a, which gives b < a < 6.
A simple calculation shows that (a, b) E {(2 , 1), (5,3)}. We get P(x)
(a, b) E {(1, 2), (2, 1)} and P(x) = x for (a, b) E {(3, 5), (5, 3)}.
So, (a, b) E {(1, 2), (2, 1), (3, 5), (5, 3)}.
= 2x for
Example 7. 7.
Let P( x) be a polynomial with integer coefficients such that
for all positive integer n , P(P(n)) gives the remainder n - l when divided by
n. Prove that the polynomial has no integer roots.
Solution. Since P(n) _ P(O) (mod n), we get that
n - 1 _ P(P(n)) _ P(P(O)) = -1
(mod n).
Thus 1 + P(P(O)) is divisi ble by all n and P(P(O)) = -1.
Assume that P(r) = O for some integer r. Then P(x) = (x - r)Q(x) for some
polynomial Q(x) with integer coefficients, which means that P(O) = -rQ(O).
Moreover,
-1 = P(P(O)) = (P(O) - r)Q(P(O)) = -r(l
+ Q(O))Q(P(O)),
hence 1 + Q(O) = ±1. This implies that Q(O) is even and that P(O) = -rQ(O)
is also even.
Assume that P(O) = b. Then Q(P(O)) = Q(b) = ±1, but Q(b) = Q(O)
(mod 2), hence 1 = O (mod 2), contradiction. So, the polynomial P(x) has
no integer roots.
~
Example 7.8.
Let P1(x), P2(x), ... , Pn(x) be non-constant polynomials
with integer coefficients. Prove that there are infinitely many positive integers
a such that P1(a), P2(a), ... , Pn(a) are all composite.
Solution.
Choose a positive integer no for which Ci = IPi(no)I > 1 for
i= 1, 2, ... , n. Set a= no+ Tc1c2 · ... · Cn. Then, Pi(a)
O (mod ci) for all
i . Thus cilPi(a). By taking T sufficiently large, we obtain IPi(a)I > ci for all
i, and hence Pi(a) is composite.
=
129
Number theory and polynomials
Example 7.9.
Prove that there is a polynomial P(x) of degree d > 2 with
integer coefficients such that the terms of the following sequence are pairwise
coprime for all integers n:
P(n),
P(P(n)),
Solution.
Put P (x) = x (x - 1) Q (x) + 1, where deg Q(x) = d - 2, a
polynomial with integer coefficients. Let p(k>(x) denote the k-fold composition of P with itself. We will prove that for any integers n and k > 1 we
have gcd(n , P(k)(n)) = 1 (in fact, we will actually prove the stronger statement that p(k)(n) = 1 (mod n)). Applying this to p(m)(n) we conclude that
gcd(P(m)(n), p(m+k)(n)) = 1 and hence any two terms in the given sequence
are relatively prime.
Choose an arbitrary n. We will prove by induction on k that p(k) (n) = l
(mod n) for all k > l . For k = l we have
P (n)= n(n - l)Q (n)+ l = 1 (mod n).
For the inductive step we have
p(k+l)
(n)
p ( p(k) (n))
p(k) (n) (p(k) (n)-
1)Q (p(k) (n))+ 1
0+1
1 (mod n).
This completes the proof.
Example 7.10. (Tournament of Towns) Find all positive integers n such
that for all polynomials P( x) with integer coefficients of degree n, there are
infinitely many pairs (a, b) of distinct positive integers such that P(a) + P(b)
is divisible by a + b.
Solution. We prove that all the even n satisfy the given condition. Let us
first prove that if n is odd, then n does not satisfy the problem condition.
Take P (x) = 1 + xn, where n is odd. Then
P (a)+ P (b) = 2 +an+ bn = 2
(mod a+ b).
Thus a + b < 2, and we see that there are only finitely many pairs (a, b). Thus
n does not satisfy the conditions of the problem. Now, let n be even. Write
P(x) = Q(x) + R(x), where
Q (x) = P(x) + P(-x)
2
and R (x) = P(x) - P(-x)
2
.
Number theory and polynomials
130
Since R(a) + R(b) is divisible by a+ b, it remains to consi~er Q(a) + Q(b).
Note that Q(x) is of degree n > 1. Without loss of generahty, assume that
the leading coefficient of Q(x) is positive (otherwise consider the polynomial
-Q( x)). Then, there are infinitely many positive integers m such that
Q(m)>2m.
Now, assume a= m and b = Q (m) - m > m. Th en, a+ b = Q(m). Since
Q(x) is an even polynomial , we can find that
Q (a)+ Q (b)
Q(m)+Q(Q(m)-m)
Q(m)+Q(-m)
2Q(m)
O (mod Q(m)).
We find infinitely many pairs (a, b) satisfying the problem statement.
Example 7.11. Let an be a geometrie progression cons isting of integers with
a common ratio which is not ± 1. For any non-constant polynomial P( x) with
integer coefficients prove that there are infinitely many positive integers n such
that IP(an)I composite.
Solution. Assume the contrary, i.e., assume that for all n > M, IP(an)I is
prime. Choose m > M and let IP(am)I = q be a prime number. Assume that
r is the ratio of the progression. Then
(amąt)k = (aormqtl = a~rkmąt = a~rkm
(modą).
Hence, we get that (amąt)k = (am)k (modą) for all positive integers k,r.
lt follows that
P(amąt) = P(am) = O (modą).
Taking t sufficiently large, we obtain IP(amąt) I > IP( am) I = q.
Hence IP( amąt) I must be composite .
Example 7.12. Let J(n) = 1 + 2n + 3n 2 + 2016n 2015and (to, t 1, ... , t 2016),
(so,s1, .. . , s2016) be two permutations of (O, 1, 2, . .. , 2016). Prove that there
are two distinct numbers in
A= {sof (to), s1f (t1), . . . , s2016!(t2016)}
such that their difference is divisible by 2017.
Solution . Note that 2017 =pis a prime number. Consider the polynomial
J(n) = 1 + 2n + ... + (p _ l)nP-2_
131
Number theory and polynomials
Now, (n - 1) 2 f (n) = pnP - pnP-I - (nP - 1), so
(n - 1)2 f (n) = -(nP - 1) (mod p).
By Fermat's Little Theorem (FLT), if gcd(n,p)
= 1, then
(n - 1)2 f (n) = -(n - 1) (mod p).
lf n 't 1 (mod p) , then f(n) = -Hence , if f(m) - f(k)
1
n-l
-
(mod p).
(mod p) and k , m t 1 (mod p), we obtain
k- m
(mod p).
Finally,
f (1) = 1 + 2 + ... + p - l - O (mod p).
This proves that {f(O) , ... , f(pp. Now, consider the set
l)} is indeed a complete residue system mod
A= {sof(to), s1f(t1), ... , sp-1f(tp-1)}
and assume the contrary. This set forms a complete residue system modulo
p. Thus, we find that there must be only one O remainder. Without loss of
generality, assume that so = O, to = 1. Then,
{s1f(ti) , . . . , Sp-1f(tp-i)} = {1, 2, ... ,P - 1} (mod p).
Hence ,
s1f(t1) · ... · Sp-1J(tp-1) = 1 · 2 · ... · (p- 1) = (p- 1)! = -1
(mod p).
Moreover,
{J(t 1), .. . , J(tp-i)},
{s1, ... , Sp-d
= {1, 2, ... ,p-
1} (mod p).
Hence,
s1f (t1) ·.. .·Sp-1J(tp-1) - s1·... ·Sp-I· f (t1)·... ·f (tp-1)
=((p-1)!) 2 =1 (mod p ),
which gives a contradiction.
Example 7.13.
n such that
Let P(x) = 20x 2 -llx+2016.
2016
22016
Prove that there is an integer
I P(n).
Solution. We will prove that for all positive integers n there is an integer
en such that 2n I P(cn)- Let n = l and c1 = 2. Assume that the claim is
true for all positive integers less than or equal to n. In particular, we have
2n I P(en) = Dn. Let Cn+l = Cn+ Dn. Then,
P(Cn+1) = P(cn + Dn) = l0Dn(2Dn + 4Cn- 1).
Hence 2n+l I P(cn+1).
Number theory and polynomials
132
Solution 2. Let n = 2k for some positive integer k . We will prove that
P(l) , .. . , P(n) have different remainders modulo n. Assume the contrary.
Then, there is 1 :S a< b < n such that P(a) - P(b) is divisible by n, i.e
P(a) - P(b) = (a - b)[20(a + b) - 11).
Since 20( a + b) - 11 is odd, we get that a - b must be divisi ble by n . But
a - b < n . So, there is a positive integer 1 < m < n such that P( m) = O
(mod n). Now, take n= 220
162016
.
An important result that you will use many times , during your life of fighting with mathematical olympiads problems , is called Schur's Theorem. This
invaluable theorem states the following fact.
Schur"s Thcorcm
Let P(x) be a non-constant polynomial with integer coeffi.cients. Then
the set of prime divisors of the sequence Xn = P(n) is infinite.
There are so many approaches to proving this theorem . An easy one is to
write the polynomial P(x) = adxd + · · · + a0 , and consider the sequence
P (aox) = ao(adag- 1xd + · · · + 1). Then we apply Euclid's method for proving
the infinitude of prime numbers. We assume there exist only finitely many
prime numbers Pl , ... , Pm and take x = Pl · . .. · Pm to obtain a contradiction.
Example 7.14. (Adrian Beker) Find all the polynomials P(x) with integer
coeffi.cients such that P(O) i- O and
p(n)(m). p(m)(n)
is a square of an integer for all nonnegative integers m and n.
(Note that we define the iterated composition p(k)(n) by p(o)(n) = n, and
p(k)(n) = P(P(k-l)(n)) for all the integers k > O).
Solution. Clearly , the polynomial P(x) = 1 + x satisfies the statement of the
problem. Now we prove that there isn't any other polynomial satisfying this
property. Let an= p(n)(O). Setting m = O in the hypothesis, we see that nan
must be a square for each n> O. We now prove the following lemma.
Claim. Suppose Pis a polynomial with integer coeffi.cients and P(O) i- O and
define an = p(n)(O). If nan is a square for all n, then for all suffi.ciently large
primes p the sequence an (mod p) is periodic with the minimal period length
p.
Proof. Take p > max{2, IP(O)I}. By the Pigeonhole Principle it is elear that
there must be some j > i such that aj = ai (mod p). Hence if we let t = j -i,
we see that ak+mt = ak (mod p) for all m > O and k > i. Thus the sequence
an (mod p) is eventually periodic. Let T be the length of the smallest (i.e.,
133
Number theory and polynomials
fundamental) period. Clearly T < p. We need to prove that T = p. Assume
the contrary. Since O < IP(0)I < p, this cycle of values modulo p cannot be
all O (mod p). (If ar
O (mod p) , then ar+l
P(O) t= O (mod p).) Thus
we can choose an ar on the cycle with ar t= O (mod p). Take an arbitrary
positive integer n. There is an integer k such that n+ kp= r (mod T). Since
(n + kp )an+kp is a square, we get that (n + kp )an+kp must be a quadratic
resid ue mod p. But n+ kp n (mod p) and an+kp ar (mod p), so this says
that nar must be a quadratic residue mod p for all n. This is a contradiction,
thus T = p. This forces the cycle to go through every congruence class modulo
p, in particular , to include O (mod p). Hence the sequence is periodic, and not
just eventually periodic.
Before preceding with the solution of the problem we give an alternative approach to proving this Claim.
Proof 2. As above the sequence an (mod p) is eventually periodic with minimal period T. It is elear that any other period length must be divisible by
T. (If M is another period length, write M = Tq + r where O < r < T.
Since M is a period, we have as+M = as (mod p) for all sufficiently large s.
Since T is also a period, we find that as+r = as (mod p) for all large s. But
then minimality of T implies r = O and hence M is a multiple of T.) Now,
choose a prime p > IP(0)I- Since pap must be a square, we get that ap is
divisible by p. Therefore, for all n> O we have an+p = pn(ap) = pn(o) _ an
(mod p). Hence an is periodic mod p with a period pand this implies that T
must divide pand therefore TE {l,p}. If T = 1, then P(an) = an (mod p)
for all sufficiently large n. Thus eventually the sequence an (mod p) takes on
only one value. But we have already seen that akp = O (mod p), the value
P(O) (mod p), a
O must be on the cycle. However this gives O P(ap)
contradiction since p > IP(0)I > O. It follows that T = p.
Now we return to the problem. Pick a prime p > 2IP(0)I , then the sequence
an (mod p) is periodic with length p. Since the cycle an (mod p) must include
every residue class modulo pand applying P just cycles these values, it follows
that {P(O), P(l) , ... , P(p-1)} must be a complete residue system modulo P.
We claim that this implies <legP(x) < 1. Assume the contrary and consider
the polynomial Q(x) = P(x + 1) - P(x) of degree <legP(x) - 1. Then, by
Schur 's theorem, there are infinitely primes dividing Q(x ). Take q a sufficiently
large prime dividing Q(x). Then P(n + 1) - P(n) (mod q) for some n E
{O, 1, ... , q - 1}, which contradicts the surjectivity property proven above. It
follows that P(x) = b + ax for some integers a, b. After some calculations
and number theoretic considerations (which are left to the reader) we obtain
a=b=l.
=
=
=
=
=
=
Example 7.15. (Iranian Mathematical Olympiad 2010) Find all polynomials P(x) with integer coefficients such that for all primes q and positive integers
134
Number theory and polynomials
u,v we have that if q I (uv -1), then q I (P(u)P(v) -1).
Solution. We know that the set of primes dividing the sequence Xn = P( n)
is infinite. Assume that P(x) = a0 + ... + adxd is non-constant and suppose
that ao # O. Let q > laolbe a prime dividing Xu = P(u). Since q does not
divide ao, it follows that ą does not divide u. Since gcd(u, q) = 1, there is an
integer v such that q I (uv - 1). Now P(u)P(v) - l - -1 (mod q), contrary
to the hypotheses. This implies that a 0 = O. Assume that P(x) = xkQ(x),
where Q(O) # O. Then it is easy to check that Q still satisfies the hypothesis
of the problem. Indeed if uv = l (modą), then
P(u)P(v) - 1 = (uvlQ(u)Q(v)
- 1 = Q(u)Q(v) - 1
(modą).
Hence by the argument above Q must be a constant polynomial (i.e., Q(x) =
c). In this case Q(u)Q(v) - 1 = c2 - 1 = O (modą) for all sufficiently large q.
Then c = ±1 and we obtain P(x) = ±xd.
Solution 2. Let R(x) = xdP(:)
where dis the degree of Pand
Jety be the
multiplicative inverse of x mod q. We get that
P(y) =PG)
(mod q).
Then, since q divides P(x)P(y) - 1, we obtain
xd = xdP(x)P(y)
= xdP(x)PG)
= P(x)R(x)
(mod q).
Then q I P(x)R(x) - xd for any x with gcd(x, q) = l. Now taking ą large
enough, we find that
Since P(x) divides xd it follows that P(x) = ±xk for some k, and since P has
degree d we find that P(x) = ±xd.
Jntroductory problems
8
135
Introductory
problems
l. (High School Math Journal 2015) Let x and y be positive integers such
16
that 2014 I
Lx
16
16
- i y i . Prove
that 201416 I
i =O
L x l6- iyi.
i=O
2. (Chinese Western :tvlathematic al Olympiad 2010) Let a 1, . .. , an and
b1, . . . , bn be non-negative real numbers such that
n
L(ai
n
+b i)= 1,
i =l
n
L i (ai - bi) = O, L i (ai + bi) = 10.
2
i =l
i=l
10
Prove that max(ak, bk) < ---.
2
- k + 10
3. If
P(x) = (1 - x)(l + 2x)(l - 3x) • ... • (1 + 14x)(l - 15x) ,
find the absolute value of the coeffi.cient of x 2 in P(x).
4. (Aleksander Khrabrov - Saint Petersburg Mathematical Olympiad 2001)
Let f (x) and g(x) be quadratic polynomials such that for all integers n,
then f(n)g(n) is an integer. Is it true that for all integers n, then also
f(n), g(n) and J(n) + g(n) are integers?
5. (Murray Klamkin - Crux Mathematicorum) Let P(x) and Q(x) be monie
polynomials with non-negative real coefficients such that
P(x)Q(x)
where degP(x)
= 1 + x + x 2 + ... + xm+n,
= m and degQ(x) = n. Prove that:
(i) if m and n are odd , there areno such polynomials;
(ii) if m = n , there are no such polynomials;
(iii) for every m there are infinitely many n such that there are such
polynomials ;
(iv) the coeffi.cients of P(x) and Q(x) are O or 1.
6. (Moscow Math ematical Olympiad 2015) Are there two polynomials with
int eger coeffi.cients such that each of them has a coeffi.cient whose absolute value is greater than 2015, but for the product of these two polynomials the absolute values of all the coeffi.cients do not exceed 1?
Introductory problems
136
7. (Moscow Mathematical Olympiad 1997) Three functions
fi (x) = x + .!:_, h (x) = x 2 ,
h (x) = (x - l) 2
X
are written on the blackboard. You are allowed to add, subtract and
multiply these functions (and so also you can square them, cube them
and so on), multiply them by an arbitrary number, add an arbitrary
number to them and perform the same operations with the expressions
thus obtained. Construct the function .!:_by means of these operations.
X
Prove that if we erase any of the functions fi (x) , h (x) , h (x) from the
blackboard, the function .!:_will no longer be constructible.
X
8. (Mathematics and Youth 2006) Let a, b, c be non-negative real numbers
such that a+ b + c = l. Find the greatest and least values of the
polynomial
f (a, b, c) = a(b - c)3 + b(c - a) 3 + c(a - b)3 .
9. (Ivan Borsenco - Mathematical Reflections J124) Let a and b be integers
such that lb- al is an odd prime. Prove that the polynomial
P (x) = (x - a) (x - b) - p
is irreducible over Z[x] for any prime p.
10. (Vietnamese Mathematical Olympiad) Find all polynomials P(x) with
integer coefficients such that
11. (Vietnamese Mathematical Olympiad 1997)
(i) Find all polynomials P(x) of the lowest degree with rational coefficients that satisfy P( ~ + ijg) = 3 + ~(ii) Does there exist a polynomials P(x) with integer coefficients that
satisfies P( ~ + ijg) = 3 + ~?
12. (~ongolian Mathematical Olympiad 2014) Find all polynomials P(x)
w1th the property that for all positive integers k there is a positive integer
m such that P (2k) = 2m.
137
Jntroductory problems
13. (łv1oscow Mathematical Olympiad 2008) Let k > 6 be a natural number.
Prov e that if some polynomial with integer coefficients takes values at k
integ er point s th at are among numb ers from 1 to k - 1, then these values
are equal.
14. (Andy Liu - Tournam ent of Towns 2009) Consider the lattice points
(x , y) where O < y < 10. We construct a polynomial of degree 20 with
integ er coefficients. What is the maximal number of those lattice points
located on the graph of the polynomial?
15. (G. Zhukov - Kvant M2427) There are N real numbers written on the
blackboard. At each step we construct a polynomi al with the numb ers on
the blackboard and write its real roots (if any) to the blackboard. After
a finite number of operations we see that among the numbers written on
the blackboard there are all the integers from -2016 to 2016. What is
the minimal value of N?
16. (Vietnamese Mathematical Olympiad 2015) Define a family of polynomials by fo( x) = 2, fi( x) = 3x and for all n> 2,
fn (x) = 3xfn-l (x) + (1 - x - 2x 2)fn-2 (x).
If Jn (x) is divisible by x 3 - x 2 + x, find n.
17. (Mongolian Mathematical O lym pia d 2016) Th e polynomials P (x) and
Q(x) satisfy P( x) 2 = 1 + Q(x)3. Prove that both polynomials must be
constant.
18. (Bogdan Enescu - Mathematical Reflections S40) Let f and g be irreducible polynomials with rational coefficients and let a and b be complex
numbers such that f (a) = g(b) = O. Prove that if a+ b is a rational
number , then f and g have the same degree.
19. (Czech-Polish-Slovak Match 2012) The polynomial P(x) with integer
coefficients has the following property:
For all polynomials F(x), G(x) , Q(x) with integer coefficients such that
P(Q(x)) = F(x)G(x)
one of the polynomials F(x), G(x) is constant.
be constant.
Prove that P(x) must
20. (Aleksander Khrabrov - Saint Petersburg Mathematical Olympiad 2013)
The quadratic polynomials J( x) and g(x) are given. It is known that
the equation J(x)g( x) = O has exactly one real root and the equation
J(x) + g(x) = O has exactly two real roots.
Prove that equation J(x) - g(x) = O has no real roots .
138
Introductory problems
21. (Fedor Petrov - Kvant M2433) Let f(x) be a third degree polynomial.
We call (a, b, c) a cycle if we have
f (a) = b, f (b) = c, J (c) = a.
We know that there are 8 cycles containing 24 different real numbers
and for each cycle (ai , bi , ci), where i= 1, 2, . .. , 8, we compute the value
of ai +bi+ Ci• Prove that between the 8 values , we have
(a) at least three distinct num bers;
(b) at least four distinct numbers.
22. (D. Petrovsky - Ukrainian Mathematical Olympiad) Consider the polynomials P(x) , Q(x) with real coefficients. We know that the polynomial
S(x) = P(x)Q(x) has only positive coefficients. If P(O) > O, prove that
for all x > O we have
23. (P. Kozhenskov - Kvant M2438) Let 9o(x) be a polynomial such that
deg9o(x) = n and with n distinct real roots x 1, ... , Xn. We construct
polynomials 91(x) , 92(x) , ... , 9n (x) in this manner
9o(x)
91(x)
aoxn + a1xn-l + · ··+an-IX+
an
2
a1xn + · · · + an-1X + anx + ao
9n(x)
anxn + aoxn-l
+ · · · + an-2X + an-1•
Define bi = 9i(x1) for i = 1, ... , n. Prove that if b1 =f.O, the polynomial
J(x) = b1xn-l + · · ·bn-IX + bn has roots X2, .. . , Xn,
24. (Dorin Andrica - Mathematical Reflections S81) Consider the polynomial
P(x)
n
1
k=O
n+ k + 1
= I:---xk
with n> 1. Prove that the equation P (x 2) = (P(x)) 2 has no real roots.
25. (I. Bogdanov - Russian Mathematical Olympiad 2011) The nonzero real
numbers a, b, c are such that any two of the three equations
11
ax
4
+ bx + c = O, bx 11 + cx 4 + a = O, cx 11 + ax 4 + b = O
have a common real root. Prove that all three equations have a common
real root.
139
Jntrodu ctory problems
26. (Mat hemat ics and Youth) Let n be a posit ive int eger and let P >n + 1
be a prime. Prove that the equat ion
1 + --
x
n +1
+ --
x2
2n + 1
+ .. · + --
xP
pn + 1
= o.
has no int eger solution.
27. (Baltic Way 2016) Find all quadruples (a, b, c, d) of rea l numb ers satisfying t he system of equat ions
a 3 + c3
a 2 b + c2 d
b3 + d 3
2
O
ab + cd
-6
2
2
l
(Hint: define the polynomial P (x) = (ax + b)3 +( ex + d)3).
28. (A. Golovanov - Russian Mathematical Olympiad 2012) Given a polynomial P( x) and the numbers a1, a2, a3, b1, b2, b3 such that a1a2a3 -:/-O.
Suppos e that for every real number x we have
Prove that th e polynomial P(x) has at least one real root.
29. (Czech-Slovakia Mathematical Olympiad 1995) Find all polynomials f
with real coefficients such that for every real number x the following
inequality hold s:
x f(x)f(l
- x) + x 3 + 100 > O.
30. (I. Bogdanov - Russian Mathematical Olympi ad 2010) Find all polynomial s P( x) of odd degree with real coefficients such that for all real
numb ers x we have P(P (x)) < (P(x)) 3 and the coefficient of x 2 in P( x )
is zero.
31. Let P( x) = xd + ad-1xd-I + ... + ao have all of its roots on the interval
[-1 , l]. Prov e that if jP(x) I > 1 for all x E [O, 1), then IP(x )I < 1 for all
x E(-1 , 0].
32. (Polish Mathematical Olym piad 2013) Let b, c be integers and let
f(x) = x 2 + bx + c
and let k1, k2, k3 be integ ers such that n I f (k1), n I f (k2), n I f (k3 ).
Prove that n I (k1 - k2)(k2 - k3)(k3 - k1).
Jntroduetory problems
140
33. Let J(x) = a2016x2+ bx + a2016e _ 1, where a, b, c are integers. Suppose
that the equation J(x) = -2 has two positive integer roots. Prove that
(f(l))
2
+ (f(-l))
2
2
is composite.
34. Find the number of polynomials P(x) = ax 2 + bx + c with integer coefficients such that P(l) < P(2) < P(3) and
(P(1)) 2 + (P(2)) 2 + (P(3))
2
= 22.
35. (Edward Barbeau) Let f(x) be a quadratic polynomial. Prove that there
are quadratic polynomials h(x) and g(x) such that
J(x)f(x
+ 1) = g(h(x)).
36. (I. Robanov - Russian Mathematical Olympiad 2003) Let
Q(x) = x 2 +ex+ d
P(x) = x 2 + ax +band
such that P(Q(x))
= Q(P(x)) has no real roots. Prove that b =I-d.
37. (A. Khrabrov - Saint Petersburg Mathematical Olympiad 2001) Let f(x)
and g(x) be quadratic polynomials with integer coefficients such that
f(x) > O and g(x) > O. If
:i:iv'2
>
:i:iv'2
>
for all real numbers x , prove that
for all real numbers x.
38. (A. Kanel-Belov - Moscow Mathematical Olympiad 2010) The sum of
any two of the three trinomials x 2 + ax + b, x 2 +ex+ d, x 2 +ex+ f does
not have real roots. Can the sum of all these trinomials have real roots?
39. (P. Kozhlov - Russian Mathematical Olympiad 2010) Let the polynomial
(x 2 + 2Oax+ 10b)(x 2 + 2Obx+ IOa) have no real roots, where a and b are
distinct real numbers . Prove that 2O(b- a) fJ.Z.
40. (Saint Petersburg Mathematical Olympiad 2005) In the figure below, we
have four points which are part of the graph of the polynomials
f (x) = x 3 + bx 2 + ex + a,
g (x) = x 3 + ax 2 + bx + c.
141
Jntroductory problems
N
p
•M
Q
Does there exist values of a, b, c such that the points M , P, Q belong to
the graph of f(x) and points M, N belong to the graph of g(x)?
41. Find all integ ers m such that the equation
x 3 + (m + 1) x 2 -
( 2m -
1) x - ( 2m 2 + m + 4) = O
has an integer root.
42. (V. Brayman) Do there exist integers a, b, c, d with a -=I-O such that the
equat ions have as many distinct positive integer roots as their degrees?
ax 3 + bx 2 + ex + d = O, bx 2 + ex + d = O, ex + d = O.
43. (Saint Petersburg Mathematical Olympiad 2012) If the numbers a, b, c
are distinct, prove that the system of equations
{
x 3 - ax 2 + b3
x 3 - bx 2 + c3
x
3
-
cx
2
+a
3
O
O
O
has no real roots.
44. (Saint Petersburg Mathematical Olympiad 2012) The real numbers a, b, c
are such that among the three equat ions x 3 -a x 2 +b = O, x 3 - bx 2 + c = O,
x 3 - cx 2 + a = O, any two of them have a common root. Prov e that
a= b = c.
45. The polynomial x 3 +a x 2 +bx +c = Ohas three real roots . If a 2 = 2(b+l),
prove that la - cl :S:2.
Jntroductory problems
142
·
46. (Belarusan Mathemat1cal
Olympia· d 2011) Let a , b, c be nonzero integers
.
f
(
)
_
ax2
+
bx
+
c
has
two
real
roots
m
.
such t h at t h e po 1ynom1a1 x F" d b
common with the polynomial g (x) = x 3 + bx 2 + ax + c. m a, , c.
2
3
4 7. Let a # O and let the polynomial P (x) = ax 4 + bx + cx - 2bx + 4a have
2
2
two real roots x 1, x 2 such that x1 x2 = 1. Prove that 2b + a_c = 5a ·
48. (Moldova Mathematical Olympiad 2008) The polynomial
P (x) = x 4 - 4x 3 + 4x
2
+ ax + b
has two positive roots x 1, x2 such that x1 + x2 = 2x1x2.
Find the maximum of a + b.
49. If the polynomials
P (x) = x 4 + ax 3 + bx2 +ex+ 1,
Q (x) = x 4 + cx 3 + bx
2
+ ax + 1
have two common roots, solve the equations P (x) = O and Q (x) = O.
50. Find all polynomials Q(x) of degree at most n-3
of the polynomial
P (x) = xn + nxn-l
n (n - 1)
+ ---xn2
such that all the roots
2
+ Q(x)
are real.
51. A polynomial with integer coefficients is said to be economical if its
leading coefficient is 1 and the set of all its other coefficients, including
the constant term, coincides with the set of its roots taken with their
multiplicities, i.e. if exactly m of the coefficients are equal to a, then
a is root of the polynomial with multiplicity m . Find all economical
polynomials of degree n in the cases:
(a) n= 2;
(b) n= 3;
(c) n= 4.
52. (Ukrainian Mathematical Olympiad 2016) A polynomial
p (x) = x2016+ 2O16x2015+ a2014x2014+ a2013x2013+ ... + 1
can be expressed as P (x) = (x - x 1 ) • ... . (x - x 2016), where among
t~e _numbe~s x1, ... ,x20rn at least 2015 are negative (not necessarily
d1stmct). Fmd all the coefficients of P(x).
Jntroductory problems
143
53. (Belarusan Mathematical Olympiad 2009) Let P(x) and Q(x) be nonconstant polynomials with integer coeffi.cients such that the polynomial
R(x) = P(x)Q(x) - 2009 has at least 25 distinct integer roots. Prove
that degP(x) > 2 and degQ(x) > 2.
54. Let P(x) be a polynomial with real coeffi.cients and assume that
(P(x))
3
= x9 + asx8 + a7 x7 + · · · + 15x + 1.
Find the sum of the squares of the roots of the polynomial P( x), knowing
that the sum of the coeffi.cients of the polynomial (P(x)) 3 is 216.
55. (N. Aghakhanov - Russian Mathematical Olympiad 2004) Let
P (x) = xd + · · · + ao
be a polynomial with integer coeffi.cients having d different integer roots.
Prove that if any two of the roots are relatively prime , then
gcd(ao, a1) = 1.
56. Let x 1, ... , x 48 be the roots of the polynomial P (x) = 18x48 + 3x + 2006.
48
Find the value of the sum
I: 1 +x· x . .
i
i=l
i
57. (Canadian Mathematical Olympiad 2010) Let P(x) and Q(x) be polynomials with integer coefficients. Let an= n! +n. Show that if
~[::i
is
P(n) .
.
c
.
an integer for every n, then Q(n) 1s an mteger 1or every mteger n such
that Q(n) =/=O.
58. (Canadian Mathematical Olympiad 2016) Find all polynomials P(x)
with integer coefficients such that P(P(n) + n) is a prime number for
infinitely many integers n.
59. (Ki.irschak Competition 2004) Find the smallest positive integer different
from 2004 with the property that there exists a polynomial f (x) with
integer coefficients such that the equation f (x) = 2004 has at least one
integer solution and the equatio n f(x) = n has at least 2004 distinct
integer solutions .
145
Advanced problems
9
Advanced
problems
1. The polynomial
akx k + ak+lX k+l + . .. + an-k-lX n-k-1
+ an-kX n-k
is said to be palindromie if O < ak, ai = an-i and ai < ai+ 1 for every
k < i < ; . Prove that the product of any two universal polynomials is
a palindromie polynomial.
2. (Mathematics and Youth Journal 2002) Let n be an even natural number. Find the number of polynomials Pn(x) of degree n such that:
(i) all the coefficients of Pn(x) belong to the set {-1, O,1} and
Pn(0)-/- O;
(ii) there is a polynomial Q(x) whose coefficients belong to the set
{-1, O,1} and Pn(x) = (x 2 - l)Q(x).
3. (Vietnamese Mathematical Olympiad 2015) Let a be a positive root of
equation x 2 + x = 5 and let co,c1, ... , Cn be non-negative integers such
that co+ c1a + · · · + enan = 2015.
(i) Prove that co+ c1+ · · · + Cn = 2 (mod 3).
(ii) Find the minimal value of co+ c1+ · · · + Cn·
4. (Czech-Polish-Slovak Match 2005) Find all values of n> 3 such that the
polynomial P(x) = xn - 3xn-l + 2xn- 2 + 6 is reducible over Z[x ].
5. (China Training Camps) Let n > 3 and let p be an odd prime. Prove
that the polynomial
f (x) = xn + p2xn-1 + ... + p2x + p2
cannot be represented as the product of two nonconstant
with integer coefficients.
polynomials
6. (Mongolian Mathematical Olympiad 2010) Let P(x) be a monie irreducible polynomial with integer coefficients such that IP(0)I = 2010.
22 0 10
Prove that the polynomial Q (x) = P(x
) is irreducible.
7. (Mircea Becheanu - Mathematical Reflections 0134) Let p be a prime
and let n be an integer greater than 4. Prove that if a is an integer that
is not divisible by p, then the polynomial
f (x) = axn - px 2 + px + p 2
is irreducible over Z [x].
Advanced problems
146
8. (Erdos) If J (x) = (x _ x 1 ) ..... (x - xn), where Xi E [-1, 1], prove that
there doesn't exist a E (-1, O) and b E (O,1) such that IJ(a)I > 1 and
IJ(b)I> 1.
9. (Oleksandr Rybak- Ukrainian Mathematical Olympiad 2008) Let n> 2.
Consider the polynomials Po(x), P1(x), . .. , Pn(x) such that
<legPi(x) = n - i and Pn(x) f=O.
For all 2 <i<
n there is a polynomial Qi(x) such that
Pi=
Pi-2
+ Pi-IQi(x).
Prove that if P0 (x)R(x) + P1 (x)S(x) = 1, then <legR(x) > n - 2 and
<legS (x) > n - 1.
10. (USATST TST 2014) Let P(x) and Q(x) be two polynomials with real
coefficients and <legP (x) = d. Prove that there exist polynomials A (x),
B(x), and C(x) such that
(i) degA(x), degB(x) <
d
2;
(ii) at most one of them is zero;
•..) A(x) + Q(x)B(x) = C( )
( 111
P(x)
x .
11. (Mircea Becheanu - Romanian TST 1981) Consider the polynomial
P(x) = xP-l + xP- 2 + •· •+ x + 1,
where p > 2 is a prime number. Prove that for any even positive integer
n, the polynomial
n-1
-1 +
ITP( xPk) is divisi ble by x + 1.
2
k=O
12. (Aleksander lvanov - Bulgarian Mathematical
Olympiad
2014)
Find all natural numbers n for which there exist polynomials
fi (x), . .. ,fn (x) ,g(x) with integer coefficients such that
(x
2013
+ n) I g(x) and ((fi(x)) 2 - 1) · ... · ((fn(x)) 2 - 1) = (g(x)) 2 - 1.
13. (Navid Safaei - Mathematical Reflections U448) Let p > 5 be a prime
number . Prove that the polynomial 2xP-p3Px+p 2 is irreducible in Z[x].
147
Advanced problems
14. (George Stoiea - AMM 11822) Let P(x) and Q(x) be polynomials with
eomplex eoeffieients sueh that all the coeffi.cients of the polynomial
P( Q(x)) are real. Prove that if the leading eoeffi.cients and the eonstant
term of Q(x) are real, then both P(x) and Q(x) have real eoefficients.
15. (Kiirsehak Competition 2017) Let p (P (x)) = (Q(x) )2 . Prove that there
is a polynomial R(x) sueh that P(x) = (R(x)) 2 .
16. (M. Dadarlat and G. Eckstein - Romanian TST 1989) Find all monie
polynomials P(x) and Q(x) with integer eoefficients sueh that
Q(O) = O and P(Q(x)) = (x - 1) • ... • (x - 15).
17. (Belarusan Mathematieal Olympiad 2017) If k > 2 and 55k =an .. . ao,
prove that the polynomial P(x) = anxn + ... + a0 has no rational roots.
18. Find all monie polynomials with integer coeffi.eients sueh that
P (O)= 2017
and for all rational numbers r, the eąuation P(x) = r has a rational
root.
19. Do there exist four polynomials P1(x), P2(x), P3(x), P4(x) with real
coefficients sueh that the sum of any three of them always has a real
root, but the sum of any two of them has no real roots?
20. Find all polynomials with real eoefficients such that if
P (x) + P (y) + P (z )= O
for real numbers x, y, z, then x + y +z=
O.
21. (Czeeh-Slovak Mathematieal Olympiad 1998) A polynomial P(x) of degree n 2: 5 with integer eoefficients has exaetly n distinet integer roots
and P (O) = O. Find all integer roots of P(P(x)) in terms of the roots of
P(x).
22. (O. N. Kochikhin - Moscow Mathematieal Olympiad 2016) Let
P(x) = xd + ad - lXd-l
+ ... + ao.
For some m > 2 all the real roots of the polynomial
P(P( ... P(x)) . . .) = p(m\x)
m times
are positive. Prove that all the real roots of the polynomial P(x) are
positive.
148
Advanced problems
23. (Putnam
2014) Prove that for all positive integers n, all roots of the
n
polynomial P(x) =
L 2k(n-k)xk are real.
k= O
24. (Chinese TST 2017) Prove that there exists a polynomial
P( x ) = x 58 + a1x 57 + ... + a5s
having exactly 29 positive and 29 negative roots and log 2017 lailare positive integers .
25. (Polish Mathematical
by:
Olympiad 1977) Let Wn be polynomials defined
Wn+1(x) = Wn(x) 2 - 1,
w1(x) = x 2 - 1,
(n= 1, 2, ... )
and let a be a real number. How many different real solutions does the
equation wn(x) = a have?
26. (Alexander Khrabrov - Tuymada 2005) Let f(x) = x 2 + ax + b be a
polynomial with integer coeffi.cients such that for all real numbers x
9
f(x) > -10·
Prove that for all real num bers x we have f (x) > 27. Let p and q be natural numbers such that
function
.
l
p2
8
1
4
.
< q < p 2 . Consider the
f (x) = x 2 - px + q and let a and b be coprime integers in
the mterval
[ą
J(a) and -;f(b) are mtegers.
.
p,p such that -bProve that
f(a) + f(b) = q.
28. (Cristinel Mortici) Let f(x) = ax 2 + bx + c, where c is integer.
infinitely many natural numbers n we have
f
For
(n+~)
> n n+ f (n+
n: < n+ n
2
-
I,
1
)
2
- 1.
Find the polynomial f.
29. Find all the pairs of real numbers (a, b) satisfying the following property:
for any pair of real numbers ( c, d) such that both of the equations
x
2
+ ax + 1 = c,
x 2 + bx + 1 = d
have real roots, then the equation x 2 + (a + b)x + 1 = cd also has real
roots.
149
Advanced problems
30. (German Mathematical Olympiad 2004) Let x 0 be a nonzero real root
of the polynomial ax 2 + bx + c, where a, b, c are integers and at least one
of b, c is nonzero . Prove that
1
lxol > lal+ Ibi+ lei- 1
31. Find all a such that the following system of equations has a solution
(a,b , c) which are distinct numbers in (-1 , 1]:
a3
b3
b+c+a
a+c+a
c3
b+a+a
32. (Czech-Slovak Mathematical Olympiad 2008) Find all the real numbers
a, b,c with the following property: Each of the equations
x 3 +(a+ l)x 2 + (b + 3)x + (c + 2)
x 3 + (a + 2)x 2 + (b + 1)x + (c + 3)
x 3 +(a+ 3)x 2 + (b + 2)x + (c + 1)
O
O
O
has three distinct real roots, but the total number of distinct roots is
five.
33. (Hong Kong Mathematical Olympiad 2015) Let a, b, c be distinct nonzero
real numbers. If the equations ax 3 + bx + c = O, bx3 + ex + a = O and
cx 3 + ax + b = O have a common root, prove that at least one of these
equations has three real roots (not necessarily distinct).
34. Let
P(x) = ax 3 + (b - a)x 2 - (c + b)x + c
and
Q(x) = x 4 + (b - l)x 3 + (a - b)x2 - (c + a)x + c,
where a, b, c are non zero real numbers and b > O. Let P( x) have three
distinct real roots xo, x1, x2 such that they are also roots of Q(x).
Prove that abc > 28.
Find all possible integer values of a, b, c.
35. (United Kingdom - Romanian Masters of Mathematics 2016) Let
an = n
3
+ bn2 +en+ d, where b, c, d are integers.
(i) Prove t hat t here is a sequence whose only terms which are perfect
squares are a2015 and a2016 •
Advanced problems
150
.
f
(ii) Determine the poss1ble values o a2015 • a2016
for the sequences sat-
isfying point (i).
36. Let a, b, c, d be positive real numbers such that the polynomial
ax 4 - ax 3 + bx 2 - ex + d
has four roots in the interval ( O, ~) . Prove that
21a + 164c > 80b + 320d.
37. (Mathematics Magazine) Find all rational numbers r1, r2, • · ·, r5 such
that
r 1r 5 = 1, r 1r4 + r2r5
r 1r 2 + r2r3 + r3r4 + r4r5
= 2, r1r3 + r2r4 + r3r5 = 3,
= 4 and r12 + r22 + r32 + r42+ r52_5
- ·
38. (Alexandru Lup~) Let
p (x) = ax4 + bx3 + cx2 +
4\1'2 - b
8 - a - 2c
2
x+
8
.
For all x E [-1, 1], we have P(x) > O. Find the value of a,b,c.
39. The coefficients of a polynomial ax 4 + bx 3 + cx 2 + dx + e are such that
a, e > O and ad 2 + b2 e - 4ace < O. Prove that this polynomial has no
real roots.
40. (Nikolai Nikolov - Bulgarian Mathematical Olympiad 2012) Let a# O,1.
Jim and Tom play the following game. Starting with Jim, and proceeding
alternately, each player replaces one * in the expression below with an,
where n E Z. Jim wins the game if the resulting polynomial has no real
roots, otherwise, Tom wins. What is a winning strategy?
41. (A. Golovanov - Tuymada 2013) Prove that for any fourth degree polynomial A(x), there are quadratic polynomials P(x), Q(x), S(x), R(x)
such that
A (x) = P (Q (x)) + R(S (x)).
151
Advanced problems
42. (Bulgarian Mathematical Olympiad 1995) Let
P(x) = xd + ad-IXd-l + ... + ao
be a polynomial with integer coefficients, where a0 # O. Suppose the
roots of P taken with multiplicity are the same as the coeffi.cients ai for
i= O, 1, 2, ... , d - 1 also with multiplicity. Find P(x) .
43. (Feng Zhigang - Chinese Western Mathematical Olympiad 2009) Let M
be a subset of 1Robtained by deleting finitely many real numbers from
lR. Prove that for any given positive integer n, there exists a polynomial
f(x) with degf(x) = n such that all its coeffi.cients and its n real roots
are in M.
44. (Ye. Malinnikova - Russian Mathematical Olympiad 1996) Does there
exist a finite set M of nonzero real numbers such that for any n E N
there is a polynomial of degree at least n with coeffi.cients in M , all of
whose roots belong to M?
45. Do there exist 2000 real numbers (not necessarily distinct), not all zero,
such that if we put any 1000 of them as roots of a monie polynomial
of degree 1000, its non-leading coeffi.cients are a permutation of the remaining numbers?
46. (Russian Mathematical Olympiad) If n> 3 and X1 < x2 < . .. < Xn are
the roots of an n-th degree polynomial P(x). Further assume that
X2 - XI<
X3 -
X2 < · · · < Xn - Xn-1·
Prove that IP(x)I on the interval [x1, Xn] attains its maximum value
between the two largest root (i.e., in the interval [xn-1, Xn]).
47. (Polish Mathematical Olympiad 1998) Let g(k) denote the greatest
prime divisor of an integer k if lkl > 2, and g(-1) = g(O) = g(l) = 1.
Find if there exists a non-constant polynomial W(x) with integer coefficients such that the set {g(W(x)) I x EZ} is finite.
48. (Marian Tetiva) Let f be a non-constant polynomial with integer coefficients and let k be a positive integer. Show that there are infinitely
many positive integers n such that J(n) can be written in the form
d1d2 · .. . · dk+I, where
152
Advanced problems
49. (Titu Andreescu - Mathematical Reflections U450) Let P be a nonconstant polynomial with integer coeffi.cients. ~rove t?~t f?r each
positive integer n there are pairwise relatively pnme positive mtegers
k1 , k2 , ... , kn such that k1k2 . .. kn = JP(m)I for some positive integer m.
50. (Crux Mathematicorum) Let P(x) be a polynomial with integer coefficients such that for all positive integers n we have P (n) > n. For all positive integ ers m, th ere exists a term in the sequence P (1) , P (P (1)), .. .
which is divisible bym . Prove that P (x) = 1 + x .
51. (Vlad 11atei) Find all polynomials P(x) with integer coeffi.cients such
that
(a 2 + b2 + c2 ) (P(a) + P(b) + P(c)) for all integers a, b,c.
J
52. Let P(x) be a polynomial with integer coeffi.cients such that
(r
22011 _
82 2011)
I (P(r) _ P(s))
for any positive integers rand s. Prove there is a polynomial Q(x) with
2017
integer coefficients such that P (x) = Q(x2
) .
53. (Polish Mathematical Olympiad 2009) The sequence of integers
Jo,fi, h, ... is defined by the conditions: Jo= O, fi = 1 and
Jn = Jn-1+ Jn-2, n= 2,3, ....
Find all the polynomials W with integer coefficients having the following
property: for each natural number n there is an integer k such that
W(k) = Jn54. (Korean Mathematical Olympiad 2008) Find all polynomials 'P(x) with
integer coefficients such that there are infinitely many positive integers
a and b such that gcd (a, b) = 1 and (a+ b) I (P(a) + P(b)).
55. (Fedor Petrov - Saint Petersburg Mathematical Olympiad 2002) Let
P (x) be a polynomial with real coefficients such that for all integers
n, k > O the number P(n + l) · · · · · P(n + k)
P(l) ... . . p (k)
is an integer. Prove that
P(0) = O.
3
2
56. ~et P (x) = ~ + 3x + 6x + 1975. Find the number of integers a in the
mterval [1, 3 17] such that P(n) is divisible by 32011.
°
Advanced problems
153
57. Let Q (x ) = (p - 1) xP - x - l, where pis an odd prime number . Prove
that there are infinitely many positive integers a such that Q(a) is divisible by pP.
58. (Oleksiy Klurman - Ukrainian Mathematical
Xn
Olympiad 2016) Let
= 2016P(n) + Q(n),
where P(x) and Q(x) are non-constant polynomials with positive integer
coefficients. Prove that there are infinitely many primes p for which there
exists a square-free integer m such that p I Xm.
Solutions to introductory problems
10
155
Solutions to introductory
problems
1. (High School Math Journal 2015) Let x and y be positive integers such
16
that 2014 I
L
16
x
16
-iyi.
Prove that 2014 16 I
i=O
L
x 16-iyi.
i=O
Solution. We prove the stronger lemma.
Lemma. Let p = 2 + k(2n + 1) be a prime, where k, n are non-negative
2n
integers and
L
2
x n-iyi
is divisible by p. Then x, y are divisible by p.
i=O
Proof . It is obvious that if one of x and y is divisible by p, then also the
another one is divisible by p, so assume gcd(p, xy) = l.
2n
From p I
L
x 2n-iyi
we find that p I (x 2n+l
- y 2n+l ), thus
i=O
P I xk(2n+l)
_ yk(2n+l).
Furthermore by Fermat's Little Theorem we find that
p I xp-1
_ yp-1
= xk(2n+l)+l
_ yk(2n+l)+l
_
Hence,
P I (xk(2n+l)+l
_ yk(2n+l)+l)
_ y(xk(2n+l)
L
x 2n-iyi
= (x _ y)xk(2n+l)
.
Thus x = y (mod p) which leads to
Since p f x, it follows that p I x -y.
2n
_ yk(2n+l))
=(2n + l)x =O (mod p),
2
n
i=O
which is a contradiction since gcd((2n + l)x,p) = 1. So, x and y are
divisible by p.
Back to our problem, note that 2014 = 2 · 19 · 53. Thus, 19 = 17 · 1 + 2
and 53 = 17. 3 + 2. By the lemma we find that 2, 19, 53 divide both x
and y, thus 2014 divides both of them and we are done.
2. (Chinese Western Mathematical Olympiad 2010) Let a1, ... , an and
b1 , ... , bn be non-negative real numbers such that
n
L(ai
i=l
n
n
+ bi) = 1, Li(
ai - bi) = O,
2
i=l
i=l
10
Prove that max(ak, bk) < k 2 +
L i (ai +bi)= 10.
10
•
Solutions to introductory problems
156
Solution. For all k = l, 2, . . . , n we have
(t b,)
(tibi)< (t i2b,)
2
i=l
i-1
i-1
(10-t,i
a,)(1-t,a,)
2
< (10 - k 2 ak)(l - ak)
10 - (10 + k 2 )ak + k 2 a~.
10
10
Thus ak < 2
and by the same argument bk < k 2 IO and we are
- k + 10
+
done.
3. If
P(x) = (1 - x)(l + 2x)(l - 3x) • . .. • (1 + 14x)(l - 15x),
find the absolute value of the coeffi.cient of x 2 in P( x).
Solution. Note that the coeffi.cient of x 2 is
L (-1/i·(-l)jj
15i < j515
15i<j515
1
2(64 - 1240)
-588 .
So, the absolute value of the coeffi.cient of x 2 is 588.
Solution 2. Note that
P(x)
1+(-1+2-3+···+14-15)x+ax
1- 8x + ax 2 + ... .
2
+-··
We have
P( -x)
= 1 + 8x + ax 2 + ....
Then
P(x)P(-x)
(1 - x)(l + x) · . .. · (1 - 15x)(l + 15x)
(1 - x 2 )(1 - 4x 2 ) · ... • (1 - 225x2 )
1 - (1 + 4 + · ••+ 225)x 2 .
Solutions to introductory problems
157
On the other hand the eoeffi.cient of x 2 in the expression is equal to
2a - 64. Thus, 2a - 64 = -1240, which gives a = -588.
4. (Aleksander Khrabrov - Saint Petersburg Mathematieal Olympiad 2001)
Let f(x) and g(x) be quadratic polynomials sueh that for all integers n,
then f(n)g(n) is an integer. Is it true that for all integers n, then also
f(n), g(n) and f(n) + g(n) are integers?
Solution.
Then
Let f(x)
= x 2 +x+a and g(x) = x 2 +x+b, where a+b = 1/2.
f(x)g(x)
= (x2 + x) 2 + x(x + l) + ab.
2
Choose a and b sueh that ab = -1. Now a and b are the roots of the
trinomial x 2 - ; - 1. This trinomial has two real irrational roots, thus
for all integers n we have that f(n)g(n) is an integer, but neither f(n),
g(n), nor f(n) + g(n) are integers.
Solution 2. Take
f(x)
=
(x - v'2)(x - v'3)and g(x) = (x + v'2)(x + v'3).
Thus f(x)g(x) = (x 2 - 2)(x 2 - 3) and for all integers n we have that
f(n)g(n) is an integer but neither f(n), g(n), nor f(n) + g(n) are integers.
5. (Murray Klamkin- Crux Mathematieorum) Let P(x) and Q(x) be monie
polynomials with non-negative real eoeffieients sueh that
P(x)Q(x) = 1 + x + x 2 + ... + xm+n,
where degP(x) = m and degQ(x) = n . Prove that:
(i) if m and n are odd, there areno sueh polynomials;
(ii) if m = n, there are no sueh polynomials;
(iii) for every m there are infinitely many n sueh that there are sueh
polynomials;
(iv) the eoefficients of P(x) and Q(x) are O or 1.
Solution. Let Rt(x) = 1 + x + · · · + xt. Henee (x - l)Rt(x) = xt+I - 1.
We know that all roots of Rt(x) are roots of unity. If t is odd, then
Rt (x) has one real root, namely -1. If t is even, then all the roots of
Rt(x) are complex and non-real. Let P(x) be a real monie polynomial
dividing Rt(x). Then all the roots of P(x) are t-th roots of unity. If z
is a root of P(x), then the complex eonjugate z is also a root of P(x),
and since z is a root of unity, the complex eonjuagte is 1/z. Thus P(x)
158
Solutions to introductory problems
and xd (
!)
are polynomiaJs of degree d with the same roots , thus they
differ by a constant multiple . Suppose P( x ) has degree d and write
P(x) = ao + a 1x + ••• + adxd. The roots of P( x ) come in complex
conjugate pairs (with product equal to 1) and -1 if d is odd . Thus
the product of the roots is ( -1 )d. This gives ad = ao and we see that
P(x) = xd P (
!)
and hence aj = ad-j
for j = O, 1, . .. , d. So, returning
to the problem statement, if we write P(x) = ao + · · · + amxm and
Q (x) = bo + · · · + bnxn , the n ao = am = bo = bn = 1.
(i) If m and n are odd, then both have a real root, but m + n is even
and Rm+n (x) has no real root, contradiction.
(ii) Let m = n.
m+n
Then the coefficient of xm is 1 =
ajbm-j > aobm + boam = 2,
L
j=O
contradiction.
(iii) Set n= r(m + 1) for some integer r. Then P(x) = 1 + x + •••+ xm
and Q(x) = 1 + xm+l
P(x)Q(x)
+ x 2(m+l) + · · · + xr(m+l).
= 1 + x + · · · + xr(m+l)+m
Now
= Rm+n(x)
and we are done.
(iv) Without loss of generality assume m < n. Then the coefficient of
xm is 1 =
m
m
j=O
j=l
L am-jbj = 1 + L am-jbj. Hence for all j = 1, ... , m
at least one of the components of the pair (am-j, bj) is zero. Since
am-j = aj it follows that at least one component of the pair (aj , bj)
is zero. Note that this also holds for m < j < n if we interpret
aj = Ofor j > m . Let 1 < k < n and assume that for all O < i , j < k
each aj, bi is either O or 1.
The coefficient of xk is 1 =
k
k-1
j=O
j=l
L ajbk-j = ak + bk + L ajbk-j.
k- 1
By our assumption
L ajbk - j is an integer, hence it is either O or
j=l
1. Thus ak + bk E {O,1}. If ak + bk = O, then ak = bk = O. If
ak + bk = 1, then since akbk = O one of them is 1 and the other one
is O. Now by induction we obtain the desired conclusion.
·6. (Moscow Mathematical Olympiad 2015) Are there two polynomials with
integer coefficients such that each of them has a coefficient whose abso-
Solutions to introductory problems
159
lute value is greater than 2015, but for the product of these two polynomials the absolute values of all the coefficients do not exceed 1?
Solution.
The answer is yes. We call a polynomial good if all its coefficients are O or 1. Note that the product of a good polynomial of degree
n and a polynomial xm + 1, where m > n, is again a good polynomial.
Starting with the polynomial x + l and multiplying it by 2019 polynok-1
mials xn
1
+ 1, xn + 1, ... , xn
2
2019
+ 1, where nk >
L
nj
and nk is odd
j=l
for each k = l, 2, . . . , 2019, then we obtain a good polynomial f(x) that
is divisible by the polynomial (x + 1)2020 . So,
f(x) = (x2020+ 2O2Ox2019
+ ... + l)(xk + axk-1 + ... + 1).
The coefficient of x 2020+k of the polynomial f (x) is 2020 + a and 2020 +
a < l, so a < -2019, which gives the desired conclusion.
Solution 2. Consider the polynomial
and
g(x) = h(x)(l + x + · · · + x 18).
Since deg h( x) = 18, we find that the coefficient of x 18 in g(x) is the sum
of the coefficients of h(x) , which is h(l) = 2160 > 2015. Now consider
the polynomial
g(x)g(-x)
= (1 - x6)(1 - xlO)(l - x18)(1 - x38).
We have that this polynomial has all the coefficients with absolute values
less than or equal to 1.
Solution 3. Consider the product of the form
P( x) = (1 - x)(l - x 2 )(1 - x 4 ) · ... · (1- x
22016
).
The coefficient of any monomial in this product is ±1 since the binary
representation of any positive integ er n is uniqu e .
where k1 < k2 < ... < km then the
I nd ee d ' 1.f n -- 2k1 + 2k2 + ... + 2km'
k1
2km
C
11
22011th
· 1± x n· is equ al to (-l)mx2 . · · · •X
• Thus 1or a n <
monom1a
.e
absolute value of the coefficient of all the monomials xn of the polynom1al
P( x) is 1. Since deg P( x) = 22017- 1 we are done . Now
22016_1)
2
P(x) = (l-x)2011(l+x)(l+x+x2+x3)·
... •(l+ x+x +··•+x
Solutions to introductory problems
160
and it is obvious that both
20 7
2
3
2
22016 1
( 1 - x) 1 and ( 1 + x) ( 1 + x + x + x ) · ... · ( 1 + x + x + .. · + x
- )
have a coefficient with absolute value greater than 2015.
Solution 4. Let P1(x) = Q1(x)R 1(x) such that absolute values of all the
coefficients of P 1(x) are less than or equal to 1. Assume that Q1 (x) has
a coefficient with absolute value greater than 2015 . Let deg(P1(x)) < n.
Now consider Q(x) = Q1(xn)R 1(x) and R(x) = R1(xn)Q1(x). Then,
Q(x)R(x) = P1(xn)P1(x) = P(x) . All the coefficients of P(x) have
absolute value less than or equal to 1. Moreover, both Q(x) and R(x)
have at least one coefficient with absolute value greater than 2015. Now,
set T8 (x) = 1 + x + •••+ x 8 - 1. We know that Tm(x)Tn(x ) I (xmn - 1) if
gcd(m, n) = 1 and Tm(x)Tn(x) = 1 + •••+ min(m, n)xmin(m,n)-l + •••.
Set m, n> 2015 and gcd(m, n)= 1. Then
and use the above procedure.
7. (Moscow Mathematical Olympiad 1997) Three functions
1
fi (x) = x + - , h (x) = x 2 , h (x) = (x - 1) 2
X
are written on the blackboard. You are allowed to add, subtract and
multiply these functions (and so also you can square them , cube them
and so on), multiply them by an arbitrary number, add an arbitrar y
number to them and perform the same operations with the expressions
thus obtained . Construct the function .!_by means of thes e operations.
X
Prove that if we erase any of the functions fi (x) , h (x) , h (x) from the
blackboard , the function .!_will no longer be constructible.
X
Solution.
Note that h (x) - h (x) = 2x - 1, so 2x is constructible and
x=
1
1
2(h (x) - h (x) + 1) = · 2x
2
is constructible. Then we can construct
1
x = fi (x) -
1
2 (h (x) - h (x) + 1).
It is obvious that .!_is not constructible by using only h (x) , h (x)
X
sin ce we could only every construct polynomials in x . Moreover, since
Solutions to introductory problems
161
ff (1) = f~ (1) = O, the derivative of any function constructed by using fi (x) , /3 (x) is zero at x = 1, but ! does not have this property!
(One can avoid using calculus for this c:Se though it is a little trickier.
The statement that g'(l) = O says that g(x) = g(l) has a double root
at x = 1. Thus one can use this phrasing instead, but the required
checking is a little more involved.) Finally, any function constructible
by use of fi (x), h (x) is of the form f(x) for some non-negative integer
xn
n and some polynomial f(x) and we can write them as /~:) where k is
x
a positive integer . Now define
A= { f(x)
x2k
f (x)-f(-x)
1 + x2
E JR[x]}.
x(l + x 2 )
1
EA.
Moreover
h
(x)
EA,
but
(/.A. Now we
Then fi (x) =
X2
X
prove that the sum and the product of any two elements in A are in A.
Let f (x) g(x) E A Obviously, the sum of any two elements in A is in
x2k ' x2l
·
A. Now
f(x) g(x)
f (x) g(x)
x2k . x2l = x2k+2l
and
f(x)g(x) - f( -x)g( - x) = f(x)(g(x) - g(-x))
+ g(-x)(J(x)
- J(-x)).
Since both g (x) - g (-x) and f (x) - f (-x) are divisible by 1 + x 2 we
are done. Thus ! cannot be represented by elements in A.
X
Solution 2. Proceeding as in the first solution until the construction
by fi and h, note that ·since fi (i)= O, h (i)= 1, for any function h(x)
constructed by fi (x), h (x), then h(i) E JR,but ~ (/.JR.
i
8. (Mathematics and Youth 2006) Let a, b, c be non-negative real numbers
such that a + b + c = 1. Find the greatest and least values of the
polynomial
f (a, b, c) = a(b - c)3 + b(c - a) 3 + c(a - b)3 .
Solution.
First we factor the polynomial f(a, b, c) . Note that
f (a, a, c) = f (a, b, b) = f (a, b, a) = O,
Solutions to introductory problems
162
thus the polynomial is divisible by (a - b) (b - c) (c - a). Then we can
write
f (a, b,c) = (a - b) (b - c) (c - a) g(a, b, c),
where <legg(a, b, c) = 1 and g(a, b, c) is symmetric, so the only choice is
g ( a, b, c) = K (a+ b + c) for some constant K. Looking at the coefficient
of a3 , we see that K = 1. It follows that
f (a, b,c) = (a - b) (b - c) (c - a) (a+ b + c).
Since a+ b + c = 1, then f(a, b, c) = (a - b)(b - c)(c - a). Now we use
two different approaches .
Approach I: Observe that f (a, c, b) = -f(a, b, c). We can assume that
max( a, b, c) = a. Suppose the maximal value is attained at (a, b, c).
It is easy to see that this maximum must be positive, thus a > c > b
(if any two of a, b, c are equal, we get f = O and if a > b > c, then we
get f < O). Now, set a+ b = d. Then d + c = 1 and
f(a,b,c)
(a-b)(b-c)(c-a)
(a- b) (c- b) (a - c)
< (a+b)c(a+b-c)
dc(d - c),
SO
f(a,b,c)
1
< -ud
UV
· vc · (d - c)
< ~ (ud+vc+d-c)
UV
3
3
)c)
~ ( (u + 1) d + (v - 1
UV
3
3
Now we search for values of u, v such that u+ 1 = v - 1 (since this will
make the upper bound independent of c) and for which we are in the
equality case of the above use of AM-GM, that is, with ud = vc = d _ c.
This gives
1
1
d
c
---=-----=1
u v
d-c
d-c
'
thus v - u = 2, uv = 2 then u = -1 + v'3,v = 1 + J3.
Plugging in these values for u and v, we get
Solutions to introductory problems
163
The equality is attained when
d=-1-=
2-u
1
3-v'3-
_3+v'3 _
6 'c-1-d-
b=O
'
a=---
_3-v13
6 '
3+ v13
6
Approach II: Assume that the maximal value is attained for (a, b, c)
and take c = min(a, b, c). Since f (a, b, c) = f (a - c, b - c, O), we can
assume c = O and thus we are reduced to maximizing
f (a, b, O) = ba(b - a)
over non-negative a, b with a+ b < 1. Since we are looking for a maximum, we may assume b > a. In this region we have an increasing
function of b, so we may further assume a+ b = 1. Thus we need to find
the maximal value of the following expression ,
ab (b - a) = (1 - b) b (2b - 1) = -2b 3 + 3b2 - b = g(b)
for b > 1/2. Since g(l/2) = g(l) = O this maximum will occur at a
critical point. We compute that g' (b) = -6b 2 + 6b - 1, which in the
3 + v13.Thus we·can find that the
6
. value 1s
. g (3 + v'3) = v'3, and we are done.
max1mal
interval [1/2, 1] vanishes only at b =
6
18
9. (Ivan Borsenco - Mathematical Reflections J124) Let a and b be integers
such that lb - al is an odd prime. Prove that the polynomial
P (x) = (x - a) (x - b) - p
is irreducible over Z[x] for any prim e p.
Solution. The quadratic polynomial P (x) = (x - a) (x - b) - p will be
reducible if and only if it has a rational (hence an integer root since P(x)
is monie) root. Let n be such an integ er root. Then (n - a) (n - b) = p,
so {n - a, n - b} = {1,p} or {n - a, n - b} = {-1, -p}. In either case
lb - al= l(n - a) - (n - b)I = p - 1
is even and hence cannot be an odd prime as required by the statement.
Solution 2. We want to show that
(x - a)(x - b) - p = x 2 - (a+ b)x + ab - p = O
164
Solutions to introductory problems
has no solutions in integers when lb- al is an odd prime and Pis prime.
The discriminant of this quadratic is
2
(a+ b)2 - 4ab + 4p = (a - b)2 + 4p = q + 4p
where q = lb - al , so the problem is equivalent to showing that ą 2 + 4p
cannot be a square. Suppose to the contrary that for some integer x we
have ą2 + 4p = x 2 . Hence , 4p = (x - q)(x + q).
In order for (x - q)(x + q) to be divisible by 4, x must have the same
parity as q, so letting x = q + 2y for some positive integer y, we now
have 4p = 2y(2ą + 2y), i.e. p = y(q + y). Since q is an odd prime, we
cannot have y = 1 (since then p = q + 1, which is impossible for p , q
prime with q odd) , therefore y > l, which is also impossible since th en
the prime p would factor. It follows that the discriminant cannot be a
square, therefore the polynomial is irreducible over the integers.
Solution 3. Assume by contradiction that P(x) = (x - a)(x - b) - p
is indeed reducible over Z[x] for some prime p. Then, since P(x) is a
quadratic polynomial , there must be some integers r and s such that
P(x) = (x - r)(x - s). Thus , P(r) = O or (r - a)(r - b) = p. However,
since l(r - a) - (r - b)I = la - bi is odd, their product p must be even,
so p = 2. Let x = r - a and y = r - b and without loss of generality, let
x < y. Since xy = p > O, either x , y are both positive or x , y are both
negative . Without loss of generality let them both be positive . Then
since lx-yl = la-bi, which is an odd prime, then xy > 1(1+3) > 2 = p,
contradiction.
10. (Vietnamese Mathematical Olympiad) Find all polynomials P(x) with
integer coefficients such that
Solution. Let Q (x) = P (x) - x. Clearly , Q (1 + ~) = O. We construct a polynomial with integer coefficients of minimal degree such that
1 + ~ is one of its roots. Let x = 1 + ~. Then, (x - 1)3 = 2, i.e.
2
x 3 - 3x + 3x - 3 = O. We claim that x 3 - 3x 2 + 3x - 3 divides Q( x ).
Indeed
Q (x) = (x 3 - 3x 2 + 3x - 3) H (x) + ax 2 + bx + c,
where H (x) has integer coefficients and a, b, c are integers.
can easily find that a= b = c = O. Then
Setting x = l +~'we
Q (x) = (x 3 - 3x 2 + 3x - 3) H (x) .
Solutions to introductory problems
165
Q (1 + v'5) = 1 - 2\/'5, then Q (1 _ v'5) = 1 + 2\/'5 . Let
2
R( x) = x - 3x + 3x - 3. Since R ( 1 + v'5) = 5\/'5 - 2, we find that
R ( 1 - v'5) = -5\/'5 - 2. Then
Furthermore,
3
Q(l-v'5)Q(1+v'5)=R(l-v'5)R(i+v'5)H(1-v'5)
H(1+v'5).
We can find that H ( 1 + v'5) = a + by'5 for some integers a, b. Then
H (1 - v'5) = a - b\!'5. Finally we have -19 = -121(a 2 - 5b2 ), contradiction!
11. (Vietnamese
Mathematical
Olympiad 1997)
{i) Find all polynomials P(x) of the lowest degree with rational coefficients that satisfy P( ~+W)=
3 + ~-
(ii) Does there exist a polynomials P(x) with integer coefficients that
satisfies P( ~ + W) = 3 + ~?
Solution.
We first prove the following lemma .
Lemma. If w= u~+
vW with u,v ,w E (Q, then u= v =w=
O.
Proof. By cubing both sides and using the identity
a
3
+ b3 + c3 - 3abc =(a+ b + c) (a 2 + b2 + c2 - ab - ac - be),
we have
w 3 - 3u 3 - 9v 3 + 9uvw = O.
Let N be the least common multiple of the denominators of u , v, w. By
multiplying the above equation by N 3 we can assume that u, v, w EZ.
Then 3 I w, so w= 3w1 for some w1 EZ. We find 3 u and then 3 I v. By
continuing in this way, we find by infinite descent that u = v = w = O.
Our proof is complete.
J
(i) Consider a polynomial P( x ) = ax + b with a, b E (Q.
. If P(x) satisfies the problem, then a(~+ W)+ b = 3 + ~' i.e.
(a - l) ~ + a W = 3 - b E (Q and hence a = a - l = O, impossibl e.
So, there doesn't exist such a linear polynomi al. Now consider a
quadratic P(x) = ax 2 + bx + c with a, b, c E (Q. If P(x) sat isfies the
problem, then
(a+ b)~
+ (3a + b)ij3 + 6a + c = 3 + ij3_
It follows that
a+b
3a+b 6a + C
O
1
3.
Solutions to introductory problems
166
Solving this system, we get a=
1
1
2, b = - 2, c = O. Therefore,
1
2
P(x) = -(x 2 - x)
is the only solution to the problem.
Cubing both sides we get s 3 = 9s+12, so that s is
Assume that there
the root of the polynomial Q (x) = x 3 -9x-12.
exists a polynomial P(x) of degree n > 3 with integer coefficients
such that P( s) = 3 + ~ - Then , dividing the polynomial P (x) by
Q(x), we have
P(x) = Q(x)T(x) +R(x),
(ii) Let s =~+W.
where T(x) and R(x) are polynomials with integer coefficients and
degR(x) < 2. So, .
3 + ~ = P(s) = Q(s)T(s) + R(s) = R(s).
By point (i) we know that there is a unique polynomial of degree
less than or equal to 2 with rational coefficients satisfying the given
condition. So there doesn't exist sucha polynomial P(x) .
12. (Mongolian Mathematical Olympiad 2014) Find all polynomials P(x)
with the property that for all positive integers k there is a positive integer
m such that P (2k) = 2m.
Solution. Assume P (x) = adxd + · · · + ao. Then ad> O.
Let 2N < ad < 2N+l - 1. Assume jaki < M for all k =O , ... , d.
Set x > l + 2M. Then by triangle inequality
Iad-1x
d-1
xd - 1 xd - 1
+ · · · + aoI <
M-< -- 2 ·
x -1
For k such that 2k > 1 + 2M , we have
p ( 2k) < ad2kd+
2kd _ 1
2kd
2
< (2N+l - 1) 2kd +
2- 1 < 2N+dk+l.
Moreover,
k)
p (2
> ad2kd - 2kd 2- 1 > 2N. 2kd - 2kd 2- 1 > 2N+dk-l_
~en~e P (2k) = 2N+dk = 2N · 2kd_ The equation P (x) = 2N . xd has
mfimtely many solutions. Thus P (x) = 2N. xd for all x .
Solutions to introductory problems
167
13. (Moscow Mathematical Olympiad 2008) Let k > 6 be a natural number .
Prove that if some polynomial with integer coefficients takes values at k
integer points that are among numbers from 1 to k-1, then these values
are equal.
Solution. Let x1 < · · · < Xk be the integers at which the polynomial P
takes values in 1, 2, ... , k - 1. Note that xk - x 1 > k - 1. Now
(xk - x1) I (P(xk) - P(x1))
while P (xk) - P (x1) < k - 2 implies that P (xk) = P (x1).
Then we can write
P (x) = P (xi) + (x - x1) (x - xk) Q(x)
where Q (x) has integer coefficients.
If P(xi) i- P(x1) for some i= 3, ... , k - 2 then IQ(xi)I i- O, so
IP(xi) - P(xi)I > l(xi - x1) (xi - xk)I > 2 (k - 2) > k - 2,
contradiction. Thus, P(xi) = P(x1) for all i = 3, .. . , k - 2.
We can write the polynomial P(x) as
P (x) = P (x1) + (x - x1) (x - x3) · ... · (x - Xk-2) (x - xk) R(x).
Now if P(x1) i- P(x2), then R(x2) i- O. Thus
.IP(x2) - P(x1)I > (k - 4)! (k - 2) > k - 2.
Then P(x2) = P(x1) and similarily P (xk-1) = P(x1).
14. (Andy Liu - Tournament of Towns 2009) Consider the lattice points
(x , y) where O < y < 10. We construct a polynomial of degree 20 with
integer coefficients. What is the maxima! number of those lattice points
located on the graph of the polynomial?
Solution.
We must consider the integer solutions of the equation
P (x) = c, where O < c < 10 is an integer. We prove that the answer is 20. If there are x1 < . . . < x21 that satisfy the conditions of
the problem, then we have a = x1, b = X21and thus b - a > 20. But
IP(b) - P(a)I < 10. Since
(b - a) I (P(b) - P(a)) ,
we find that P(b) = P(a) = C , where O< C < 10. Then set
P (x) = C + (x - a) (x - b) Q(x) ,
168
Solutions to introductory problems
where Q(x) has integer coefficients and deg Q(x) = 18. Now for all Xk
with 2 < k < 20, we have (xk - a) (b - Xk) > 19. But
IP(xk) - P(a)I < 10,
which forces Q (xk) = O for k = 2, ... , 20. But then P(x) - C is a
polynomial of degree 20 with 21 distinct roots, a contradiction . To
achieve 20 lattice points, we take P(x) = x(x - 1) · · · (x - 19).
Solution 2. With the same notation as before, for i = 1,. • • , 10 we
have x 21 - Xi > 11 and for i = 12, ... , 21 we have Xi - x1 > 11. Since
x 21- x 1 > 11, we have P (x 21) = P (x 1) = r. Now, all the Xj except xu
are the roots of P (x) - r. Then,
P (x) = r + a (x - x1) · ... · (x - x10) (x - x12) · ... · (x - x21).
Now, IP(x 11 ) - rl > (10!)2 , contradiction. Finally we construct a polynomial of degree 20 with integer coefficients satisfying the given conditions as in the previous solution.
15. (G. Zhukov - Kvant M2427) There are N real numbers written on the
blackboard. At each step we construct a polynomial with the numbers on
the blackboard and write its real roots (if any) to the blackboard. After
a finite number of operations we see that among the numbers written on
the blackboard there are all the integers from -2016 to 2016. What is
the minimal val ue of N?
Solution. Note that we need to start with O on the blackboard, otherwise we can never get O. Further we must have at least one nonzero
number on the blackboard. Therefore we must start with at least 2
numbers on the blackboard. We will show that if we start with the two
numbers O and 2016!, then we can construct all the numbers from -2016
to 2016. Thus the minimal value of N is 2. Let a = 2016!. Then we first
1
construct -1 by means of ax + a. We next construct ±
by means
va
of ax 2 - 1. Then we construct 1 by the polynomial
_!_x - 2--. Now
va va
we see that if the number b ever appears on the blackboard, then by
means of x + b we can construct -b. In particular , we can construct -a.
Now suppose the numbers O, 1, ... , M - 1 are on the blackboard, where
M < 2016, we claim we can construct M. To do this , we write 2016! in
base Mas 2016! = akMk + · · ·+a1M +ao, where ai E {O, 1, ... , M - 1}.
Since M I 2016! and M I (akMk + · · · + a1M), then a0 is divisible by M .
So, ao = O. Now we consider the equation
akxk + · · · + a1x - 2016! = O.
169
Solutions to introductory problems
All the coefficients are on the blackboard and x = M is a solution. Thus
by an easy induction on M we see that 1, 2, ... , 2016 can all be written
on the blackboard and hence also -1, -2, . . . , -2016.
16. (Vietnamese Mathematical Olympiad 2015) Define a family of polynomials by fo(x) = 2, fi(x) = 3x and for all n> 2,
fn (x) = 3xfn-l (x) + (1 - x - 2x 2)fn-2 (x) •
If fn (x) is divisible by x 3 - x 2 + x, find n.
Solution. We have
fn (x) - (x + 1) fn-1 (x) = (2x - 1) fn-1 (x) - (2x - 1) (x + l)fn-2 (x)
then,
fn (x) - (x + 1) fn-1 (x) = (2x - 1) Un- 1 (x) - (x + 1) fn-2 (x))
We find that
!n (x) - (x + 1) fn-1 (x) = (2x - 1r-1(x - 2),
whence,
r
f n (X) - (2x - 1
(x + l)Un-1(x) - (2x - 1r- 1)
(x + 1ruo (x) - (2x - 1)0 )
(x + 1r .
Thus , fn (x) = (2x - 1r + (x + l)n. Now we must find all n such the
remainder of (2x - 1r- 1 + (x + 1r when divided by x(x 2 - X + 1) is
zero. First, we must have fn (O)= O, which implies that n must be odd .
Moreover fn (- 2) = -(1 + 5n), which must be divisible by 7. Then n=
3+6m for some non-negative integer m. It is easy to check that for such
n the polynomial fn (x) is divisible by (2x-1)3+(x+1) 3 = 9(x 3 -x 2 +x)
and we are done.
17. (Mongolian Mathematical Olympiad 2016) The polynomials P( x) and
Q(x) satisfy P(x) 2 = 1 + Q(x)3 • Prove that both polynomials must be
constant .
Solution. Wtite the equality as (P (x) - 1) (P (x) + 1) = Q(x)3 . Since
the polynomials P (x) - 1 and P (x) + 1 have not common root s, we find
that
P(x) ·- 1 = A(x)3, P( x) + 1 = B(x)3.
Solutions to introductory problems
170
for some polynomials A(x), B(x). We have
B(x) 3 - A(x) 3 = 2.
Thus the polynomials B (x)-A (x) , B(x) 2 +A (x) B (x)+A(x)
constant. Since
2
ar e both
2
B(x) 2 + A (x) B (x) + A(x) 2 = (B (x) - A (x)) + 3A (x) B(x),
we find that A (x) B( x) must be constant. Then both A(x) and B (x)
must be constant and so P(x) and Q(x) are constant.
18. (Bogdan Enescu - Mathematical Reflections S40) Let f and g be irreducible polynomials with rational coeffi.cientsand let a and b be complex
numbers such that f(a) = g(b) = O. Prove that if a+ b is a rational
number, then f and g have the same degree.
Solution.
Consider h(x) E Q[x) such that h(x) = g(a + b- x). Clearly ,
h(a) = g(b) = f(a) = O.
Since f(x) is irreducible, then f(x) is the minimal polynomial of a and
then h(x) must be divisible by f (x).
Thus deg f (x) < deg h(x) = deg g(x). Analogously , we can prove that
degg(x) < deg f(x). It follows that f and g have the same degree.
19. (Czech-Polish-Slovak Match 2012) The polynomial P(x) with integer
coeffi.cientshas the following property:
For all polynomials F(x) , G(x), Q(x) with integer coeffi.cients such that
P(Q(x)) = F(x)G(x)
one of the polynomials F(x), G(x) is constant. Prove tha t P(x ) must
be constant.
Solution. Assume by contradiction that P(x) is nonconstant. First
suppose P(x) = ax +band a-=/-O. Taking Q(x) = ax2 + (b + l) x, we
have
P(Q(x)) = (ax + b)(ax + 1).
Now, let P (x) = adxd+ · · · + ao, where d > 1 and ad -=f.O.
Set Q(x) = x + P(x). Then
P(Q(x))
P(x + P(x))
ad(x + P(x))d + · · · + ao
P(x)T(x) + adxd+ · · · + ao
P(x)(l + T(x)).
It is elear that 1 + T(x) has degree d(d - 1) and hence is nonconstant .
Solutionsto introductory problems
171
20. (Aleksander Khrabrov - Saint Petersburg Mathematical Olympiad 2013)
The quadratic polynomials f(x) and g(x) are given. It is known that
the equation f(x)g(x) = O has exactly one real root and the equation
f (x) + g(x) = O has exactly two real roots.
Prove that equation f (x) - g(x) = O has no real roots.
Solution. We have two cases.
(i) Both polynomials f(x) and g(x) have only one real root xo, which
is in common. Then
f(x) = a(x - xo)2, g(x) = b(x - xo)2.
Then, f (x) + g(x) = (a+ b)(x - x 0 ) 2 . If a+ b = O, then f(x) + g(x)
has infinitely many roots. Thus, a+ b =JO and f (x) + g(x) has only
one real root, which contradicts the assumptions.
(ii) The polynomial f (x) has a double root and the polynomial g(x) has
no real roots . Without loss of generality, assume that the graph of
f (x) is upward. Then the graph of f (x) intersects the graph of
-g(x) in exactly two points. It is obvious that the graph of the
parabola -g(x) is upward and strictly above the x-axis. If not , the
graph of the parabola will intersect the x-axis. Thus , the graph of
g(x) is located in the lower half plane and it doesn't intersect the
graph off (x). This implies that the equation f(x) = g(x) has no
real roots, as desired.
y = f(x)
I
\
\
''
''
'
Solutions to introductory problems
172
Solution 2. As in the previous solution, we can reduce to the case
where f has one real root, g has no real root, and f + g has two real
roots. In terms of the discriminants this says that DJ = O, D 9 < O, and
D f +g > O. Since one easily computes that
DJ+g + DJ-g = 2(DJ + D 9 ),
it follows that D1_9 < O and hence f (x) - g(x) = O has no real roots .
21. (Fedor Petrov - Kvant M2433) Let f(x) be a third degree polynomial.
We call (a, b, c) a cycle if we have
J(a) = b, f (b) = c,
J(c) = a.
We know that there are 8 cycles containing 24 different real numbers
and for each cycle (ai, bi, Ci), where i= 1, 2, ... , 8, we compute the value
of ai + bi + Cż. Prove that between the 8 values, we have
(a) at least three distinct numbers;
(b) at least four distinct numbers.
Solution.
(a) Suppose that the sum S occurs for several cycles. Then, for any x
in the cycle we have
x + f(x) + f (f (x)) = S.
The polynomial on the left-hand side is of degree 9 and hence has
no more than 9 real roots. Each cycle consists of three of these
roots, so the sum S does not occur more than 3 times. Now we are
done .
(b) Write the above equality as S - x - f (x) = f(f (x)). Thus
f(S-x-f(x))
=f(f(f(x)))
=x.
Hence,
f (S - x - f (x)) + f (x) + f (f (x)) = S.
The degree of the above equation is 7. Indeed, let the coefficient of
x 3 in the polynomial f (x) be A. Then,
f (S - x - f (i))
+ f (x) + f (! (x))
A(S - x - f (x)) 3 + Ax 3 + Af(x) 3 + ...
-3Axf(x) 2 + ....
Thus, any sum S occurs no more than
done .
l;J =
2 times and we are
173
Solutions to introductory problems
22. (D. Petrovsky - Ukrainian Mathematical Olympiad) Consider the polynomials P(x) , Q(x) with real coefficients. We know that the polynomial
S(x) = P(x)Q(x) has only positive coefficients. If P(O) > O, prove that
for all x > O we have
Solution. If for some y > O we have P(y) < O, then the polynomial
S(x) must have a root in the interval (O,y), but this is impossible since
all the coefficients of the S(x) are positive. Thus for all x > O, we have
P(x) > O and similarly Q(x) > O. Now set
A (x)
2P (x),
B (x)
2Q (x),
C (x)
A (x) B (x) = 4P (x) Q (x) = 4S(x).
Write the original inequality as
4C(x 2) ::;;(C(x)) 2 + (A(x 3 )) 2 + 2B(x 3 ).
Then by the AM-GM Inequality, we have
(C(x)) 2 + (A(x 3 )) 2 + 2B(x 3 )
(C(x)) 2 + (A(x 3 )) 2 + B(x 3) + B(x 3)
> 4{!(C(x)) 2(A(x 3 )) 2(B(x 3 ))2
4JC(x)A(x 3)B(x 3) = ✓r-C-(x-)C_(_x-3).
Now, by the Cauchy-Schwarz's inequality
C (x) C (x 3 )
(ao + a1x + · · · + adxd)(ao + a1x3 + · · · + adx3d)
~ (ao + a1x 2 + · · · + adx2d)2 = (C (x 2)) 2
and then our proof is complete.
23. (P. Kozhenskov - Kvant M2438) Let 9o(x) be a polynomial such that
deg9 0 (x) = n and with n distinct real roots x1, ... , Xn- We construct
polynomials 91(x), 92(x ), ... , 9n(x) in this mann er
9o(x)
g1(x)
aoxn + a1xn-l + · · · + an-1X + an
a1xn + · · · + an-ix 2 + anx + ao
9n(x) -
anxn + aoxn-l + · · · + an-2X + an-I•
Define bi = 9i(x 1) for i = 1, .. . , n. Prove that if b1 =/O, the polynomial
f(x) = b1xn-l + · · ·bn-IX + bn has roots x2, ... , Xn.
Solutions to introductory problems
174
Solution. Set h(x) = (x _ x 1)J(x). Then, we prove that h(x) has roots
x1, x2, ... , Xn- Note that
h (x) = b1xn + (b2 - x 1 b1 ) xn-l
+ · · · + (bn - x1bn-1) X - x1bn.
Since
then
Now, for 1 < k < n - 1 we have
(ak+1X1 + ... + ak) - (akx~+l + ... + ak-1x1)
ak (1 - X1n+l).
bk+l - x1bk
-
Furthermore,
-x1bn
-anXl n+l - (aox1 n+· · ·+ an-1X1)
an (1- X1n+l) - 90 (x1)
-
an (1- X1n+l) .
Therefore the coefficients of h( x) are obtained from the coefficients of
g 0(x) by multiplying 1 - x~+l and we are done .
24. (Dorin Andrica - Mathematical Reflections S81) Consider the polynomial
n
l
k=O
n+ k + 1
P (x) = :E---xk
with n > I. Prove that the equation P ( x 2 ) = (P( x) )2 has no real roots.
Solution.
Suppose there exists a real root t to the equation. Since
p (t2) > 1 ~ n > O,
it follows that P (t2 ) = (P(t)) 2 > O. By Cauchy-Schwarz Inequality we
get
Solutions to introductory problems
175
which implies that
However, we have
n
Ln
k=O
a contradiction.
Solution
1
+k
.
1
+ 1 < (n+ 1) -n+ 1 = 1,
It follows that P (x 2 ) = (P(x)) 2 has no real roots.
2. Consider the polynomial Q(x) = P(x 2 ) - (P(x)) 2 . Since
Q(O) =
1
(n+ 1)2 > O,
1
n+l
the problem is equivalent to proving the inequality P(x 2 ) > (P(x)) 2 •
Define the vector
V(x) = (
✓2:n+ 1' ~,
. .. , v'nl+1).
Using the dot product, we have P(x 2 ) = V(x) • V(x)
= jV(x)j 2 and
P(x) = V(x) · V(l) = IV(x)I · jV(l)I · cosa ,
where a is the angle between V(x) and V(l). Since cos2 a< 1, if we can
show that jV(l)j 2 < 1, then we have the conclusion. Indeed,
1
1
1
1
IV ( 1) 12= V (1) . V ( 1) = 2n + 1 + 2n + ... + n + 1 < (n + 1) . n + 1 = 1.
25. (I. Bogdanov - Russian Mathematical Olympiad 2011) The nonzero real
numbers a, b, c are such that any two of the three equations
ax 11 + bx 4 + c = O,
bx 11 + cx
4
+ a = O,
cx
11
+ ax 4 + b = O
have a common real root. Prove that all three equations have a common
real root.
Solution. Assume that x = r is the common root of the polynomials
ax 11 + bx 4 + c, bx 11 + cx 4 + a. It is elear that r -=J.O, otherwise a = c = O
a contradiction. Then,
br 11 + cr 4 + a = O
ar 11 + br4 + c = O.
Solutions to introductory problems
176
Multiplying the first equation by a and the second by b and then subtracting, then multiplying the first equation by b and the second by c
and subtracting,
we get
r 4 (b2 - ac) - (a2 - be) = r 11 (b2 - ac) -
(c2 - ba) = O.
Now if b2 - ac= O then a2 - be= c2 - ba= O, which gives a= b = c.
'
'
.
.
2
Then, the statement of the problem 1s obv10us. Assume now b - ac # O.
Then,
1a2- bel
4
Ir I = Jb2- acJ'
rlll = ic22 - bal.
l
lb - acl
Now without loss of generality, assume that lb2 - acl is the intermediate among numbers la2 - bel, lb2 - acl, 2 - bal. Thus, one between
Ir 4 I , Ir 11 I is less than 1 and the other is greater than 1, a contradiction
lc
unless lrl = 1. Then,
Ia2 - beI = Ib2 - ac I = Ic2 - baI.
Likewise, the other two common roots s and t have absolute value equal
to 1, i.e. Jsl= ltl = 1. Since r, s, t are real, at least two of them are
equal, say r = s and then the three equations have a common root.
26. (Mathematics and Youth) Let n be a positive integer and let p >n+
be a prime. Prove that the equation
x2
x
1
xP
1 + -+ -+ · · · + -= o.
n + 1 2n + 1
pn + 1
has no integer solution.
Solution. Multiply both sides by (1 + n) (1 + 2n) • ... • (1 + pn). Then,
we get the polynomial P (x) = apxP + ••• + a0 , where
a· _ ( 1 + n) ( 1 + 2n) • ....
1 + in
i -
( 1 + pn)
·
Since gcd (n,p) = 1, we find that there exists a unique i E {1, ... ,P - 1}
such that 1 + in is divisible by p. Furthermore, since p > n+ 1, we see
that i # 1. However, for such i we have
1 + in < 1 + (p - 1) n < 1 + (p - 1) 2 < p2.
Thus 1 + in is not divisi ble by p 2 . This says that all the coefficients
ao,... , ap except the coefficient of xi are divisible by p but not by p 2.
Now, assume there is an integer s such that P (s) = O. Then,
apsP
+ · · · + a 0 = O.
Solutions to introduetory problems
177
But then the term aisi must be divisible by p. Since p fai, we must have
P I s. Then, the all the terms aksk are divisible by p 2 (note that since
i =I-I, a1 is divisible by p), but a0 is not. Thus the left hand side is not
2
a multiple of P and we have a contradiction. Thus there is no integer
solution.
27. (Baltic Way 2016) Find all quadruples (a, b, e, d) of real numbers satisfying the system of equations
a3 + e3
a 2 b + e2 d
b3 + d3
ab2 + ed2
2
O
I
-6
(Hint: define the polynomial P (x) = (ax + b)3 +(ex+ d)3).
Solution.
P(x)
Let
(ax + b)3 +(ex+ d)3
3
(a + e3)x 3 + 3(a2 b + e2 d)x 2 + 3(ab2 + ed2 )x + b3 + d3.
Then, P(x) = 2x 3 - I8x + I. Since P(O) > O, P(l) < O and P(3) > O,
the polynomial has three distinct real roots. But
P(x) = [(a+ e)x + b + d][(ax + b)2 - (ax + b)(ex + d) +(ex+ d)2].
The second factor is positive unless ax + b = ex+ d = O. In this case the
polynomial has at most two real roots. Otherwise, the polynomial has
one real root which is the root of the first factor. Thus the system has
no real solutions.
28. (A. Golovanov - Russian Mathematical Olympiad 2012) Given a polynomial P(x) and the numbers a1, a2, a3, b1, b2,b3 such that a1a2a3 =f.O.
Suppose that for every real number x we have
Prove that the polynomial P(x) has at least one real root .
Solution. Let P(x) be a constant _polynomial. Then, P(x) = O and
the statement holds true . Now assume the contrary . Then, the sign of
P(x) doesn't change . Assume P(x) > O for all real numbers x (otherwise
multiply P(x) by -1). Then, there is a real number s such that for all
real numbers x we have
P(x)>P(s)=T>O.
Solutions to introductory problems
178
There is a real number r such that a3r + b3 = s. Writing th e equality
for x = r, we have
p (a 1 r + b1 ) + P (a2r + b2) > 2T > T = P (a3r + b3)
a contradiction.
Solution 2. If one of a 1 , a 2 are not equal to a3, say a1 i- a3, then there
exists a real number r such that
Setting x = r into the original equation, we find that
and we are done. Otherwise, a 1 = a2 = a3 = a. Assume that the leading
coefficient of P(x) is ad. Comparing the coefficient of xd in both sides
(deg P( x) = d), we have ad(ad + ad) = adad, contradiction!
29. (Czech-Slovakia Mathematical Olympiad 1995) Find all polynomials f
with real coefficients such that for every real number x the following
inequality holds:
xf(x)f(l
- x) + x 3 + 100 > O.
Solution. Let n = <legf (clearly f =I=-O). Write f (x) = axn + g(x ),
where degg < n - l or g = O. The leading term of xf(x)f(l
- x) is
2 2
1
(-l)na x n+ • Since the left-hand side cannot be a polynomial of odd
degree (otherwise it would have negative values either for large positive
or large magnitude negative x), then 2n + 1 = 3, i.e. n = l, and
(-l)a 2 = -1, so a= ±1. Hence, we get f(x) = x +bor f(x) = -x + b,
where b E IR. If f (x) = x + b, the left hand-si de becomes the polynomial
2
2
x +(b +b)x+l00, which is non-negative if and only if (b2 +b)2-400 < O,
2
i.e. lb +bl < 20, which gives b E [-5,4]. If f(x) = -x+b, the left handside becomes the polynomial x 2 + (b2 - b)x + 100, which is non-negative
2
if and only if (b -b )2-400 < O, i.e. lb2 - bi < 20, which gives b E [-4, 5].
In conclusion, the polynomials are f(x) = x + b, where b E [-5, 4], or
f (x) = -x + b, where b E [-4, 5].
30. (I. Bogdanov - Russian Mathematical Olympiad 2010) Find all polynomials P( x) of odd degree with real coefficients such that for all real
numbers x we have P(P (x)) < (P(x)) 3 and the coefficient of x 2 in P(x)
is zero.
Solutions to introductory problems
179
Solution. Let P (x) = a0 + a 1x + ... + adxd where ad =I-O. Then,
<leg(P (P (x))) = d2 and <leg((P(x)) 3 ) = 3d. Now, if d2 =/-3d the
polynomial (P(x)) 3 - P (P (x)) will be a polynomial of odd degree,
2
max {d , 3d}. But a polynomial of odd degree always has negative values for either large positive x or for large magnitude negative x. Thus,
this polynomial cannot assume non-negative values everywhere. Now,
2
assume d = 3d . Then, d = 3 and the leading coefficients of x 9 in both
polynomials P (P (x)) and (P(x)) 3 must be the same, whence a! = ai.
So, a3 = l. Then P (x) = x 3 + ax + b for some real numbers a, b. Now,
P (P (x)) - P(x) 3 = ax 3 + a2 x +ab+ b ~ O.
By the same argument we must have a = O. Then , we get b < O and
P(x) = x 3 + b, where b < O.
31. Let P(x) = xd + ad_1xd-I + ... + a0 have all of its roots on the interval
(-1, l]. Prove that if IP(x)I > 1 for all x E [O,1], then IP(x)I < 1 for all
xE(-1,0].
Solution.
Set
P(x) = (x - x1) · . .. · (x - xd),
where x 1 , ... ,xd E [-1 , l] . Assume there is a real number r E (O,1) such
that IP(-r)I > 1. Then ,
IP(r)P(-r)I
= l(r - x1) · . .. · (r - xd)I · l(r + x1) · ... · (r + xd)I
= l(r 2 - Xi)· .. . · (r 2 - x~)I.
Since r 2 E (O, 1) and
x;E [O,1], then lr 2 - xrl < 1.
Thus , IP(r)P(-r)I < 1. Since IP(-r)I > 1, we deduce that IP(r)I < 1,
a contradiction.
It remains the case r = O. If r = O, then IP(O)I = lx1 · • • • · xdl < 1.
32. (Polish Mathematical Olympiad 2013) Let b, c be integers and let
f (x) = x 2 + bx + c
and let k 1, k2, k3 be integers such that n I /(k1), .n I /(k2), n I f(k3).
Prove that n I (k1 - k2)(k2 - k3)(k3 - k1).
Solution.
Set
A
B
C
kf (k2 - k3) + k1(k3 - k1) + k~(k1 - k2)
bk1 (k2 - k3) + bk2(k3 - ki)+ bk3(k1 - k2)
c(k 1 - k2) + c(k2 - k3) + c(k3 - k1)
Solutions to introductory problems
180
lt is elear that B = C = O and
A+ B + C = (k 2 - k3)f(k1) + (k3 - k1)f(k2) + (k1 - k2)f(k3)
Thus, n divides A= -(k
1-
= A.
k2)(k2 - k3)(k3 - k1)-
Solution 2. Let M = (k 1 - k 2)(k 2 - k3)(k3 - k1) and let P be a prime
number dividing M .
Assume vp(k 1 - k 2) = x, vp(k 2 - k 3) = y, vp(k3 - k1) = z and vp(n) = t.
It suffices to prove
X+ y + Z> t .
If min(x, y, z) > t, we are done. So, assume max(x, y, z) < t . Note that
n I f(k1) - f(k2) = (k1 - k2)(k1 + k2 + b).
Thus,
Vp((k1- k2)(k1+ k2 + b)) =X+ Vp(k1+ k2 + b) > t.
Hence, pt-x
I (k1 + k2 + b). Likewise, pt-y I (k3 + k2 + b), so
pmin(t-x ,t-y) I (k _ ki).
3
Therefore ,
vp(k3 - k1) = z > min(t - x, t - y) = t - max(x, y),
which implies that x + y + z> z+ max(x, y) > t.
33. Let f (x ) = a 2016 x 2 + bx + a 2016 c - 1, where a, b, c are integers . Suppose
that the equation f(x) = -2 has two positive integer roots. Prove that
(!(1))2 + (f(-1))
2
2
"t
.
1s compos1 e.
Solution. The equation a 2016 x 2 + bx + a 2016 c + 1 = O has two positive
1
integer roots rand s. Then, rs = c - 2016 , which implies that a 2016 = 1.
a
Hence, we get the following equation
x 2 + bx + c + 1 = O.
Then , f (x) = x 2 + bx + c + 1 = (x - r) (x - s) .
We have f(l) = (1 - r) (1 - s) and f( - 1) = (1 + r) (1 + s). Then,
(!(1)) 2 + (f(-1))
2
(1 - r)2(1 - s) 2 + (1 + r) 2 (1 + s) 2
2
2
(1 + r )(1 + s ),
2
2
which is a composite number, since 1 + r 2 and 1 + s 2 > 1 are integers
greater than 1.
Solutions to introductory problems
181
34. Find the number of polynomials P(x) = ax2 + bx + c with integer coefficients such that P(l) < P(2) < P(3) and
(P(1)) 2 + (P(2)) 2 + (P(3)) 2 = 22.
Solution. It is elear that P(l), P(2), P(3) are integers. Moreover, since
the sum of their squares are 22, we find that
(P(1))
2
+ (P(2)) 2 + (P(3)) 2 = 4 + 9 + 9 = 22.
From the inequality P(l) < P(2) < P(3), we find
P(l) = -3, P(2) E {-2, 2} and P(3) = 3.
Then,
(P(l),P(2) , P(3)) E {(-3,-2,3),(-3,2
, 3)}.
Then, we have to solve the system of equations
a+b+c
4a + 2b + c
9a+3b+c
=
=
P(l)
P(2)
P(3).
By setting the desired values of P(l), P(2), P(3), we get two distinct
polynomials which are solutions to the problem:
P (x) = 2x2 - 5x or P(x) = -2x 2 + llx - 12.
35. (Edward Barbeau) Let f(x) be a quadratic polynomial. Prove that there
are quadratic polynomials h(x) and g(x) such that
+ 1) = g(h(x)).
f(x)f(x
Solution. Let f(x)
f(x)f(x
+ 1)
= ax 2 + bx + c. Then,
2
(ax 2 + bx + c)(a(x + 1) + b(x + 1) + c)
a 2(x 2 + x) 2 + b2(x2 + x) + ab(x(x + 1))(2x + 1)
2
+ac(x 2 + (x + 1)2 ) + 2bcx +be+ c •
We write the above expression as
a2(x 2 + x) 2
+
+
b2
(2x + 1)2 + ab (x(x + 1)) (2x + 1)
4
2c ( ax
2
+ (a + b)x + ~) + ac+ c2- ~ .
182
Solutions to introductory problems
This is equal to
(ax 2 +ax+ ~(2x+
l)r
Hence, h(x) = ax 2 +(a+
+2c (ax 2 + (a+b)x+
b)x +
b
2
D
+ac+c2-
2
2
and g(x) = x + 2cx +ac+ c -
~b2
4.
Solution 2. Assume f(x) = a(x - r)(x - s). Then ,
f(x)f(x
+ 1) = a2 (x - r)(x + 1 - s)(x - s)(x + 1 - r)
= a2 (x 2 -(r + s-l)x
+ rs-r)(x
2
-(r + s-l)x
+ rs-s) .
Now, take h(x) = x 2 - (r + s - l)x + rs and g(x) = a2 (x - r)(x - s).
36. (I. Robanov - Russian Mathematical Olympiad 2003) Let
P(x) = x 2 + ax +band
Q(x) = x 2 +ex+ d
such that P(Q(x)) = Q(P(x)) has no real roots. Prove that b # d.
Solution. Write the equation P(Q(x)) = Q(P(x)) as
(x 2 + ax + b)2 + c(x 2 + ax + b) + d = (x 2 +ex+ d) 2 + a(x 2 +ex+ d) + b.
Hence,
2(c - a)x 3 + lx 2 + mx +n= O
for some l, m, n. Now, if c ;/ a, the above equation is of third degree
and it has a real root , contradiction. Hence, c = a. If b = d, then
P(x) = Q(x) for all x and the equation P(Q(x)) = Q(P(x)) is an identity
and it has infinitely many roots. Thus b ;/ d.
37. (A. Khrabrov - Saint Petersburg Mathematical Olympiad 2001) Let f(x)
and g(x) be quadratic polynomials with integer coeffi.cients such that
f(x) > O and g(x) > O. If
:i:?
> J2
:i:i> J2
for all real numbers x, prove that
for all real numbers x.
Solution. Let f(x) = a1x 2 + b1x + c1 and g (x) = a2x 2 + b2x + c2.
Define the function h(x) = f(x) - ../2,g(x). We must have
h (x) = (a1 - a2h)
x 2 + (b1 - b2h)
x + (c1 - c2v2) > O,
so the discriminant of the polynomial h(x) must be non-positive, i.e.
183
Solutions to introductory problems
where A, B are integers .
Since v'2 is irrational , it follows that A + B v'2 = O for integer A , B
implies A= B = O. Now,
A= bi+ 2b~ - 4a1c1 - 8a2c2 = (bi - 4a 1 c 1 ) + 2 (b~ - 4a2 c2) < O
because f (x) > O and g (x) > O imply that
Hence, A+ Bv'2-/- O, so A+ Bv'2 < O. This yields h (x) > O, thus
f (x) > v'2g(x) and then
:i:i
> v'2.
38. (A. Kanel-Belov - Moscow Mathematical Olympiad 2010) The sum of
any two of the three trinomials x 2 + ax + b, x 2 +ex+ d, x 2 + ex+ f does
not have real roots . Can the sum of all these trinomials have real roots?
Solution. The answer is no. Let
Since the sum of any two of these polynomials has no real roots, then the
sign of the sum of these polynomials doesn't change. Since the coefficient
of x 2 is 2 > O, then we find that
f (x) + g(x) > O,
g(x)
+ h(x) > O, f (x) + h(x) > O.
Then , also f (x) + g (x) + h (x) > O.
39. (P . Kozhlov - Russian Mathematical Olympiad 2010) Let the polynomial
(x 2 + 20ax + 10b)(x 2 + 20bx + l0a) have no real roots , where a and b are
distinct real numbers. Prove that 20( b - a) ~ Z.
Assume without loss of generality that b > a. We prove
O < 20(b - a) < l. If this is not the case, we have 20(b - a) > l. Since
the discriminants of the polynomials x 2 + 20ax + l0b and x 2 + 20bx + l0a
1
1
. Hence, 10b2 - b +
must be negative, we have 10b2 < a < b < O.
20
20
1
Since the discriminant of 10b2 - b +
is negative and 10 > O, then
Solution.
20
10b2 - b + 2_ > O, contradiction .
20
Solutions to introductory problems
184
40. (Saint Petersburg Mathematical Olympiad 2005) In the figure below , we
have four points which are part of the graph of the polynomials
f (x) = x 3 + bx 2 + ex + a,
g (x) = x
3
+ ax 2 + bx + c.
N
p
•M
Q
Does there exist values of a, b, c such that the points M, P, Q belong to
the graph of f (x) and points M, N belong to the graph of g( x)?
Solution. Assume that such numbers exists.
Then, since N = g(0) > f(0) = P > O, we find that c > a > O. The
polynomial f(x) has a positive real root (i.e . Q), hence all its coefficients
cannot be positive. Since a, c > O, we get b < O. Finally, since the graph
of the two polynomials f(x) and g(x) intersect at M, we find that ther e
is a real number m <Osuch that f(m)
2
bm
But bm 2 <O<
= g(m) . Thus,
+ cm + a = am 2 + bm + c.
am 2 and cm< O< bm and a< c, contradiction .
41. Find all integers m such that the equation
x
3
+ (m + 1) x 2 - (2m - 1) x - ( 2m 2 + m + 4) = O
has an integer root .
Solution.
Note that the left-hand side of the equation can be written
as x ( x - 2m + 1) + m (x 2 - 2m + 1)+ x 2 - 2m + 1 = 5.
Then, (x + m + 1) (x2 - 2m + 1) = 5.
Now, if the equation has an integer root, we have
2
(x + m + I, x 2 - 2m + 1) E {(1, 5), (-1, -5), (5, 1), (-5, -1)}
We have four cases.
Solutions to introduetory problems
185
(i) x + m + l = 1 and x 2 - 2m + 1 = 5.
2
Then, x - 2m + 1 + 2 (x + m + 1) = 7.
2
Hence, x + 2x - 4 = O, which has no integer roots.
(ii) x + m + l = -1 and x 2 - 2m + 1 = -5. Then, x 2 + 2x + 10 = O,
which has no real roots.
(iii) x + m + l = 5 and x 2 - 2m + 1 = 1. Then, x 2 + 2x - 8 = O, which
gives x = 2, -4 and m = 2, 8.respectively .
(iv) x + m + l = -5 and x 2 - 2m + 1 = -1. Then, x 2 + 2x + 14 = O
which has no real roots.
42. (V. Brayman) Do there exist integers a,b, e, d with a=/ Osuch that the
equations have as many distinct positive integer roots as their degrees?
ax
3
+ bx 2 + ex + d = O,
bx 2 + ex + d = O,
ex + d = O.
Solution. If we replace (a, b, c, d) with (a, bt, ct 2 , dt 3 ), the roots of each
equation are multiplied by t. Now we search for a', b', c', d' so that the
equations have rational roots, then we multiply by the least common
multiple of the denominators. Assume a' = 1. If the equation
x 3 + b'x 2 + c' x + d' = O
has positive rational roots, these roots must be integers. Then,
b' = -u - v - w,
c' = uv + vw + uw,
d' = -uvw
are integers. Then, the discriminant of the polynomial b'x 2 + c' x + d'
must be a square, i.e. D' = c'2 - 4b'd' must be a square. Take u = l
and v = 2. Then, D' = w 2 - 12w + 4 = (w - 6) 2 - 32 must be a square.
Set w · 12. Then, b' = -15, c' = 38, d' = -24. Hence, we have the
following equations
x 3 - 15x 2 + 38x - 24 = O,
-15x 2 + 38x - 24 = O,
38x - 24 = O
4 6 12
.
with roots 1, 2, 12, -, -, - . Now, 1ft = 19·5·3, then the value of a, b, c, d
3 5 19
are
(a, b, c, d) _(1, -15 2 · 19, 38 · 152 · 192 , -24 · 153 · 193 ) .
43. (Saint Petersburg Mathematical Olympiad 2012) If the numbers a, b, c
are distinct, prove that the system of equations
x 3 - ax 2 + b3
x 3 - bx 2 + c3
{ 3
3
x - cx 2 + a
O
O
O
186
Solutions to introductory problems
has no real roots.
Solution. It is obvious that x =/-O. Without loss of generality, assume
that a > b. We have ax 2 - b3 = x 3 = bx 2 - c3 . Then ,
b3 - c3 = x 2 ( a - b).
Hence, b3 > c3 , which gives b > c. By a similar argument, we find that
c3 - a3 = (b - c)x 2 , hence b > c >a> b, contradiction.
44. (Saint Petersburg Mathematical Olympiad 2012) The real numbers a, b,c
are such that among the three equations x 3 -ax 2 +b = O, x 3 -bx 2 +c = O,
x 3 - cx 2 + a · O, any two of them have a common root. Prove that
a= b = c.
Solution. Assume that r is a common root between the first and the
second equation. Then,
(b-a)r
2
=c-b.
Analogously for the common roots of the second and third equation and
the first and third equation s, t (in this order) we can find that
(c - b) s 2 = a - c,
(a - c) t 2 = b - a.
Now, assume b > a. From the first equality, we find that c > b. Then,
from the second equality we find that a > c, so a > b, which gives
a = b = c. The same argument holds if a > b.
45. The polynomial x 3 +ax 2 +bx+c = Ohas three real roots. If a2 = 2(b+l),
prove that la - cl < 2.
Solution. Let r, s, t be roots of the polynomial x 3 + ax 2 + bx + c. It is
easy to find that
r 2 + s 2 + t 2 = a 2 - 2b = 2.
Hence,
4-(a-c)
2
2 - 2b + 2ac - c2
2 (1 - rs - st - tr + rst (r + s + t)) - (rst) 2
2 (1 - rs) (1 - rt) (1 - ts) + (rst) 2
Note that 2rs, 2rt, 2ts < r 2 + s 2 + t 2 = 2. Hence, rs, rt, ts < l and we
conclude that 4 - (a - c) 2 > O, so we are done.
Solutions to introductory problems
187
46. (Belarusan Mathematical Olympiad 2011) Let a, b, c be nonzero integers
such that the polynomial f (x) = ax 2 + bx + c has two real roots in
common with the polynomial g (x) = x3 + bx 2 + ax + c. Find a, b, c.
Solution.
Let
h(x)
-
g(x) -f
(x)
3
x + (b - a) x 2 + (a - b) x
x(x-r)(x-s).
Since ci=- O, O is not a root off and hence r, s are the common roots of
J(x) and g(x). Now, from Vieta's formulas for h(x) and f(x), we can
find that
b
C
b-a=-(r+s)=-,
rs = a - b = -.
a
a
b
C •
b
Hence, -- = - , 1.e. b = -c. Moreover, b - a= -, thus
a a
a
a2
b= -a-l
=a+l+--.
1
a-l
Since b is an integer, it follows that a E {O,2}. Hence, a = 2, b = 4,
c = -4. We obtain the polynomials
f (x) = 2x 2 + 4x - 4,
g (x) = x 3 + 4x 2 + 2x - 4.
These polynomials indeed have common roots at x = - l ± v'3so they
are a solution to the problem.
4 7. Let a i=-O and let the polynomial P (x) = ax 4 + bx 3 + cx 2 - 2bx + 4a have
two real roots x1, x2 such that x1 x2 = 1. Prove that 2b2 + ac = 5a 2 .
Solution.
We have
P(x)
Let ax 4 +bx 3 +cx 2 -2bx+4a
= (x-x1)(x-x2)(ax
2
+mx+n).
ax 4 + bx 3 + cx 2 - 2bx + 4a
ax 4 + (m - ap) x 3 + (a - mp+ n) x 2 + (m - pn) x + n,
where p = x 1 + x 2 . By comparing the coefficients, we find that
n = 4a, m - pn = -2b, a - mp+ n = c, m - ap = b.
Then , we get m - 4ap = -2b, 5a - mp=
which gives 5a 2 = map + ac = 2b2 + ac.
c. So, b = ap and m = 2b,
Solutions to introductory problems
188
48. (Moldova Mathematical Olympiad 2008) The polynomial
P (x) = x 4 - 4x 3 + 4x 2 + ax + b
has two positive roots x1, x2 such that x1 + x2 = 2x1x2.
Find the maximum of a + b.
Solution. Assume x 1x 2 = c > O. Then x1 + x2 = 2c. Moreover, since
x1 + x2 > 2Jx 1x 2, we get c > l. The polynomial P(x) can be written
as
P(x)
= x 4 - 4x 3 + 4x 2 + ax + b = (x 2 + a1x + a2)(x 2 - 2cx + c).
We find that -4 = a 1 - 2c, 4 = c + a2 - 2ca1, hence a1 = 2c - 4 and
a2 = 4c2 - 9c + 4. Then, a+ b = c(a1 - a2) = -4c 3 + llc 2 - 8c = f(c).
We have to find the maximal value of the function f(c) for c > 1. Set
c = ~~ + d (this is Cardano's old trick). Then,
3
g (d) = f
2
(11
+ d) + 11
+ d) - 8 (11 + d) .
(1211+ d) = -4 (11
12
12
12
Therefore,
g
(d) = -4d3
25 d - 253
+ 12
216 "
So, it is enough to find the maximal value of -4d 3 +
Define y = d
Inequality,
c:-
2
d ).
2y2 <
- (
Then, 2y 2 = 2a' (
2d2+25_d2+2
48
.
3
!!-
5 _d2)
48
3
25
25
d = 4d (
_
12
48
d2
r.
d2).
By the AM-GM
253
- 723 •
25
25
The equahty holds whenever 2d2 = - - d2 , that is d2 = - . Therefore ,
5
16
4
48
144
d = 12 · Hence, c =
= 3• Now, one can find that the maximal value
12
16
of f(x) is f
.
(i)=3
27
Note. We could also have found the maximal value of f(c) by using
calculus. We have
J' (c) = -2 (6c 2 - llc + 4) = -2(2c - 1)(3c - 4).
189
Solutions to introductory problems
Hence, c =
4
3 is the point of relative maximum of the function.
Since
lim f (x) = -oo, we obtain that the maxima! value of f(c) for c > 1
x-++oo
occurs at c = ~, thus max {a + b} = f ( ~) = - ~~ .
49. If the polynomials
4
P (x) = x + ax 3 + bx 2 +ex+
Q (x) = x 4 + cx 3 + bx 2 + ax + l
1,
have two common roots , solve the equations P (x) = O and Q (x) = O.
Solution. Note that P (x) -Q (x) = (a - c)x(x 2 - 1), thus the common
roots must be among -1 , O, 1. Since Q (O) = P (O) = 1, we obtain
that x = l, -1 are common roots. Hence P (-1) = P (1) = O and
Q (1) = Q (-1) = O. This implies that b = -2 and a+ c = O. Hence,
P (x) = x 4 + ax 3 - 2x 2 - ax + l = (x 2 - 1) (x 2 + ax - l)
and
Q (x) = x 4 - ax 3 - 2x 2 + ax + l = (x 2 - l)(x 2 - ax - 1).
Then, the roots of P(x) are 1, -1 ,
·
are 1, -1,
a± J4 + a 2
2
-a± J4 +a 2
2
and the roots of Q(x)
.
50. Find all polynomials Q(x) of degree at most n-3
of the polynomial
n (n - 1)
P (x) = xn + nxn-l + ---xn2
such that all the roots
2 + Q(x)
are real.
Solution. Let r 1 , ...
formulas, we have
L
, rn
Ti
be the roots of the polynomial.
= -n,
L
TiTJ
= n(n 2- 1) .
Now by, AM-GM Inequality, we have
I::r/
> r1r2+r2r3+···+rnr1
I::r/
> r1r3 +r2r4 + ·· · +rnr2
By Vieta 's
Solutions to introductory problems
190
Hence,
(n-1)
Note that
Lr?>
= n(n-1).
2Lrirj
Lr?= (L ri)
2
-
2
L rirj = n,
hence the equality case occurs. Thus r1, ... , rn are all equal.
Hence, r1 = ... = rn = -1. Hence
P(x) = (x + 1)" and Q(x) =
I:
(~)xk.
k=O
51. A polynomial with integer coefficients is said to be economical if its
leading coefficient is 1 and the set of all its other coefficients, including
the constant term, coincides with the set of its roots taken with their
multiplicities, i.e. if exactly m of the coefficients are equal to a, then
a is root of the polynomial with multiplicity m . Find all economical
polynomials of degree n in the cases:
(a) n= 2;
(b) n= 3;
(c) n= 4.
Solution. lf P(x) is economical, then so is xP(x) since xP(x) has one
more zero coefficient and one more root at zero . Conversely, if xP(x)
is economical, then so is P(x). Also note that P(x) = x is the only
linear economical polynomial. Thus it suffices to look for economical
polynomials of degree at most 4 with nonzero roots and coefficients.
(The more generał problem of all degrees is solved in Advanced problem
44.)
If r 1 , .. . , rn are the nonzero roots of P(x), then the constant coefficient
is ( -1
r2 ... r n and this must be one of the roots. It follows that all
but at most one of the roots of P(x) is ±1. Thus all the coefficients of
P(x) except for the constant coefficient are ±1 and the product of these
coefficients is (-1
rr1
r.
If n= 2, then we need a quadratic polynomial P(x) = x 2 + x + b which
has roots 1 and b. Plugging in we get 2 + b = b2 + 2b = O and hence
b = -2. Thus we get the solution P( x) = x 2 + x - 2.
If n= 3, then we need a cubic polynomial P(x) = x 3 ± x 2 =f x + c which
has roots -1, 1, and c. In either case, plugging in x = 1 gives 1 + c = O
and hence c = -1. Thus the three roots must be a double root at -1
Solutions to introductory problems
191
and a single root at 1 and hence P( x) = (x + 1)2(x - 1) = x 3 + x 2 - x - 1,
and this is an economical polynomial.
If n= 4, then we have four cases
+ x 3 + x 2 + x + d,
P( x) = x 4 - x 3 - x 2 + x + d,
P( x) = x 4 - x 3 + x 2 - x + d, or
P(x) = x 4 + x 3 - x 2 - x + d.
P( x) = x
4
In all four cases 1 must be a root. In the last three cases, this gives
d = O, which is not allowed. Thus we must have the first case. This
gives d = -4 and we must have P( x) = x 4 + x 3 + x 2 + x - 4. But this
polynomial does not have -4 as a root or a double root at x = 1. Thus
we get no examples.
Adding back in the polynomials with zero as a root, we see that we have
two economical quadratics, x 2 and x 2 + x - 2, three economical cubics
x 3 , x 3 + x 2 - 2x, and x 3 + x 2 - x - 1, and three economical quartics x 4 ,
x 4 + x 3 - 2x 2 , and x 4 + x 3 - x 2 - x.
52. (Ukrainian Mathematical Olympiad 2016) A polynomial
p (x) = x2016+ 2016x201s + a2014X2014
+ a2013x2013+ ... + 1
can be expressed as P (x) = (x - x1) · ... · (x - X2016), where among
the numbers x 1, ... , x2016 at least 2015 are negative (not necessarily
distim ;t). Find all the coefficients of P( x).
Solution. Without loss of generality, we can assume that X 1, ... , x2015 <
O. Since x 1, .. . , x2016 = 1, we find that x2016 < O. By Vieta's formula
we have
X1 + · · · + X2Q16= -2016.
Hence , lx1I + lx2I + · · · + lx2016I= 2016 and lx1I · lx2I · ... · lx2016I= 1.
Then by AM-GM Inequality, We have
Hence, the equality case occurs and lx1l = lx2l = · · · = lx2016I = 1 and
then x1 = · · · = x2016 = -1. Hence P (x) = (x + 1)2016.
53. (Belarusan Mathemati cal Olympiad 2009) Let P(x) and Q(x) be nonconstant polynomials with integ er coefficients such that the polynomial
Solutions to introduetory problems
192
R(x) = P(x)Q(x) - 2009 has at least 25 distinct integer roots. Prove
that degP(x) > 2 and degQ(x) > 2.
Solution.
Then,
Let r 1 , ...
, r 25 be the
integer roots of the polynomial R(x) .
P (ri) Q (ri) = 2009.
Note that 2009 = 72 • 41. This number has exactly 12 distinct integer
divisors . Then, there are at least three numbers , say r1, r2, r3 such that
P (r1) = P (r 2) = P (r 3). Hence the degree of the polynomial P(x) is at
least 3. The same argument holds for the polynomial Q( x).
54. Let P(x) be a polynomial with real coefficients and assume that
(P(x)) 3 = x 9 + asx 8 + a1x7 + · · · + 15x + 1.
Find the sum of the squares of the roots of the polynomial P(x), knowing
that the sum of the coefficients of the polynomial (P(x)) 3 is 216.
It is easy to see that P(x) is a cubic polynomial , so write
P (x) = ax + bx2 + ex + d. Then the coefficients of x 9 and x 0 are a3
and d3 and so a3 = d3 = 1, i.e., a = d = 1. Thus
Solution.
3
P (x) = x 3 + bx2 + ex + 1.
Hence, the coefficient of x in (P(x)) 3 is 3c, so e = 5. Hence,
P (x) = x 3 + bx2 + 5x + 1
Note that sum of the coefficients of (P(x)) 3 is (P(1)) 3 = 216, so P (1) = 6
and b = -1. Hence, P (x) = x 3 - x 2 + 5x + 1. Let r, s, t be roots of the
above polynomial. By Vieta's formulas we find that
r + s + t = -1,
r s + st + tr = 5.
Then, the sum of the squares of the roots of the polynomial P( x) is
(r + s + t )2 - 2 (r s + st + tr) = -9.
55. (N. Aghakhanov - Russian Mathematical Olympiad 2004) Let
P (x) = xd + · · · + ao
be a polynomial with integer coefficients having d different integer roots .
Prove that if any two of the roots are relatively prime, then
Solutions to introductory problems
193
Solution. Assume there is a prime number p such that p I a 0 and p I a1.
Then, ao = pa, a1 = pb for some integers a, b. Let r 1, . .. , r d be the roots
of the polynomial. By Vieta's formulas we have
and
r2 · • • • · r d
+ · · · + r1 · ... · r d-1
Since p I r1 · ... · r d, then at least one of r 1, . .. , r d is divisi ble by p, say r1.
Since p divides the above expression, then r 2 • . .. • r d must be divisi ble
by p and so at least one of r 2 , . .. , rd is divisible by p. Then, there are
at least two roots which are divisible by p and hence are not coprime, a
contradiction. Thus gcd(ao, a1) = 1.
56. Let x1, . . . , X43be the roots of the polynomial P (x) = 18x48+ 3x + 2006.
48
· ·.
Find the value of the sum """'x
L...Jl ix
+
i=l
i
Solution. It is easy to see that -1 is not a root of P and hence Xi -=I=- l.
. we get Xi=--. Yi
H ence t he Yi are t he
Let Yi = -- Xi , then so1vmg
1 + Xi
1 - Yi
solutions to P
(-y-)=
l-y
O. Thus the Yi are the roots of polynomial
S(y) = (l-y)48p
(1 y y).
We compute that
18y48 + 3y(l - y) 47 + 2006(1 - y) 48
2021y48 - 96147y47 + · · · .
S (y)
Hence by Vieta's formulas, we get
48
48
L =L
Yi
i=l
i=l
96147
1 + Xi = 2021 .
Xi
57. (Canadian Mathematical Olympiad 2010) Let P(x) and Q(x) be połyI
Sh
h "f P(an) .
nomials with integer coefficients. Let an= n . +n.
ow t at 1 Q(an) 1s
Solutions to introductory problems
194
P(n) .
. t
c
. t
h
an integer for every n, then Q(n) 1s an m eger ior every m eger n suc
that Q(n) =IO.
Solution. Dividing P(x) by Q(x), there exist two polynomials A(x)
and R(x) with rational coefficients such that P(x) = Q(x)A(x) + R(x)
and degR(x) < degQ(x) i.e.
P(x)
R(x)
Q(x) = A(x) + Q(x),
degR(x)
( )
< degQ x.
Let b be the smallest positive integer such that B(x) = bA(x) is a polynomial with integer coefficients. Then,
bP(x) = B( ) bR(x)
Q(x)
x + Q(x) '
degR(x) < degQ(x).
Suppose that R(x) is not the zero polynomial.
bR(an)
Since degR(x) < degQ(x), if n is large enough, we get O< Q(an) < l.
bR(an)
But we also have Q( an)
bP(an)
( )
. .
= Q(an) - B an E .Z, contrad1ct10n. So,
bP(x)
.
R(x) = O and Q(x) = B(x). Now, let n be an mteger. Then there
are infinitely many integers k such that n = ak (mod b) (just take any
k > b with k
n (mod b)). Then, B(n)
B(ak) (mod b), which gives
=
B(n) = O (mod b) and therefore
=
~i:!
is an integer.
58. (Canadian Mathematical Olympiad 2016) Find all polynomials P(x)
with integer coefficients such that P( P( n) + n) is a prime number for
infinitely many integers n.
Solution. We prove that P( x) = p, where p is a prime number or
P(x) = -2x + b, where bis an odd integer. Since P(P(n) + n) = O
(mod P(n)), then P(P(n) + n) is divisible by P(n). As P(P(n) + n) is
prime for infinitely many integers n, then P(n) = ±1 or P(P(n) + n) =
±P( n) is a prime. Since the first case gives the constant polynomial
P(x) = ±1, which is not a solution, we must have P(P(n) +n)= ±P(n)
for infinitely many integers n. So, if we consider the polynomial Q(x) =
P(P(x) + x) ± P(x), we have that Q(x) has infinitely many roots, and
this gives P(P(x) + x) = ±P(x). Let k = degP(x). Comparing the
degrees, we get k 2 = k, i.e. k = O or k = l. If k = O, then P(x) = P,
where pis prime. If k = l, then P(x) = ax + b, where a =IO. We have
two cases.
Solutions to introductory problems
195
+ x) = P(x), i.e. P( (ax + b) + x) = ax + b, which gives
a(a + 1)x + ab + b = ax + b. Comparing the coefficients, we get
a( a + 1) = a and ab+ b = b, impossible.
{ii) P(P(x) + x) = -P(x), i.e. P((ax + b) + x) = -ax - b, which gives
a(a + l)x +ab+ b = -ax - b. Comparing the coefficients, we get
(i) P(P(x)
a(a+ 1) = -a and ab+b = -b. We get a= -2 and b arbitrary. So,
P(x) = -2x + b. Since P(n) is prime for infinitely many integers
n, then b must be odd.
In conclusion, we get P( x) = p, where p is a prime number or
P(x) = -2x + b,
where bis an odd integer.
59. (Kiirschak Competition 2004) Find the smallest positive integer different
from 2004 with the property that there exists a polynomial f (x) with
integer coefficients such that the equation f (x) = 2004 has at least one
integer solution and the equation J(x) = n has at least 2004 distinct
integer solutions .
Solution. Let 9(x) be a polynomial with the _desired property for the
number n in the statement of the problem. Thus, there is an integer
a such that 9( a) = 2004. We define the polynomial 91(x) = 9( x + a).
It is obvious that 91(0) = 9(a) = 2004, that is, the constant term of
9 1(x) is 2004 and the equation g1 (x) = n has distinct integer roots
a1, a2, ... , a2004, none of which is O since 91(0) = 2004. It follows that
91(x) - n= (x - a1)(x - a2) · . .. · (x - a2004)92(x),
where the polynomial 92(x) has integer coefficients and 92(0) = c E Z.
Setting x = O in this equality, we get
Since c =/O, we get
From this we can see that O < n < 2004 is not possible, so the above
inequality gives n > (1002!) 2 + 2004.
We provide an example for n= (1002!) 2 + 2004. Let
(x+ l) (x-2) (x+ 2) · ... · (x-1OO2) (x+ 1002)+(1002!) 2+2004.
We have 9(0) = 2004 and 9(±k) = (1002!) 2 + 2004 if 1 < k < 1002.
In conclusion, n= (1002!) 2 + 2004.
g(x) = -(x-1)
Solutions to advanced problems
11
197
Solutions to advanced problems
1. The polynomial
a k Xk + a k+lX k+l
+ · · · + an-k-1X ·n-k-1 + an-kX n-k
is said ton be palindromie if O < ak ,i a· = ani _ • and a·i_i < a ·+i for every
k < i < 2 • Prove that the product of any two universal polynomials is
a palindromie polynomial.
Solution.
Let Pi,n-i(x)
= xi + xi+ 1 + . . . + xn-i where O < i < ; .
Thus every palindromie polynomial is a linear combination of Pi,n-i (x).
Indeed ,
L Ci~,n-i(x),
k'.5:.iS.n/2
where Ck= ak > O and whenever k + l <i<
; , then Ci= ai - ai-1 > O.
On the other hand, every polynomial ·of the form
L Ci~,n-i where
kS,iS,n/2
Ci > O and Ck > O is palindromie. So, we have to prove that the set of
polynomials of the form
qPi,n-i is closed under multiplication.
k5:,i5:,n/2
Notice that
L
L CiPi,n-i(x) L djPi,m-j(x)= L L Cidj~,n-i(x)Pi,m-j(X).
If n - i + j < m - j + i then
Pi,n-i(x) Pi,m-j(x)
= (xi + xi+l + · · · + xn-i)(xi + xi+l + •••+ xm-j)
n-i
= L xk(xi + xi+l + ... + xn+m-j)
k=i
n-i+j
=
L Pk,n+m-k(x).
k=i+j
Then the product of ~,n- i(x)Pi,m-j(x)
can be written as a sum of polym +n
. Thus the set is closed
nomials Pk,n+m-k(x) where O < k <
2
under multiplication, as desired.
2. (Mathematics and Youth Journal 2002) Let n be an even natura! number. Find the number of polynomials Pn(x) of degree n such that:
Solutions to advanced problems
198
(i) all the coefficients of Pn(x) belong to the set {-1, O, 1} and
Pn(O) # O;
(ii) there is a polynomial Q(x) whose coefficients belong to the set
{-1, O,1} and Pn(x) = (x 2 - l)Q(x).
Solution. Let M = {-1, O, 1} and let n= 2m for some positive integer
m and set Q(x) = boxn- 2 + · · · + bn-2· Then
2
Pn(x) = (x2 - l)Q(x) = boxn + b1xn-l + (b2 - bo)xn- + · · ·
+ (bn-2 - bn-4)x 2 - bn-3X - bn-2·
Now we must find all (bo, b1, .. . , bn-2) such that bi E M for all O <
i < n - 2 and such that bi+2 - bi E M for all O < i < n - 4. Note
that bn_2 = -P(O) # O. We know that for all 1 < i < n - 2 we have
bi+2 - bi E {O,±1}. If bi = 1, then bi+2 E {O,1}, if bi = -1, then
bi+2 E {O,-1} and if bi= O, then bi+2 E {O,±1}.
Let Yk be the number of k-tuples (b1, b3, ... , b2k-1) such that b2i-1 E M
for all 1 < i < k and bi+2 - bi E M for all i= 1, 3, ... , 2k - 3. Given one
of the (k -1)-tuples (b1,b3i .. . ,b2k-3) counted by Yk-1 , we can always
extend it to a k-tuple counted by Yk· If b2k-3 = ±1, there are two
possible extensions (add a O or a b2k-3) and if b2k-3 = O, there are three
extensions. Thus Yk - 2Yk-1 is the number of (k - 1)-tuples that end in
a O. But if a (k - 1)-tuple ends in a O, then we can remove the zero and
get a (k - 2)-tuple. Conversely, for any (k - 2)-tuple counted by Yk- 2
we can append a O to get a (k - 1)-tuple counted by Yk-l ending in a O.
Thus Yk = 2Yk-1 + Yk-2· Since we can check that Y1 = 3 and Y2 = 7, we
find that
(1 + J2)k+l + (1 _ y'2)k+l
Yk = --
- -----
2
--
By construction Ym-1 is the number of possibilities for the odd coefficients of Q(x). We could similarly let xk be the number of (k + 1)-tuples
(bo, b2, ... , b2k) such that b2i E M for all O < i < k, bo, b2k # O, and
bi+2 - bi E M for all i = O,2, ... , 2k - 2. Then Xm is the number of
possibilities for the even coefficients of Q(x) and we could derive a similar recursion for Xk. However, we do not need to do this. To derive Xk
from Yk, we notice that Yk+l counts (k + 1)-tuples that might begin or
end with zero. We already saw that Yk is the number of (k + 1)-tuples
that end in a zero and we want to exclude these . A symmetric argument
would show Yk is also the number of (k + 1)-tuples that begin with a
zero and need excluded. However, if we take Yk+l - 2yk then we have
199
Solutions to advanced problems
subtracted off (k + 1)-tuples that begin and end with a zero twice . Since
the same argument shows that the number of these is Yk-1, we see that
Thus, the total number of possibilities for Q(x) is 2Ym-1Ym which is
equal to
(1 + J2)2m-l
+ (1 _ v'2)2m-l + 2(-l)m - l
2
(1 + v12r- + (1 - v12r-1
n
2
+ (-1)2-1
.
1
3. (Vietnamese Mathematical Olympiad 2015) Let a be a positive root of
equation x 2 + x = 5 and let co, c 1 , . . . , Cn be non-negative integers such
that CO+ C1 a + · · · + enan = 2015.
(i) Prove that co+ c1 + · · · + Cn = 2 (mod 3).
(ii) Find the minimal value of co+ c1 + · · · + Cn·
Solution.
(i) Define P (x) = CnXn + · · · + co - 2015. Then P (a) = O. Hence
P(x) is divisible by the minimal polynomial of a. This minimal
polynomial is x 2 + x - 5 as you might expect (if this polyno mial
factored it would have two linear factors and hence two rational
roots and this is easy to exclude using the rational root theorem).
Thus , we have
P( x) = (x 2 + x - 5)Q(x)
for some polynomial Q(x) with integer coefficients. Set x = 1.
Then, P(l) = -3Q(l) is a multiple of 3. Thus
co+ c1 + · · · + Cn = 2015 = 2 (mod 3),
and we are done.
(ii) If co+ c 1 + · · · + Cn is minimal, then O ~ Ci ~ 4 for all i, otherwise
replace the (co,c 1, ...
, en) with
(CO,···, Ci- 1, Ci - 5, Ci+l + 1, Ci+2 + 1, Ci+3, Ci+4, . . . , Cn)
which has smaller sum . Set
Q(x) = an- 2Xn-2 + · · · + ao.
Solutions to advanced problems
200
By comparing the coefficients, we find that
-5ao
co - 2015
-5a1 + ao
-5a2 + a1 + ao
-5a3 + a2 + a1
Then, since O< Ci< 4, we find that co= O and ao = 403.
Then c1 = 3 and a1 = 80.
Continuing in this way, since Ci+l = -5ai+l + ai + ai-l, we have
Ci+l = ai + ai-1
and
(mod 5)
ai + ai-1 J
·
5
An easy calculation shows that n = 11 and
ai+l =
l
(co, ... , en)= (O,3, 3, 1, 1, 1, 3, 4, O,O,3, 1) and co+c1 +···+en
=20.
4. (Czech-Polish-Slovak Match 2005) Find all values of n> 3 such that the
polynomial P(x) = xn - 3xn-l + 2xn- 2 + 6 is reducible over Z[x].
Solution. The case n = 3 is trivial. For n = 4, assume
x 4 - 3x 3 t 2x 2 + 6 = (x 2 + ax + b)(x 2 + ex + d),
where a, b, c, d E Z. Comparing the coeffi.cients, we can find that
a + c = -3,
ac + b + d = 2,
bd = 6.
Then, a and c have different parity, so ac is even and this implies that
b + d is even. Since bd = 6, then b and d must be even , but then
bd = 6 must be divisible by 4, contradiction. Assume now n > 5 and
P(x) = Q(x)R(x), where Q(x), R(x) are non-constant polynomials with
integer coeffi.cients. Assume
Q (x) =
k
n
i=O
i=k
L aixi, R (x) = L bn-iXn-i,
l;J
where k <
< n - 2. Nowak= bn-k = ±1.
Then aobo = 6, aob1+ a1 bo = O and aobk + · · · + akbo = O.
We prove inductively that ao I a1, ... , ak. Note that for all Z< k, we have
O= ao (aobz+1+ · · · + az+1bo)= aibz+1 + aoa1bz+ · · · + aoboaz+1•
~
6
Solutions to advanced problems
201
Then 6az+1 = -(ao bz+1 +aoa 1 b1+ • • •+a 0 b1az) since ao I a1, ... ,al, then
2
a5 I 6a1+1 = aoboaz+1 ==}
ao I boaz+l·
As gcd (ao, bo) = 1, we find that a 0 I az+i and we are done.
Now ao I ak = ±1, then a 0 = ±1 and bo = ±6. Analogously, we can
prove that
bo I b1, · · · , bn-3
(if necessary, set bz = O for all l > n - k). Since bn-k = ± 1, then the last
relation is true if n - k > n - 3. Thus, it remains two cases.
(i) If k = 2, then the coefficient of xn- 2 is aobn-2 +a1 bn-3 + a2bn-4 = 2.
--...,._,
±1
We know that 6 I a1bn-3 + a2bn-4 from above, contradiction.
(ii) If k = 1, the polynomial P(x) must have an integer root which is
a divisor of 6. If n is even the polynomial hasn't any integer root .
Let n be odd. Then, x = -1 is a root of the polynomial. Therefore,
odd n are the desired case.
5. (China Training Camps) Let n > 3 and let p be an odd prime. Prove
that the polynomial
cannot be represented as the product of two nonconstant polynomials
with integer coefficients.
Solution.
First we prove the following lemma.
Lemma . Let P (x) = cdxd +···+co , where Ci EZ and there is a prime
p such that p ł cd and p I ck for all O < k ::; d - 1. If
cdxd+ ···+co=
(arxr + · · · + ao) (bsX8 +···+bo)
where d = r + s and ar, ... , ao, bs, ... , bo E Z, then p I ai for all O < i <
r - 1 and p I bj for all O < j ::; s - 1.
Proof If all ar, ... , ao, bs, ... , bo are divisible by p, then cd = arbs must
be divisible by p, contradiction. Now, let O < k < r be the first index
such that ak is not divisible by p and let O < l < s be the first index
such that bz is not divisible by p . If k + l < d = r + s, the coefficient of
xk+l is equal to
Ck+l= akbt +
L
L
aibj +
aibj.
i+j=k+l, O::;;i<k
i+j=k+l, o::;;i<l
Solutions to advanced problems
202
Since for i < k, we have p I ai and for j < l we have P I bj, we find that
p divides the two sums. As p I Ck+l, then p I akbt, contradiction! Thus
d = k + l = r + s implies that k =rand
l = s. Then we can write
arXr + · · · + ao
arXr + p · g1 (x)
= bsx8 +p·h1(x)
b8 x 8 +···+bo
for some polynomials 91 (x), h 1 (x) with integer coefficients, where
deg 9 1 ( x) < r - l and deg h 1 ( x) < s - l. Our proof is complete.
Back to our problem, we know that ar = b8 = ±l. Without loss of
generality, assume ar = b8 = l. Moreover, by the Lemma each of ao
and bo is a multiple of p and hence ao = bo = ±p. This implies that
h1 (O) = 91 (O) = ±1. Then
(xr + p · 91 (x)) (x 8 + p · h1 (x))
= xd + pxrh1 (x) + px 91 (x) + p 291 (x) h1 (x) •
8
Assume r < s. Then, taking the coefficient of xr mod p 2 we find that
h1 (O) must be divisible by p, contradiction! If r = s, taking again the
coefficient of xr mod p 2 we find that h1 (O)+ 91 (O) must be divisible by
p, contradiction!
6. (Mongolian Mathematical Olympiad 2010) Let P(x) be a monie irreducible polynomial with integer coefficients such that IP(0)I= 2010.
22010
Prove that the polynomial Q (x) = P(x
) is irreducible.
Solution.
First we prove the following lemma.
Lemma. Let f(x) be a monie polynomial with integer coefficients. Then
the following statements are equivalent:
(i) The polynomial f (x2 ) is reducible over Q[x].
(ii) The polynomial f(x) is reducible over Q[x] or there are polynomials
G(x), H(x) E Z[x] such that
±J (x) = (G (x)) 2 - x (H (x)) 2 .
Proof. Let ±J(x) = (G(x)) 2 - x (H(x)) 2 • Then
±f (x2 ) = (G (x2 ) )2- x2 (H (x2 ) ) 2
= (G (x2 ) - xH (x2 )) (G (x2 ) + xH (x2 )),
so the polynomial f (x2 ) is reducible. Now assume that J (x2 ) is reducible. Then there are polynomials 9(x ), h(x) with integer coefficients
such that
f (x2) = 9(x)h(x).
Solutions to advanced problems
203
W~ ca_nfurt~er assume that g is irreducible (otherwise replace g by one
of its irreduc1ble factors and move the other factors to h). Write
2
g(x) = G (x ) + xL (x 2 ),
h(x) = H (x 2 ) + xT (x 2 ).
Then
f (x 2 ) = g(x)h(x)
= G (x 2 ) H (x 2 ) + x 2 L (x 2 ) T (x 2 )
+ xG (x 2 ) T (x2 ) + xL (x 2 ) H (x 2 ) ,
hence
2
2
f (x ) - G(x )H(x 2 ) - x 2 L(x 2 )T(x 2 ) = xG(x 2 )T(x 2 ) + xL(x 2 )H(x 2 ).
Since the left-hand side is an even polynomial and the right-hand side is
an odd polynomial, we see that both sides must be zero. Thus
f(x)
= G(x)H(x) + xL(x)T(x)
and
G(x)T(x)
.
+ L(x)H(x) = O.
Now if LT = O, the first of these reduces to f (x) = G (x) H (x), which
leads to the reducibility of f(x). Thus we may assume LT =JO.Since g
is irreducible G(x) and L(x) are relatively prime . Then from the second
equality, we find that T(x) is divisible by L(x).
Writing T(x) = M(x)L(x), we get that H(x) = -M(x)G(x) and hence
f (x) = M (x) (G (x)2 - xL (x)2).
lf M is non-constant, then we get that f is reducible and if M is constant,
then _since f(x) is monie, we have M = ±1 and we are done.
Then we get the following corollary.
Corollary. Let f(x) be a monie irreducible polynomial with integer coefficients such that lf(0)I is not a square. Then the polynomial f (x 2 ) is
irreducible .
Proof. Note that by the above lemma, if f(x 2 ) is reducible then either
f(x) is reducible or ±J (x) = G (x) 2 - xH(x) 2 . Since f(x) is irreducible
then the latter must be true. If x = O, we find that ±J (O) = G (0) 2 ,
then lf (0)I must be a square.
Back to our problem, we can easily find that Q 1 (x) = P(x 2 ) is irreducible. Now we define a sequence of polynomials by the following
recursive relation Qn+I (x) = Qn(x 2). lt is obvious that IQn(0)I= 2010,
thus by our lemma and corollary we conclude that Qn (x) is irreducible.
204
Solutions to advanced problems
7. (Mircea Becheanu - Mathematical Reflections 0134) Let p be a prime
and let n be an integer greater than 4. Prove that if a is an integer that
is not divisible by p, then the polynomial
f (x) = axn - px 2 + px + p 2
is irreducible over Z [x].
Solution. Assume the contrary, hence there exist nonconstant polynomials
B (x) =
m
n-m
i=O
j=O
L bixi and C(x) = L CjXj
with integer coefficients such that f (x) = B (x) C(x), where clearly 1 :::;
m < n - l. Since boco= p 2 , assume without loss of generality bo, co> O
(since we may change the sign of B(x) and C(x) simultaneously without
altering the problem) . Then either bo = co = p or without loss of
generality bo = p 2 , c0 = l.
In the first case we have boc1+ b1co= p, i.e. b1 + c1 = 1, hence b1 and c1
cannot be multiples of p simultaneously . Now, boc2 + b1c1 + b2co= - p,
which gives p I b1c1. Without loss of generality, we can assume that b1
is divisible by p. For all integer j such that n > j > 3, by comparing
the coefficient of xJ we find that
bocJ+ b1cJ- 1 + · · · + cobJ= O
Hence, b1Cj-l + b2Cj-2 + · · · + c1bj-l is a multiple of p for all n >
j > 3. Taking j = 3, then p I b2c1, hence p I b2. By induction , if
p I b1, b2, · · · , bk, taking j = k + 2, we obtain that bk+1c1 is divisible by
p, hence bk+I is divisible by p . Now take j = n - l. Since bo is divisible
by p we find that p I bo, b1, .. . , bn- 2 (with bm+l = bm+2 = · · · = bn-2 = O
if m < n- l). Since a = bmCn-m is not divisi ble by p, then m = n-1 and
a
.
C (x) = c1x + p . Set -c1 = q, so bn-I = -- and for all n - l > J > 3,
q
we have
0 = bj-1C1 + bjCO= -qbj - 1 + pbj.
pn-3a
Then, by induction b2 = - qn- 2 . Now, b1c1 + b2co = -por
(b2 + 1)p
pn- 2a + pqn-2
b1 = --=
-----qn-1
q
Since gcd (p, q) = 1 (otherwise a would be divisible by p), then qn- I I a.
Note also that f ( ~)
= O, hence
!!.= pn-2. _ a_+
q
qn-1
q + 1 EZ,
Solutions to advanced problems
205
so q = 1, which implies that p is a root of the original polynomial.
We conclude that p 4 < lapnl = lp2 (p- 2)1 < p 3 , contradiction.
In the second case, since cois not a multiple of p, but
PI (bocj + b1cj-1 + · · · + cobi) for j = O,1, ... ,n -1,
after trivial induction we conclude that bjCOand then bj is a multiple of
p for all j = O, 1, ... , n - l and since m < n - 1, then p I a= bmCn-m,
contradiction.
8. (Erdos) If f (x) = (x - x 1 ) • .. . • (x - Xn), where Xi E [-1, 1], prove that
there doesn't exist a E (-1, O) and b E (O,1) such that lf(a)I > 1 and
li (b)I > 1.
Let n = 2 and Let -1 < x < y < z :S 1. Then
Solution.
(y - x) (z - y) < (y-x+z-y)
= (x-z)<
2
2
2
-2-
1.
The equality case occurs for x = -1, y = O,z = 1. This implies the
non-existence of such a and b. To see this note that applying it with x
and z the two roots x1 and x2 of f and y = a or b, shows that a and
b cannot be between x1 and x2. Thus we must have a < x1 < x2 < b.
Now applying the inequality, with x = a, z = b, and y = Xi shows
that (xi - a)(b- Xi) < 1. Hence multiplying these two inequalities gives
lf(a)f (b)I < 1.
For n > 2 assume -1 < x1 < x2 :S · · · :S Xn < 1.
If such a and b exist and x1 < a < b < Xn, then
lf(a)I = (a - x1) · ... ·(a - Xk)(xk+l - a) · ... · (xn - a)
= (a - x1)(xn -a)(a-x2) · ... · (a-xk)(xk+l -a) · . .. · (xn-1 -a)
< (a - x2) · ... ·(a - Xk)(xk+l - a)· ... · (xn-1 - a)
= l(a - x2) · ... ·(a - Xn- i)I
Analogously, lf (b)I < l(b - x2) · ... · (b - Xn- 1)!. Hence the same a and
b work for the numbers x2, .. . , Xn-1· Thus we can throw away some of
the Xi to get the cases a < x1 or b > Xn which are indeed similar and
then we restrict ourselves to the case
-1
< a < X1 < X2 < · · · Xk < b < Xk+l < · · · < Xn :S 1.
Hence,
lf (b)I
<
(b - x1) · ... · (b - Xk) (xk+l - b) · ... · (xn - b)
(b + 1l(1 - bt-k
(l _ b)n-2k(l _ b2)k.
Solutions to advanced problems
206
Now if 2k < n, then the above expression is less than 1. Thus , 2k > n
and
2 <
+
VI! (a)I + Vlf (b)I
V(x1 - a)· . . . · (xn - a)
V(b - x1) · .. . · (b- Xk) (xk+I - b) ·. • • · (xn - b)
By Minkowski's inequality we have
V(x1 - a)· ... · (xn - a)
+ V(b - x1) · ... · (b - Xk) (xk+I - b) · ... · (xn - b)
<
'fj(b - a)k (2xk+I - a - b) · ... · (2xn - a - b)
<
'fj(b + l)k(3 - b)n-k.
< 22k-n((b + 1) (3 - b))n-k < 2n since for
Hence 2n < (b+ l)k(3-b)n-k
b < 1 we have (b + 1) (3 - b) < 4, contradiction!
Solution
2. Assume such a and b exist. Define
n
g (x) =
L lx - Xi1i=l
Since g is convex, then
g (a) < lal g (-1) + (1 - lal) g(O) and g (b) < bg (1) + (1 - b) g(O).
We know that
n
g(O) =
L 1-xil < n.
i=l
Furthermore,
n
g (a)=
L la- xil> n VI! (a)I> n
i=l
and similarly, g(b) > n. Then we must have g (-1) > n and g (1) > n.
Note that
g ( -1)
n
n
i=l
i=l
+ g ( 1) = L I1 - Xi I + L I1 + Xi I = 2n
because Xi E [-1, 1].
Thus g (-1) = g (1) = n, which implies g(a) = g(b) = n and this implies
la- Xkl = lb- xkl = 1, so a - Xk = -l and b - Xk = 1, which yields
b - a = 2, contradiction. The statement of the problem ensures that for
all a E (-1, O) and b E (O, 1) we have min (lf (a)I, lf (b)I) < 1.
Solutions to advanced problems
207
9. (Oleksandr Rybak- Ukrainian Mathematical Olympiad 2008) Let n> 2.
Consider the polynomials Po(x), P 1 ( x), ... , Pn(x) such that
<legPi(x) = n - i and Pn(x) =JO.
For all 2 <i<
n there is a polynomial Qi(x) such that
Prove that if Po(x)R(x) + P1(x)S(x) = 1, then degR(x) > n - 2 and
degS(x) > n-1.
Solution.
find that
Since degPi(x) < deg~_
degPi-IQi(x)
2 (x) and~=
~-2
+Pi-IQ i (x) we
= n - i+ 2.
Then for all i= 2, ... , n, we have degQi(x) = 1.
Now we prove by induction that there are polynomials Ę(x), Si(x) such
that degĘ(x) = i - 2 and degSi(x) = i -1 and
Pi(x) = Po(x)Ę(x) + P1(x)Si(x).
The case i= 2 is trivia!, set R2(x) = 1 and S2(x) = Q2(x). Now, since
by the induction hypothesis we find that
Pk+I = Pk-1 + PkQk+1(x)
= Po(x)Rk-1(x) + P1(x)Sk-1(x) + Qk+1(x)(Po(x)Rk(x) + P1(x)Sk(x))
= Po(x)(Rk-1(x) + Rk(x)Qk+1(x)) + P1(x)(Sk-1(x) + Sk(x)Qk+1(x))
= Po(x)Rk+I (x) + P1(x)Sk+1(x),
where we have set
Our proof is complete.
Take i= n. Then, Pn= PoRn+P1Sn. Since Pn is nonzero, we can define
Rn,0 = Rn and Sn0 = Sn . Then P0Rn0 + P1 Sn0 = l and <legRn,0 = n - 2,
Pn
Pn
<legSno= n - l. If there are Rand S such that PoR + P1S = l, then
Solutions to advanced problems
208
Let R = R- fln 0 and S = S - Bno so that PoR + P1S = O.
Now PoRfln 0 + P 1Sfln 0 = O, while PoRno = l - P1Sno we find
(1 - P1Sn0 )R + P1SRno = O.
Then, R = P 1(Sn0 R- Sfln 0 ). It follows that R is divisible b~P1.
lf R i- O, then degR > degP 1 = n - l. Then degR = degR > n - l
and similarly S is divisible by Po, which implies that
degS > degPo = n .
This completes our proof . Now, assume that R = O.
Then deg R = deg Rno = n - 2 and a similar argument works for <legS.
10. (USATST TST 2014) Let P(x) and Q(x) be two polynomials with real
coefficients and degP(x) = d. Prove that there exist polynomials A(x) ,
B(x), and C(x) such that
(i) degA(x), degB(x) < ~;
(ii) at most one of them is zero;
... ) A(x) + Q(x)B(x)
( u1
P(x)
Solution.
= C( )
x .
We prove the following lemma.
Lemma. For all nonzero polynomials P(x) and Q(x) and any positive integer l < min( deg P( x), deg Q(x)), there exist polynomials
A(x), B(x), C(x) which are not all the zero polynomial such that
degA(x) < l, degB(x) < degP(x)-l-l
and degC(x) < degQ(x)-l-1
such that
A(x)
+ B(x)Q(x) = C(x)P(x)
.
Proof We argue by induction on degP(x) + degQ( x ). Without loss of
generality we may assume deg(P(x)) < deg(Q(x)).
Set Q(x) = D(x)P(x) + R(x) , where <legR(x) < deg P(x).
If degR(x) < l, set A= -R, B = l, C = D. Then we can check that
our conditions are satisfied.
Now, if <legR(x) > l, write R(x) = Q(x) - D(x)P(x) .
Then by the induction hypothesis, there are polynomials A', B', C' (not
all zero) such that A'(x) + B'(x)P(x) = C'(x)R(x) and degA'(x) < l ,
<legB'(x) < deg R(x)-l - land deg C'(x) < deg P(x) -l - l. Note that
A' and C' cannot both be zero , since this would imply B' = O as well.
Now A'(x) + B'(x)P(x) = C'( x )(Q(x) - D(x)P( x )) and then
A'(x)
+ (B'(x) + C'(x)D(x))
P(x)
= C'(x)Q(x).
Solutions to advanced problems
209
It remains to check the degree condition:
+ C'(x)D(x))
deg(B'(x)
max ( deg B'(x), deg C'(x) + deg D(x))
< max (deg Q (x) - l - 1, deg Q (x) - l - 1 + deg P (x) - l - 1)
degQ(x) - l - 1.
Now set A= -A', B = C', C = B' + C' D and we are clone.
From the lemma the problem is immediate. We take l = ld/2 J. Then
the degree conditions of the lemma imply (i). If any two of A , B, and C
are zero, then it is easy to see that the third is, hence this cannot occur,
proving (ii). Finally, it is easy to see that the conclusion of the lemma
gives (iii).
11. (Mircea Becheanu - Romanian TST 1981) Consider the polynomial
P( x) = xp-l + xP- 2 + •••+ x + l,
where p > 2 is a prime number. Prove that for any even positive integer
n, the polynomial
n-1
ITP( xPk) is divisi ble by x + 1.
-1 +
2
k=O
n-1
Solution.
Let Q(x)
= - 1 + ITP(xPk).
It 's enough to prove that
k=O
Q(±i) = O, where i = -1. We have
2
xP - 1
P( x) = xp-l + xP- 2 + · · · + x + l = --,
x-l
SO
n-1
IJ
k=O
k
n-1 (xPk)P- 1
n-1 (xPk+l)P- 1
xPn - l
P( xP ) =
xPk - l =
xPk - 1 = x - l ·
IT
IT
k=O
k=O
xPn - X
. If p > 2 is a prim e,
X- 1
1
then p i~ o~d. Since n is even, then pn = 1 (mod 4), and iPn = i, so
Th erefore, Q (x) = -1
xPn - 1
+ X-
=
Q(i) = ~ - i = O. Likewise, we get Q( -i) = O, so Q(x) is divisible by
x2
+ 1.
i-1
Note 1. In the solution, it is not necessa ry that pis prime. It is sufficient
that p be odd. We can prove the following generalization.
Let P (x) = ad_1xd-l + · · · + ao, where dis an odd positive integer. Let
A= ad-l - ad-3 + ad-5 - · · · and B = ad-2 - ad- 4 + ad-6 - · · ·.
Solutions to advanced problems
210
(i) If d = 1 (mod 4) and B = O, then for all n even, the polynomial
n-1
Q (x) =-An+
II P(xdk)
k=O
is divisible by x 2 + 1.
(ii) If d = 3 (mod 4) and A= O, then for all n even, the polynomial
n-1
R (x) = -Bn
+ II P(xdk)
k=O
is divisible by x 2 + 1.
Proof.
=
(i) Since d
1 (mod 4), then dk
Hence, for all k we have
=l (mod 4) for all k, thus idk = i.
p (idk) = P(i) =-A+
hence Q (i)= -An+
Bi=
-A,
An= O. The same is true for Q(-i).
(ii) Since d = 3 (mod 4), we find that dk = l (mod 4) for all even k
and dk = 3 (mod 4) for all odd k. Therefore, idk = i for all even
k and idk = -i for all odd k. So, P (i) = -A+ Bi = Bi and
P (-i) = - Bi. Thus,
n-1
II P(idk) = (Bi)1ł · (-Bi)1ł = (-B
2 • i 2 )1ł
= (B 2 )1ł = Bn.
k=O
So,
n-1
R (i)= -Bn +
II P(idk) = -Bn + Bn = O
k=O
and likewise R( -i)
= O.
Note 2. A similar problem was proposed in the Turkish Mathematical
Olympiad 2009. We leave the solution to the reader.
Find all primes p for which there exist an odd integer n and a polynomial
Q(x) with integer coefficients such that the polynomial
2p-2
1 + pn2 +
IJ Q(xi)
i=l
has at least one integer root.
211
Solutions to advanced problems
12. (Aleksander Ivanov - Bulgarian Mathematical
Olympiad 2014)
Find all natural numbers n for which there exist polynomials
fi (x) , ... , f n (x) , 9(x) with integer coefficients such that
(x
2013
+ n) I 9(x) and ((f1(x)) 2 - 1) · ... · (Un(x)) 2 - 1) = (9(x)) 2 - l.
Solution. Note that x 2013 _ x (mod 3). Then, there is an integer s
such that s 2013+ n is divisi ble by 3. Then, 9( s) is divisi ble by 3 and
then all the fi (s) are divisible by 3 (otherwise (Jk(s)) 2 -1 = O (mod 3),
contradiction). Hen ce,
(-lt
- -1
(mod 3).
Thus, n must be odd. Next, we give an example for all odd n. Set
3
91 (x) = x 2013+ n and 9i+l (x) = 4 (9i(x)) - 39i (x). Then for all i~ O,
we have that 9i+l (x) is divisible by 91 (x). Moreover
(9i+i(x))
2
2
- 1 = (4(9i(x)) 2 - 1) ((9i(x)) 2 - 1).
Hence,
(9t+1(x))
2
2
2
- 1 = (4(91(x)) 2 - 1) · ... · (4(9t(x)) 2 - 1) ((91(x)) 2 - 1)
= ((f1(x)) 2 - 1) · ... · ((ht+1(x)) 2 - 1),
where fi = 29i for i= 1, 2, ... , 2t and ht+l = 91·
Note. This problem, has these two variants which were proposed for
Balkan Mathematical Olympiad 2013.
Variant 1. Find all n such that there exist polynomials fi (x), ... , f n ( x),
9(x) with integer coefficients such that
1 + ((f1(x)) 2 - 1) · .. . · (Un(x)) 2 - 1) = (x 2 + 2O13)2(9(x)) 2.
For this we use another approach. Taking x = iv'2013,
we find that
1 + (U1(iv'2013))
2
- 1) · ... · (Un(i✓2O13)) 2 - 1) = o.
Then,
l(U1(iJ2013)) 2 -1) · . . . · (fn(iJ2013)) 2
2
-1)1= 1.
But it is obvious that fk (iv'2013)
= a±iv'b, where a, b are integers and
2
2
b ~ O. If we set z = (a± iv'b) - 1, then lzl = (a 2 - b - 1)2 + 4a 2 b is
an integer. Since the product of n such integer factors is 1, we see that
Solutions to advanced problems
212
each factor must be 1. This forces a 2 b = O and a 2 - b- 1 = ±1. If a = O,
we get b + 1 = =fl, then b = O. lf b = O, then a 2 - 1 = ±1, hence a= O.
Thus we see that a = b = O and hence · fk(iv2Dł3) = O. Thus our first
equation becomes
1 + (-lr
=O
and n must be odd. The example is constructed in the same way as the
example above.
Variant 2. Find all n such that there exist polynomials fi (x), ... , f n ( x),
9(x) with integer coefficients such that
1 + ((fi(x)) 2 - 1) · .. . · ((fn(x)) 2 - 1) = (x + 2013)2 (9(x)) 2 .
Taking x = -2013, we will get the conclusion.
13. (Navid Safaei - Mathematical Reflections U448) Let p > 5 be a prime
number. Prove that the polynomial 2xP-p3Px+p 2 is irreducible in Z[x].
Assume the contrary: 2xP - p • 3Px + p 2 = f(x) • g(x), where
f(x), 9(x) are polynomials with integer coefficients .
Denote deg f (x) = d, <legg( x) = e. Because not all the coefficients of
xP - p · 3Px +p 2 are divisi ble by p, we find that the same statement holds
true for the polynomials f (x ), 9(x ). Th,at is, one can write
Solution.
f (X) = XSfi (X) + p h (X)
and
9 (X) = Xe91(X) + P92(X)
where e, c are the least monomials in f (x ), 9(x) such that their coefficients are not divisi ble by p. Therefore, the constant terms of fi (x), g1 ( x)
are not divisible by p. Hence,
f (X)g (X) = xe+sfi (X)91 (X) + P ( Xsfi (X)g2 (X) + Xe91 (X)h (X))
+ p 2h(x)g2(x).
It is easy to find that c + s = p. Thus, c = e, s = d. Hence,
This implies that
2xP - p · 3Px + p
2 = adbexP + p ( adxdg2(x) + bexef2(x)) + p 2f2(x)g2(x) .
Therefore, comparing the coefficients of x, one can find that min(d , e) < 1.
Therefore, one of f (x), g(x) must be linear. In this case, the polynomial
Solutions to advanced problems
213
2xP - p · 3Px + p 2 must have a rational root. Using the Rational Root
Theorem, one can find that this root must be of the form of ±p, ±p 2 or
p
p2
± 2, ± 2 • N ow, we consider four cases:
Case 1: 2(±p)P ~ p 2 3P + p 2 = O, then ±2pP- 2 ~ 3P + 1 = O. Then, by
Fermat's littl e theorem, we find that p must divide 2, contradiction.
Case 2: 2(±p) 2P ~ p 3 3P + p 2 = O, then ±2p 2P- 2 ~ p3P + 1 = O, which is
clearly false.
pP-2
3P
p)P p2
~ - •3P+p 2 = O then ±-:r+ 1 = O. Therefore,
Case 3: 2 ( ±1 2
2
'
2P-1 2
±pP-2
= ±2p-23p - 2p- l_
Then , by Fermat 's little theorem, we find that p must divide either 1 or
5. If p = 5, the n
_53 = -23 . 35 - 24
'
contradiction.
Case 4: 2 ( ±
contradiction.
p)2p
2
~
p3
2
• 3P + p 2
p2p-2
O, then
± 22p-l
~
p3P
2 +1
o,
14 . (George Stoica - AMM 11822) Let P(x) and Q(x) be polynomials with
complex coefficients such t hat all the coefficients of the polynomial
P(Q(x)) are rea l. Prove that if the leading coefficients and the constant
term of Q(x) are real, then both P(x) and Q(x) have real coefficients .
Solution. Assume P(x) = adxd + · · · + ao and Q(x) = bzx1 + ·· • +bo
with b1,bo E JR. The coefficient of xdl in P(Q(x)), which is adb/ , is real
and then ad is real. Assume that k is the largest index where bk is not
real. Since k > 1, the coefficient of x(d-l)l+k is real and has the form
where M = M(ad, b1,bz-1, . . . , bk+1) is a polynomial with real coefficients. So, bk is real and then we conclude that all the coefficients of
Q(x) are rea l. Now, we prove that all the coefficients of P(x) are real.
Inde ed, let k be the largest index where ak is not real. Again, we can
see that akb/ + N(ad, .. . , ak+l, bo, ... , bi) is real, then ak is real and we
are done.
15. (Kiirschak Competition 2017) Let P (P (x)) = (Q(x)) 2 . Prove that there
is a polynomial R(x) such that P(x) = (R(x))2.
Solutions to advanced problems
214
Solution. It is elear that deg P(x) is even. Write P(x) = (f (x) )2g(x),
where g(x) = ±(x - x 1 ) .....
(x - xd), d > 2 is an even number and
x1, ... , xd are distinct complex numbers. Now
2
P(P(x)) = (f(P(x))) 2g(P(x)) = (Q(x)) .
It follows that g(P(x)) = ±(P(x) - x 1 ) · ... · (P(x) - xd) must be the
square of a polynomial. Hence the initial sign must be a+. It's obvious
that the polynomials P(x) - Xi and P(x) - Xj (i =Ij) have no common
root, otherwise there is a complex number z such that P(z) = Xi = Xj.
Since their product is square, all of them are squares. In particular
P(x) - x1 = (R1(x)) 2 ,
P(x) - x2 = (R2(x))
2
for some polynomials R1 and R2.
Subtracting the two equations, we have
The left harid side is a non-zero constant, so both factors must be constant and then R 1 , R2 and hence P must be constant, contradiction.
Thus d = O and P(x) = (f(x))2.
Solution 2. It is obvious that the polynomial P(x) is of even degree,
say 2d. If the leading coefficient of P (x) is a2d, the n we see that the
leading coefficient of P(P(x)) is a~~+l_Since this must be a square , we
see that a2d is positive. Then we prove the following lemma.
Lemma. Let P(x) = a2dx2d + .. . + ao, where a2d > O. Then there are
polynomials R(x), S(x) with degS < d and
P(x) = (R(x)) 2 + S(x).
Proof. Set R(x) = bdxd +···+bo.
Then comparing the coefficients of
x 2d we find that b~ = a2d and we may assume that bd= .jau. We want
to find bi in such a way that for d < j < 2d the coefficient of xJ in
P(x) agrees with the coefficient of xJ in (R(x)) 2. Now assume that we
have found the coefficients bd, ... , bi+l by comparing the coefficients of
x 2d, ... , xd+i+l. Comparing coefficients of xd+i, we find that
d
ad+i =
L bkbd+i-k.
k=i
Then,
d-1
ad+i bi = ___
L bkbd+i-k
k_=_i+_l
___
2bd
_
Solutions to advanced problems
215
Thus, bi is uniquely determined and our proof is complete.
Now, set P(x) - (R(x)) 2 = S(x) and degS(x) = k < d. Then
(Q(x)) 2 = P(P(x)) = S(P(x)) + (R(P(x))) 2 .
Now
S(P(x)) = (Q(x)) 2 - (R(P(x)) 2 = (Q(x) - R(P(x)))(Q(x)
+ R(P(x))).
Without loss of generality, assume that the leading coeffi.cient of Q(x) is
positive (if not, change it to -Q(x )). Then, the degree of Q(x)+R(P(x))
is max(deg Q(x), <legR(P(x))) = max(2d 2 , 2d2 ) = 2d2 . So, the degree
of the right-hand side of the equat ion is at least 2d2 and the degree of
the left-hand side is 2kd < 2d2 , unless Q(x) = R(P(x)). In this case
S(P(x)) = O and then S(x) = O for infinitely many x. Thus, S = O and
P(x) = (R(x)) 2 •
16. (M. Dadarl at and G. Eckstein - Romanian TST 1989) Find all monie
polynomials P(x) and Q(x) with integer coeffi.cients such that
Q(O) = O and P(Q(x)) = (x - 1) • . . . • (x - 15).
Solution. Let degP(x) =pand degQ(x) = ą.
Then, <legP( Q( x)) = pq = 15, which gives
(p,q) E {(1,15), (15,1),(3,5), (5,3)}.
We have four cases.
(i) q = l. Then, Q(x) = x and thus P(x) = (x - 1) · ... · (x - 15).
(ii) p = l. So, P(x) = x + C and thus Q(x) + C = (x -1) · .. . · (x-15).
Hence,
Q(x) = (x - 1) · . . . · (x - 15) - C.
Since Q(O) = O, we find that C = -15!, thus
P(x) = x - 15!
Q(x) = (x - 1) · . . . · (x - 15) + 15!
(iii) p = 5. Then, P(x) = (x - x1) · ... · (x - x5) and
P(Q(x)) = (Q(x) - x1) · ... · (Q(x) - x5) = (x - 1) · ... · (x - 15).
Since q = 3, all the equations Q(x) = Xi for i = 1, 2, 3, 4, 5 have
3 solutions, say ai, bi, Ci that are included in {1, 2, ... , 15}. Let
Q(x) = x 3 - ax 2 + bx, where a, b are integers. Then, we find that
ai + bi + Ci = a,
aibi + biCi+ c;ai = b.
Solutions to advanced problems
216
Thus, a= ~(l + 2 + ... + 15) = 24. Finding the value of bis not
5
as straightforward , but note that a/+
b/ + Ci2 = a2 - 2b, hence
Then a/ + b/ + c/ = 248. Since for some i one of ai, bi, Ci is 1,
we find that bi2 + c? = 247. Taking this modulo 4 we obtain a
contradiction,
+ c; = 3 (mod 4).
b;
{iv) p = 3. Set P(x) = (x - x 1)(x - x2)(x - x3). Then,
P(Q(x))
(Q(x) - x1)(Q(x) - x2)(Q(x) - x3)
(x - 1) • ... · (x - 15).
Let the roots of Q(x) - Xi for i= 1, 2, 3 be ai, bi, ci, di, ei. They are
included in {1, 2, ... , 15} and then by the same argument
2
2
2
2
2
1 2
2
2
1240
ai + bi + Ci + di + ei = (1 + 2 + ... + 15 ) = - -.
3
3
Which is impossible since the right hand side is not an integer.
17. (Belarusan Mathematical Olympiad 2017) If k > .2 and 65k = an ... a0 ,
prove that the polynomial P( x) = anxn + ... + ao has no rational roots.
Solution. Assume the polynomial has a rational root. Since its coeffi.cients are positive, then the root must be negative, i.e. x = _E.,where
q
p, q > O and gcd(p, q) = 1. Then, by the Rational Root Theorem,
P I ao,
q I an,
Since ao and an are digits, then p and q must be digits . Note that
65k = anl0n + ... + ao
and
Hence
implies that
Solutions to advanced problems
217
10m - (-~)m
where Am= ---:,......-....:....__,qm-l is an integer. Thus the number lOą+p
10-(-~)
divides 65kąn, but gcd(lOq+p,qn) = 1, so lOą+p divides 65k. Since
5 · 13 = 65, we find that lOą + p E {13, 25, 65}. It is easy to find that
for all k > 2 the number 65k ends with 625 and for k = 2 we have
2
65 = 4225. Therefore, a 2 E {2, 6} and a 1 a0 = 25, thus p I aa = 5
implies that lOą + p =/ 13. Hence, lOą + p E {25, 65}, which gives p = 5,
ą E {2, 6}. Moreover, since qx + p divides P(x), one can write
anxn + ... +aa = (qx + p) (bn-1Xn-l + ... + b1x + bo)
where bn-1, ... , bo are integers. Indeed, aa= pbo and an= qbn-1·
Moreover, form= 1, 2, ... , n - 1, we have am= pbm + qbm-1• Hence
m
2
( )m ą
bm -_ -am - am-lQ
+ am-2Q
ao-"+T
- ... + -1
2
3
p
p
p
~
for m = O, ... , n - 1.
Then, it suffi.ces to prove that ampm - am-lQPm-l + ... + (-l)maoqm
is divisible by pm+l. But since
= an(-pt
qn P (-~)
1
+ an-1(-pt- ą + ... + aoqn = O,
we find that
0
=
an(-pt + an-1(-pr-lq
+ ... + aoqn
pm+l A+ (-l)mqn-m(ampm - am-1qpm-l + ... + (-l)maoqm)
(•)
for some integer A and we are done, since (*) must be divisi ble by pm+l.
Now, we find that bo= 1 and 2 = a1 = pb1 + qbo = 5b1 + q, hence q = 2
(since q = 6 is impossible) and b1 = O.
Furthermore, a 2 = pb2 + ąb1 = pb2 = 5b2, but a2 E {2, 6}, contradiction.
Note. We could have concluded also in the following way. We have to
show that -5/2 and -5/6 are not roots of P. Since ao = 5 and a1 = 2,
we have
= an(-5t + ... + a2(-5) 26n- 2 + 2(-5)6n-l + 56n
6nP(-5/6)
= 6n- 1(-10 + 30) = 20 · 6n-l
(mod 25) .
Since this is not zero modulo 25, we conclude that P(-5/6)
-5/6 is not a root. Similarly,
2nP(-5/2)
- an(-5t + .. . + a2.(-5) 22n- 2 + 2(-5)2n-l
-
a 2(-5)22n- 2 (mod 125).
=I O and
+ 52n
Solutions to advanced problems
218
Since a 2 is either 2 or 6 (and hence not a multiple of 5), we conclude
that P(-5/2) =I=O and -5/2 is not a root.
18. Find all monie polynomials with integer coefficients such that
P(O) = 2017
and for all rational numbers r, the equation P( x) = r has a rational
root.
Solution.
First we prove the following lemma.
Lemma . For all non-linear monie polynomials with integer coefficients
there is an integer a such that all the real roots of the polynomial P( x) +a
are irrational.
Proof. Set P(x) = xd+cd_ 1xd-I+ .. . +co. Choose a such that co+a = p
for some prime p which is large enough.
Specifically, let C = lc1I+ ... + jcd-il and choose p > C + l . If x = !:.is
s
a rational root where s > O and gcd(r, s) = 1, then by the Rational Root
Theorem we must have s I 1, r I p. Hence, x = ±I or ±p. But x = ±I
cannot be a root since
gives
IP(±l)I>p-(C+l)
>0.
N ow we prove that x = ±p cannot be a root of the polynomial.
Note that
Note that for /xl> 1 and d > 2, we have
Then,
IP(±p)/ > pd - (C + l)pd-l > o,
and we are done.
Applying the lemma and setting r = -a, we see that if Pis non-linear,
then there is an r such that P( x) = r has no rational root. Thus p must
be linear. In this case, since Pis monie with P(O) = 2017, we must have
P (x) = x + 2017, which clearly satisfies the problem conditions.
219
Solutions to advanced problems
19. Do there exist four polynomials Pi(x) , p 2(x), p 3(x) , P4(x) with real
coefficients such that the sum of any three of them always has a real
root , but the sum of any two of them has no real roots ?
Solution. The answer is no. Assume by contradiction that there are
four polynomials with the desired property. If a polynomial has no real
roots, then it is positive or negative everywhere . Consider a complete
graph with the four polynomials as vertices. Color the edge PiPj white
if Pi (x) + Pj (x) > O and black if Pi (x) + Pj (x) < O. We cannot have a
triangle PiPjPk with sides of the same color (otherwise Pi + Pj + Pk is
positive or negative everywh ere). By the Pigeonhole Principle at least
three of the edges must have the same color , say black. If they have a
common vertex , then in order to avoid a black triangle , th e other three
edges must be white. Then , we have a white triangle , contradiction!
So these three black edges have no common vertex . Without loss of
generality, assume that P1P2,P2P3 , P3 P4 are black. Since there is no
black triangle P1P3 and P2P4 are white. Thus,
P1 (x) + P3 (x) > O,
P2 (x) + P4 (x) > O.
P1 (x ) + P2 (x) < O,
P3 (x) + P4 (x) < O.
Add up both inequalities we get a contradiction for th e sum
20. Find all polynomials with real coefficients such that if
p (X) + p (y) + p (Z) = Q
for real numbers x, y, z , then x + y +z= O.
Solution. First suppose that P(x) has even degree and without loss
of generality assume that the leading coefficient of the polynomial is
positive (otherwise multiply it by -1). Let P( x) > O for all real numbers
x. We have
P (x) + P (y) + P (z)> O
All such polynomials satisfy the problem condition. Assume that P(x)
has a real root, say r. Set x = y =z= r. Then,
P (x) + P (y) + P (z)= 3P (r) = O,
which implies x + y + z = 3r = O, so the only possible root of the
polynomial is O. Write P (x) = xkQ(x) for some polynomial Q(x) such
Solutions to advanced problems
220
that Q(O) =I-o. Thus, Q(x) has even degree k, because ?therwise Q~x)
would have a nonzero real root and then so would P( x) • Smce the leadmg
coefficient of the polynomial is positive, we find that Q (x) > O, hence
for all real x, y, z we have
The equality occurs whenever x = y =z=
satisfy the problem condition.
O. Then, all such polynomials
Now suppose that P(x) is of odd degree. Then P has a real root and
taking x = y = z all equal to this root, we see that the root must be
at O, thus P(O) = O. Fix x, y. The equation P( z) = -P(x) - P(y)
is an equation of odd degree in z , so there is a real root t such that
P (t) + P (x) + P (y) = O. Hence t + x + y = O and
P(-x
-y)
= -P(x)
- P(y).
Setting y = O, we see that P(-x)
= -P(x), thus
+ y) = P(-x
- y) = -P(x)
-P(x
- P(y)
which implies that
P(x + y) = P(x) + P(y).
It is an easy induction to show that, letting C = P(l), this implies
P( n) = Cn for all positive integers n. Since P( x) is a polynomial,
we conclude that P (x) = Cx and clearly all such functions satisfy the
hypotheses of the problem.
21. (Czech-Slovak Mathematical Olympiad 1998) A polynomial P(x) of degree n > 5 with integer coefficients has exactly n distinct integer roots
and P (O) = O. Find all integer roots of P (P (x)) in terms of the roots of
P(x).
Solution. Assume P (x) = C (x - XI) · . .. • (x - xn), where
XI= O< lx2I< .. . < lxnland C, XI, . . . , Xn EZ , C =/-O. Now
P(P(x))
C(P(x)-
XI) · .. . · (P(x)-xn)
CP(x) (P (x) - x2)· ... · (P (x) - Xn).
Thus we see that P(x) divides P(P(x)) and hence all the integer roots
XI, .. . , Xn of P(x) are roots of P(P(x)).
Let r be an integer root of
P (P (x)) which is not a root of P(x).
Then, P (r) = Xi for some 2 <i< n . Hence O< IP(r)I< lxnl• However,
221
Solutions to advanced problems
since r is an integer not equal to any of the roots Xi, we have Ir - Xil> 1
for all i. Moreover,
since the product of two positive integers with summ is at least m - 1.
Also
since the three factors are absolute values of distinct integers, which
therefore cannot all be ±1. Multiplying these two inequalities and the
inequalities Ir - xil > 1 for 5 < i < n and ICI> 1, we get
lxnl> IP(r)I= ICr(r - x2) · ... · (r - Xn-1) (r - Xn)I > 2(lxnl- 1).
Rearranging, this gives lxnl< 2.
But it is easy to see that lxnl > f(n - 1)/21-Hence we must have
n= 5 and the roots of P must be O, ±1 , and ±2. But in this case, we
must have lrl > 3 and hence it is easy to check that we cannot have
equality in the inequalities above. Hence, we get a contradiction. Thus ,
all the integer roots of the polynomial P(P(x)) are integer roots of the
polynomial P(x).
22. (O. N. Kochikhin - Moscow Mathematical Olympiad 2016) Let
P(x) = xd + ad-lXd-l
+ ... + ao.
For some m > 2 all the real roots of the polynomial
P(P( ... P(x)) . . .) = p(m\x)
m times
are positive . Prove that all the real roots of the polynomial P(x) are
positive .
Solution. If the polynomial P( x) has no positive roots, then for all
x >Owe have P(x) > O. Thus, for all x >Owe have p(k)(x) > O, then
p(m)(x) has no positive real root. So, P(x) must have positive roots.
Moreover, if P(O) = O, then p(m)(o) = O, contradiction! Now we prove
the following lemma.
Lemma. Let P( x) have positive and negative roots . Then for all k E Z+,
the polynomial p(k)(x) has positive and negative roots.
Proof. We prove this by induction. The case k = 1 is trivial. Assume
that the statement holds true for all k < j. Let x1, x2 be the smallest
Solutions to advanced problems
222
and the largest roots of P(x) respectively and let X3,X4 be the smallest
and the largest roots of p(j) (x) respectively . Then ,
X1 < 0,
X2 > 0,
X3 < 0,
X4 > Q.
If dis odd, P(x) attains all the values from -oo to O on (-oo, x1], then
there is a real number x 5 E (-oo, x 1) such that P(xs) = x3. If dis even,
P(x) attains all the values from O to +oo on (-oo, x1], so there is a real
number x 5 E ( -oo, x 1) such that P(x 5 ) = X4. In both parity cases the
polynomial P(x) takes all values from O to +oo on [x2, oo), so there is a
real number x 6 E (x 2, +oo) such that P(x5) = X4. Now,
Then, p(J+l)(x 5 ) = O and x 5 < O, hence p(j+l)(x)
Furthermore,
and x 6 > O, hence p(j+l)(x)
has a negative root.
has a positive root .
By our lemma , if P(x) has also negative roots, then p(m)(x) has also
negative roots too, contradiction!
23. (Putnam 2014) Prove that for all positive integers n, all roots of the
n
polynomial P(x) =
L 2k(n-k)xk are real.
k=O
Solution. For n = 1, 2 the problem is trivia!. Assume n > 3 and
consider the sequence {ajh =1,...,n defined by aj = -2-n+ 2j. Now we
will compute the signs of P(ao) , ... , P(an) and we will find that P(aj)
has the same sign as ( -1 )1. It will follow that there is a root between
aj and aj+I and hence P( x) has n real roots, as desired. We have
n
2
P(ao) = I:)-1t2-k
k=O
>o
since each term with even k is positive and larger in magnitude than
the subsequent term with odd k. Hence P(a 0) has the same sign as
(-1) 0 = 1. We also have
n
2
P(an) = 2n I)
k=O
2
- 1t2 - (n-k) = (-1r2n
2
P(ao).
Solutions to advanced problems
223
For O < j < n group the
Hence P(an) has the same sign as (-l)n.
polynomial terms as
+ 2U-5)(n-j+5)xj-5
+ 2U-3)(n-j+3)xj
+ 2(j-l)(n-j+l)Xj-l
+ 2(j+2)(n-j-2)xj+2
+
·
+ 2 (j-4)(n-j+4)xj-4
- 3 + 2 (j-2)(n-j+2)xj-2
+ 2 j(n-j)Xj + 2 (j+l)(n-j-l)xj+l
+ 2 (j+3)(n-j-3)xj+3
Depending on the parity of j and n - j, there may be a single monomial
left on each end . Now for x = aj, the trinomial is O. For x = aj in
the binomials preceding the trinomial, the right-hand term, which has
the sign ( -1 )j , is greater in absolute value than the left-hand term, so
the overall sign of the binomials is (-1 )j. Similarly, in the binomials
following the trinomial, the absolute value of the left-hand term is larger
and the overall sign of the binomial is again ( -1 )j. If there are monomials
left at the ends, their signs are also ( -1 )j. Thus, P( aj) has that same
sign and the claim follows.
24. (Chinese TST 2017) Prove that there exists a polynomial
P(x) = x 58 + a1x57 + ... + ass
having exactly 29 positive and 29 negative roots and log2017 lailare positive integers.
Solution. We prove the stronger statement that for all integers m, n> O
there is a polynomial Qm,n(x) with Qm,n(O) = 1 and roots
Yl < Y2 < • •• < Ym < O < z1 < ... < Zn
and such that the absolute values of its coeffi.cients are non-negative
powers of 2017. lndeed, set
Qo,1(x) = 2017x + 1 and Q1,o(x) = -2017x
+ 1.
N ow suppose we have constructed
n
m
Qm,n(x) = crr(x -yi).
i=l
II(x -
Zj).
j=l
Now we must construct the polynomials Qm,n+1(x) and Qm+I,n(x).
Consider the polynomial xQm,n (x) and consider the intervals
[ YI
_ l
'
Y1+2 Y2]
[Y1
+ Y2Y3 + Y2]
[Ym-1
+ Ym Ym]
and
'
2
'
2
' ...'
2
' 2
[Y;,~ ], [1,z1; z2], ..., [
Zn-1 /
Zn , Zn +
i].
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Solutions to advanced problems
The polynomial xQm,n(x) has a root in all the above intervals and
lxQm,n(x)I is small in a neighborhood of the roots . Thus, there is an
c > Osuch that lxQm,n(x)I < c in a neighborhood of all the roots and
xQm,n(x) takes all the values of(-€, c) . Set the posit ive integer bm+n+I
such that 2017-bm+n+l < €. Then, we define
Then, Qm,n+1(x) has a real root on each of the above interva ls (because
there is Xk such that xkQm,n(xk) = -2017-b m+n+i ). Now we construct
the counterpa rt , i.e. Qm+I,n(x) by changing x to -x. Now, since we
need a monie polynomial, the required example is
P (x) = x 58 Q29 ,29
(1)
;
•
25. (Polish Mathematical Olympiad 1977) Let Wn be polynomials defined
by:
w1(x) = x 2 - 1,
Wn+1(x)= Wn(x)2 - 1,
(n = 1, 2, . . . )
and let a be a real number . How many different real solutions does the
equation wn(x) = a have?
Solution . Let rn(a) be the number of different solutions of the real
equat ion wn(x) = a. We will prove by induction on n that
o
Tn(a) =
n
2n
n +l
2
if a< -l
if a= -1
if - 1 <a< O
if a= O
if a> O
(2)
We establish the truth of (2) for n = 1. The equation w 1 (x) = a has the
form x 2 = 1 + a. It has O, 1 or 2 solutions when a < -1 or a = -1 or
a> -1 , respective ly. Thus , the formula (2) is true for n= 1. Suppose
that the formula (2) holds for a natura! number n. We will prove that it
holds for the number n+ 1. We have Wn+1(x) = wn(x) 2 - 1. Therefore,
the equation Wn+1(x) = ais equivalent to the equation
Wn(x) 2 = 1 + a.
(3)
If a < - 1, then of course equation (3) does not have solutions in real
numbers. Thus, rn+1(a) = O for a< -1.
Solutions to advanced problems
225
If a= -1, the eąuation (3) is eąuivalent to the eąuation wn(x) = O. By
virtue of the inductive hypothesis, the last eąuation has rn(O) = n+ 1
different solutions in real num bers. Thus, r n+1 (a) = n + 1 for a = -1.
If a > -1, the equation (3) is equivalent to the equation
(wn(x) - Ji+a)(wn(x)
= O.
+ Ji+a)
The equations
Wn(x) = JI+a,
Wn(x) = -JI+a
(4)
do not have common solutions, because by subtracting these equations,
we get the contradictory equation O= 2yT+a. Therefore, the number
of solutions of the equation Wn+i(x) = a is equal to the sum of the
numbers of solutions of each of the equations (4), i.e.
(5)
If -1 <a< O, then O< vT+a < 1 and -1 < -v'l+a
due to inductive assumption, we get
rn( v'l+a) = 2 and rn(-v'l+a)
< O. Therefore,
= 2n.
Thus, from (5) it follows that rn+1(a) = 2 + 2n = 2(n + 1) when -1 <
a< O.
If a = O, then
vT+a = 1 and -v'l+a
= -1.
Therefore, due to
inductive assumption, we get
rn( JI+a) = rn(l) = 2 and rn(-v'l+a)
= rn(-1) = n.
Thus, from (5) it follows that rn+1(a) = 2 +n= (n+ 1) + 1 for a= O.
Finally, if a > O, then vT+a > 1 and - v'l+a < -1. Therefore, we
get rn( v'l+a) = 2 and rn(-v'l+a)
= O. So, from (5) it follows that
rn+1(a) = 2 for a> O.
So we showed that the formula (2) is true for the number n+ 1. By the
principle of mathematical induction, formula (2) is true for every natural
number n.
26. (Alexander Khrabrov - Tuymada 2005) Let f(x) = x 2 + ax + b be a
polynomial with integer coefficients such that for all real numbers x
Prove that for all real n urnbers x we have f (x) > -
1
.
4
Solutions to advanced problems
226
Solution. Since the coeffi.cientsof the polynomial f (x) are integers, then
9
the polynomial takes integer values at integer points. Since f (x) > - 10
for all real numbers x, we find that f (n) > O for all integers n. Now,
either the polynomial ha no real roots (so, f (x) > O and we are clone)
or the difference of the two real roots is less than 1. If not, there must
be an integer point k between them. Thus f (k) < O, a contradiction.
2
Hence, the square of the difference of the roots is D = a -4b < l. Now,
-1
4f(x) = (2x + a) 2 - D > -D > -1 and then f(x) > 4
.
Solution 2. As above we can find that the minimal value of f(x) is
-D
4 ,
which must be greater than or equal to -~.
Since D is an integer, we
10
find that D < 3. If D < 1 we are clone. lt remains to check the cases
D = 2, 3. Note that D = a 2 = O,1 (mod 4), so D =/-2, 3 and we are
done.
p2
27. Let p and q be natural numbers such that -
8
< q < p 2 • Consider the
function f (x) = x 2 - px + q and let a and b be coprime integers in
the interval
[~,Pi
such that f(a) and f(b) are integers.
b
a
p
f(a) + f(b) = q.
Solution.
f (a)
-+b
Prove that
We have
b3 - pb2 + qb + a3 - pa 2 + qa
ab
2
(a+ b)(a - ab+ b2 - p(a + b) + q)
ab
+ 2P·
f(b)
a
Since gcd(a, b) = 1 implies gcd(a + b, ab) = 1, we obtain that
ab+ b2 - p(a + b) + q
ab
2
a -
is an integer. Then,
a
2
+ b2 - p( a + b) + q
ab
must be an integer. Now we prove that for a, b E
[!,p]
-ab < a2 + b2 - p( a + b) + q < ab.
we must have
227
Solutions to advanced problems
Indeed, in order to prove the left-hand side we compute
For the right-hand side, set
g(x) = x 2 - x(p + b) + b2 - pb + q.
We want to show that g( a) < O. Since the maximum of g on the interval
[!,
p] is attained at an end point, it suffices to show that g
(!)<
O
and g(p) < O. For the first of these we compute
= p2
-ą - p-q (p + b) + b2 - pb + q = ( p-q - b) p-q + b (b - p) < O
2
g(
-q)
p
since b E
[!,pl
·
For the second we have
g(p)=(b-p)
2
-p 2 +q:S (!-p)'-p
2
+q=qe~P')
<O.
Then, ja2 + b2 - p(a+b) +ąl < ab and ab I (a 2 +b 2 -p(a+b) +q). Hence,
a2 + b2 - p(a + b) + q = O.
Therefore,
2
J(a) + f(b) = a + b2 - p(a + b) + 2q = q.
28. (Cristinel Mortici) Let f(x) = ax 2 + bx + c, where c is integer.
infinitely many natural numbers n we have
For
Solutions to advanced problems
228
Find the polynomial f.
Solution.
The problem statement leads to the following inequalities
b
a
(a - 1) n 2 + (b + 1) n + 2a + c - 1 + n + n2
> O
b+ 2
1
(a - 1) n 2 + (2a + b - 1) n - 2a + c + b + 2 - -+ 2
< O.
n
n
Dividing both inequalities by n 2 , we get
b + 1 2a + c - 1
b
a
(a - 1) +---;- +
n2
+ n3 + n4
(a _ 1) + 2a + b - 1 _ 2a + c + b + 2 _ b + 2 + ~
n
n2
n3
n4
> O
< O.
If a - 1 < O, then the first inequality fails for large n and if a - 1 > O,
then the second inequality fails for large n. Thus we must have a = 1.
Thus the initial inequalities simplify to
(b+l)n+c+l+-+
b
n
b+ 2
(b + 1) n+ c + b - ---;-
1
2
n
1
+ n2
> O
< O.
Dividing both inequalities by n, and considering n sufficiently large , we
see that the first inequality fails if b + 1 < O and the second if b + 1 > O.
Thus we get b = -1 . So, we get the following inequalities
1
n
1
n
c+ 1- - + - 2 > O, c-1-
1
1
- + - 2 < O.
n
n
Take n sufficiently large once again and we get -1 < c < 1.. If c = -1 ,
then _ ..!:_+ ~ > Ofor infinitely many positive integers n, a contradiction.
n n
Since cis an integer, then c = O,1 and
f (x) = x 2 - x
or
f (x) = x 2 - x + 1
Both polynomials satisfy the problems conditions.
29. Find all the pairs of real numbers (a, b) satisfying the following property:
for any pair of real numbers (c, d) such that both of the equations
x
2
+ ax + 1 = c,
x 2 + bx + I = d
have real roots, then the equation x 2 + (a + b)x + 1 = cd also has real
roots.
229
Solutions to advanced problems
Solution. In terms of the discriminants, we get the following restatement of the problem. Find all pairs of real numbers (a, b) such that for
any c, d with
4c > 4 - a 2 , 4d 2 4 - b2
we have
(6)
(a + b)2 - 4 + 4cd > O.
2
We first prove that such a pair (a, b) must have 4- a 2 > O and 4- b > O.
Assume by contradiction that without loss of generality 4 - a 2 < O. Set
c=--
4- a 2
4
into inequality (6) and let d be a large positive real number. Then,
the left-hand side of inequality (6) is negative, contradiction! Hence,
4- a2 > O and 4-b 2 2 O, i.e. a,b E [-2,2]. Now, c,d 2 O and the left
hand side of inequality (6) is indeed an increasing function of c and d.
Thus the inequality holds for all c, d exactly if it holds for the minimal
values. Thus we set
c=
4- a2
4 - b2
and d =
4
4
into inequality (6). Then it becomes
This cancels down to ab(8 + ab) 2 O. Now, observe that
a2 + b2
labl<
- -- 2
<4.
-
Then, 8 +ab> 4 + ab 2 O. So, ab 2 O. It follows that a and b have the
same sign, thus we get (a, b) E [-2 , O]x [- 2, O]or (a, b) E [O,2] x [O,2].
30. (German Mathematical Olympiad 2004) Let xo be a nonzero real root
of the polynomial ax 2 + bx + c, where a, b, c are integers and at least one
of b, c is nonzero. Prove that
2
Solution. If ax + bx + c has a nonzero real root x 0 , then at least two
of a, b,c are nonzero and hence lal+ Ibi+ lcl> 2. Thus the problem is
Solutions to advanced problems
230
obvious if lxol > 1 so we assume /xol < 1. If c = O, then a, b are nonzero
b
and xo = -- . Thus
a
Ibi 1
1
lxol= ~ > ~ > lal+ Ibi- 1'
as required. If c =I-O, then since lxol< 1, we have
lei= laxJ + bxol < (lal+ lbl)lxol
hence
!cl
1
1
lxol > lal+ Ibi> lal + Ibi > lal + Ibi+ lei- 1 '
and we are clone.
31. Find all a such that the following system of equations has a solution
(a, b, c) which are distinct numbers in [-1 , 1]:
a3
b3
b+c+a
a+c+a
c3
b+a+a
a3
Let a + a + b + c = s and b
+c+a
p =I-O. Then,
Solution.
= p. It is elear that
a3 = p(b + c +a)= p(s - a).
Then,
a3 + pa - ps = O.
Analogously, a, b, c satisfy the same equation, thus a, b, c are the three
distinct roots of the polynomial x 3 +px-ps. Hence, a+b+c = O. Then ,
s = a, ab+ be+ ca= pand abc= ps. Since p =/ O, we have abc=/ O and
hence a =/ O and we compute that
1
1
p
a
s
ps
1 1
ab + ac + be l
abc
= ~ + b + ~-
First, assume two of a, b, c are positive, without loss of generality b >
a > O. Then, -1 < c = -a - b < O, which gives a+ b < l. Now,
1
1
1
1
1
a
a = ~ + b - a + b = ~ + ab+ b2 ·
This tends to infinity as a tends to O, so it will take on all sufficiently
large values. To find the smallest value it assumes, we note that it is
a decreasing function in terms of b and so takes its minimal value for
Solutions to advanced problems
231
the maxima! value of b, i.e. b = l - a. (Note that since a and b are
distinct we must have 1/2 < b < l, and conversely this guarantees that
a, b are distinct.) Hence , the minimal value of the above expression is
the minimal value of the expression
1
1
-a + -l-a
- 1 -a+l-a
> ---
4
- 1 = 3.
The equality holds if a = l - a, i.e. a = ~ and b = ~
Since a, b, c
cannot be equal , then .!._> 3, so a < ~. If two of a, b, c are negative ,
3
Q
then replacing a, b, c with their negatives, we see that .!._< -3 , hence
a
-- 1 <a< O. Thus a· E ( -- 1 -1) - {O}
3
3' 3
.
32. (Czech-Slovak Mathematical Olympiad 2008) Find all the real numbers
a, b, c with the following property: Each of the equations
x 3 +(a+l)x 2 +(b+3)x+(c+2)
x 3 +(a+2)x 2 +(b+l)x+(c+3)
x 3 + (a+3)x 2 + (b+2)x+ (c+ 1)
O
O
O
has three distinct real roots, but the total number of distinct roots is
five.
Solution. Suppose that the numbers a, b, c have the required property.
First, we notice that any two equations must have a common root, otherwise they would have at least six distinct roots. Any common root of
two of the three cubic equations is a root of the quadratic equation that
we get by subtracting them. Let's write all the three equations obtained
.by subtracting the three pairs of equations:
x 2 - 2x + 1
2x2 - x - l
x2 + X - 2
o
o
o,
i.e.
(x - 1)2
(2x + l)(x - 1)
(x - l)(x + 2)
o (2 - 1)
o (3 - 1)
o (3 - 2)
We see that first two equations of the original system have only a single
common root x = 1, so they have five distinct roots together . Therefore,
Solutions to advaneed problems
232
each of the roots of the third equation of the original system must be
a root of one of the first two equations. Thus the roots of the third
equation must all be roots of either (3 - 1) or (3 - 2). Hence the three
roots of the third equation must be 1, -2, and -1/2.
So, the third
equation is
Therefore, a = - ~, b = - ~, e = -2 and it is easy to check that these
2
2
give a solution.
33. (Hong Kong Mathematical Olympiad 2015) Let a, b, e be distinct nonzero
real numbers. If the equations ax 3 + bx + e = O, bx3 + ex + a = O and
ex 3 + ax + b = O have a common root, prove that at least one of these
equations has three real roots (not necessarily distinct).
Solution. Let r be the common root. One can add the three equations
to obtain
(a + b + e) (r 3 + r + l) = O.
If r 3 + r + 1 = O, then ar 3 +ar+ a= O. Now, (b - a) r + (e - a) = O.
By the same procedure , we can find that
(e - b)r + (a - b) = O.
As a, b,e are distinct, we find that
e-a
b-a
a- b
e-b'
hence a = b = c, contradiction . Therefore, a + b + e = O and r = 1 is a
common root . We have two cases.
{i) Two of a, b, e are positive, say a, b. Hence, for P(x) = bx3 +ex+ a,
we have P(O) = a > O. Then P(x) has at least one negative real
root. Moreover, from P(l) = O, we find that P(x) has two real
roots and so it has three real roots.
(ii) Two of a, b, c are negative , say a, b. Repeating the same argument
for Q(x) = -(bx 3 +ex+ a), we get the conclusion.
34. Let
P(x) = ax 3 + (b - a)x 2 - (e + b)x + e
and
Q(x) = x 4 + (b - l)x 3 + (a - b)x 2 - (e + a)x + e,
Solutions to advanced problems
233
where a, b, c are non zero real numbers and b > O. Let P(x) have three
distinct real roots xo,x1,x2 such that they are also roots of Q(x).
Prove that abc > 28.
Find all possible integer values of a, b, c.
Solution.
(i) First, no te that
P (x) = ax 3 + (b - a) x 2 - (c + b) x + c = (x - l)(ax 2 + bx - c).
b
C
Now , xo = 1, x1 + x2 = --, x1x2 = -- =I=O. Also,
a
a
x (x 3 + (b - a - l) x 2 + 2 (a - b) x + b - a)
x (x - l) (x 2 + (b - a) x + a - b) .
P(x)-Q(x)
Hence x1 and x2 must be the roots of the last factor .
Hence, we get
Thus,
a-b=
b
C
- - = -- .
a
a
a2
Hence, b = c and a 2 - ab + b = O imply that b = c = -. Since
a- 1
a5
.
b > O, we find that a> l. Now, abc= (a_ l) 2 • Set a - l = x > O.
We have
5
abc=
(x:}) = x + (5x +: + (10
x+~)+10.
1
Since 5x 2 + 2
X
3
2
2)
5
> 2\/'5 > 4 and lOx + - > 2v15Q= 10v'2 > 14, we
X
are done .
1
(ii) Note that b = l +a+ -a - - l , which is int eger . Then, a - 1 = ±1 ,
i.e. a E {O,2}. Since b > O, the only possible case is a = 2. So,
b = c = 4 and
P (x) = 2x 3 + 2x 2 - 8x + 4 = 2 (x - l) (x 2 + 2x - 2).
35. (United Kingdom - Romanian Masters of Mathematics 2016) Let
an = n 3 + bn2 + en + d, where b, c, d are int egers.
Solutions to advanced problems
234
(i) Prove that there is a sequence whose only terms which are perfect
squares are a2015 and a2016 .
(ii) Determine the possible values of a2015 · a2016 for the sequences satisfying point (i) .
Solution. With a sufficient shift, assume that ao, a1 are squares. Then
ao = p 2 and a 1 = ą 2 . The line y = (q - p) x + p passes through (O,p)
and (1, q). Hence it has two points in common with the curve
y 2 = x 3 + bx
2
+ ex + d.
Then, the below equation must have two real roots.
2
x 3 + (b - (q - p)2)x 2 + (c - 2(q - p)p)x + d - p •
But since it is a cubic, this means it must have three real roots . The sum
of the roots is (q-p) 2 -b, hence the third root is (q-p) 2 -b-l, which is an
integer, so the y-coordinate of this point is also an integer. This implies
that there is another square in the sequence, unless (q-p) 2 -b-l E {O,l}.
Then (q - p) 2 E {b + 1, b + 2}. Now, applying the same argument for
the points (O,-p) and (1, ą), we find that (q + p) 2 E {b + 1, b+ 2}. Since
(q + p) 2 and (ą - p) 2 have the same parity , we find that they must be
equal. Then, pq = O. Thus the answer to part (ii) is that a201s · a20rn
must be zero and in any example for part (i) one of p, q must be zero.
Since we can always reverse the sequence, we may assume p = O. Then
we find that an = n 3 + (ą 2 - 2)n 2 + n. If q = 1, then this becomes
an = n(n 2 - n+ 1). If an is a square then n and n 2 - n+ l must be
squares. But if n > l we have
(n - 1)2 = n 2 - 2n + 1 < n 2 - n+ l < n 2 ,
so n 2 - n+ 1 is not a square and if n< O, then
(-n) 2 < n 2 - n+ 1 < n 2 - 2n + 1 = (-n+ 1)2
is not a square. Thus an is only a square for n= O, 1.
Remark. The method that we have used in this problem can be applied
for finding rational solutions of elliptic equations
y2 = x 3 + bx2 + ex + d.
For example, in order to solve the eąuation y 2 = x 3 - 9x + 9, the points
(~,±¾)
!ie on the curve. Starting from the point (~, -¾),
we con-
struct a sequence (xn, Yn) of points by the following manner.
Consider the line
Solutions to advanced problems
235
which is the equation of the tangent line to the curve y 2 = x 3 - 9x + 9
at point (xn, Yn)- Then, the equation
2
- 9
X - 9x + 9 - (3xn
--(x - Xn) + Yn)
2yn
3
2
=O
must have a double root at x = Xn. Then, we have another root Xn+l
which is
Hence,
X
x~ + l8x~ ____
- 72xn + 81
_
n+l 4(x~ - 9xn + 9)
--'-'-----=-.....c.:...
Now, we can easily prove inductively that v2 (xi) < O for all i and
v2 (xn+I) < v2 (xn)- Thus , we have infinitely many rational solutions .
36. Let a, b, c, d be positive real numbers such that the polynomial
ax 4 - ax 3 + bx 2 - ex+ d
has four roots in the interval ( O,
t).
Prove that
21a + 164c > 80b + 320d.
Solution.
Let x1, x2, X3, x4 E ( O,
t)
be four real roots of the polyno-
mial. By Vieta's formulas, we find that
1
b
a
C
a
d
a
Dividing both sides of the desired inequality by a > O and using the
above relations, we must prove
+ X2X3X4 + X1X3X4 + X1X2X4)
> 80 (x1x2 + X2X3 + X3X4 + X4X1 + X1X3 + x2x4) + 320x1X2X3X4.
21 + 164 (x1X2X3
Then, we must prove
Solutions to advanced problems
236
Note that by the AM-GM Inequality, we have
1 _ 2xi + 1 - 2x 2 + 1 - 2x3)
(1 - 2x 1) (1 - 2x2) (1 - 2x3) :::; (
3
=
c+:x•r
1 _ 2x 2 + 1 - 2x 3 + 1 - 2x4)
(1 - 2x 2) (1 - 2x3) (1 - 2x4) < (
3
=
3
(1\2x2)'
1 - 2x4 + 1 - 2x1 + 1 - 2x2 )
(1 - 2x 4) (1 - 2x1) (1 - 2x2) < (
3
=
3
(1\2x1
)'
1 - 2x 3 + 1 - 2x4 + 1 - 2x1)
(1 - 2x 3) (1 - 2x4) (1 - 2x1) < (
3
=
3
3
(1\2x,y
Now by multiplying the four inequalities, we get the conclusion.
37. (Mathematics Magazine) Find all rational numbers r1, r2, . .. , r5 such
that
Solution . Define the polynomial
P (x) = x 8 + 2x 7 + 3x 6 + 4x 5 + 5x 4 + 4x 3 + 3x 2 + 2x + 1.
The above polynomial can be factored as
+ r2x 3 + r3x 2 + r4x + r5)(r5x 4 + r4x 3 + r3x 2 + r2 + r1).
Note that P (x) = (x 4 + x 3 + x 2 + x + 1)2 •
The polynomial x 4 + x 3 + x 2 + x + 1 has no real root . Furthermore, it
(r1x
4
cannot be decomposed as a product of two quadratic polynomials with
rational coeffi.cients (why?) . Thus, x 4 + x 3 + x 2 + x + 1 is irreducible
over Q, so
r1x
4
+ r2x 3 + r3x 2 + r4x + r5 = r1(x 4 + x 3 + x 2 + x + 1).
It follows that ri = ±1 for i = 1, 2, 3, 4.
237
Solutions to advanced problems
38. (Alexandru Lupa§) Let
4v'2- b + ----.
8 - a - 2c
P (x) = ax 4 + bx 3 + cx 2 + ---x
2
8
For all x E [-1, 1), we have P(x) 2: O. Find the value of a, b, c.
Solution.
Note that
Thus we must have a + 2c = 8 and hence
P(O) = P (- ~)
= O.
Further since P(x) 2: O, these must both be double roots. Thus
2
2
P(x) = ax (x + ~)
= ax 4 + v'2ax 3 + ~x'.
From the coeffi.cient of x, we conclude that b = 4v'2,hence from the
coeffi.cient of x 3 , we get a = 4, and from a + 2c = 8, we get c = 2.
Solution
2. Note that
v'2) = a + 2c - 8 > 0
p (O) = 8 - a - 2c > 0 and p (8
-
2
8
-
'
so a + 2c = 8. Thus,
P(x)=ax
4
+bx 3 +-
8 - a 2 4v'2- b
2
-x
+
2
x.
Let Q (x) = a 4 x 4 + a3x 3 + a2x 2 + a1x + ao have the property that for
all x E [-1, 1) we have Q(x) 2: O. Then, there are polynomials f(x) and
g(x) of second and first degree, respectively, such that
Q(x)
= J(x) 2 + (1 - x 2)g(x) 2.
The above representation can be proved by straightforward calculations.
Assume f (x) = mx 2 + px + q and g (x) = rx + s . Now,
P (x) = (mx 2 + px + q) 2 + (1 - x 2)(rx + s) 2 .
Solutions to advanced problems
238
Since P (O) = O, we find that ą 2 + s 2 = O, thus q = s = O and
P (x) = (mx 2 + px) 2 + (1 - x 2 )(rx) 2 = x 2 ((mx + p)
2
+ r 2 (1 - x 2 )).
2
.
(- 2v'2)= O, we get (- v'2
m + p ) + 1 r = O, thus r = O and
2
m
p = v'2.
Hence,
2
Smce P
2 2
P(x) = m x (x +
!;)'
Checking the leading coefficient, we find that m = 1. Now,
I;)'
2
p (X) = x ( X +
and then a = 4, c = 2.
39. The coefficients of a polynomial ax 4 + bx3 + cx 2 + dx + e are such that
a, e > O and ad 2 + b2 e - 4ace < O. Prove that this polynomial has no
real roots.
Solution.
Write the given polynomial as
2
2
b )
( y'ax +--x
2Ja
2
+
( v'e+--x d
2Je
)
+ ( c---- b2
4a
d2 )
4e
x2 .
Since
b2
d2
4ace - ad 2 - b2 e
c----=------>0
4a 4e
4ae
'
we have ax 4 + bx 3 + cx 2 + dx + e > O and hence it has no real roots.
Solution 2. Assume P (x) = ax 4 + bx3 + cx 2 + dx + e.
Then, P (O)= e > O. Assume x =/-O. Then,
p ~) = ax 2 + bx + c + d + e2 = ax 2 + bx + c + dy + ey 2 ,
X
where y =
X
X
.!.
. We can see the above expression as a quadratic polynomial
X
in x. Its discriminant is
f (y) = b2 - 4a (c+dy+
We prove that
ey 2 ) = -4aey 2 - 4ady + b2 - 4ac.
f (y) < O. Indeed, the discriminant
of f(y) is
2
D = 16a 2 d 2 + l6ae (b2 - 4ac) = 16a (ad 2 + b e - 4ace) < O.
239
Solutions to advanced problems
Moreover, the coeffi.cient of y 2 is -4ae < O, so f (y) < O for all y. So,
the discriminant of the polynomial is negative. Since the coefficient of
x 2 is positive, we conclude that P (x) > O for all x.
40. (Nikolai Nikolov - Bulgarian Mathematical Olympiad 2012) Let a i=O,I.
Jim and Tom play the following game. Starting with Jim, and proceeding
alternately, each player replaces one * in the expression below with an ,
where n EZ. Jim wins the game if the resulting polynomial has no real
roots , otherwise, Tom wins . What is a winning strategy?
Solution. Answer: Jim has a winning strategy when a > O or a = -1
and Tom has winning strategy when a < O or a =/=-1 . We divide the
problem into three cases.
Assume P(x) = a4x 4 +a3x 3 +a 2 x 2 +a1x+ao. At first Jim
replaces the value of a2, then Tom replaces a3 or a1 with a number
of the opposite sign and then Jim replace ao or a4 with a number
(i) a= -I.
of the same sign. Then , replacing x by _!_or multiplying by - 1, we
X
can assume that we get the polynomial
P (x) = x 4 + x 3-± x 2 - x + I.
Then,
2P (x) > 2 (x 4 + x 3 - x 2 - x + 1) = (x2 +x - 1)2 +( x 2 - 1)2+x 2 > O.
(ii) a > O. At first Jim replaces the value of a3 with any power of a,
then he tri es to rep lace a 1 if not filled or any other. If the last
unfilled coefficient is either a4 or ao, then Jim replace it with an
arbitrary large number (with positive n if a > 1 an d negative n if
O < a < l). Then P (x) > O for all x ensure that we have not any
real roots . If the last unfilled coefficient is a2, then consider the
function
2
a1
ao
f (x) = a4x + a3x + - + 2 .
X
Since lim f(x)
x➔oo
X
= xlim
f(x) = +oo, we have that the function f( x)
➔O
is bounded below. Now, choose as a2 a number large enough. Then ,
P (x) > O for all x.
(iii) a < O and a =/=-1. In this case, with sufficient changes of x to .!.,
X
we can assume that Jim replaces one of a2, a1, ao. If he replaces a0,
Solutions to advanced problems
240
then Tom replaces a 4 with a number of different sign . This ensures
that the polynomial P(x) has at least one real root and Tom wins.
Now assume Jim replaces one of a2, a1. Then, Tom replaces ao
and immediately Jim replaces a4 with a number of the same sign.
Assume without loss of generality that ao, a4 > O. Now, it remains
two numbers unreplaced: a 1, a2 or a3, a1. Consider the functions
g(x) and h(x) as follows
3
2
ao
g (X)
= a4X + a3X + a2X + -X ,
a2
a1
h (x) = a4x + -
ao
+ X2 + 3X ·
X
Tom can choose a2 in the first case or a1 in the second case, so that
g(l) < g(-2) and h(l) < h(-2). Moreover, in both cases
lim f(x) =
x➔ +oo
lim g(x) = +oo
x➔ +oo
and
lim f(x) =
x➔±oo
lim g(x) = -oo.
x➔ -oo
Moreover,
lim f(x)
x ➔O+
= lim g(x) = +oo
x➔O+
and
lim J(x) = lim g(x) = -oo,
x➔O-
x➔O-
hence the functions f(x) and g(x) assume all real values. Hence,
there exists at least one real number x such that both of the equations g(x) = -a1 and h(x) = -a3 have solutions. Then in each
case Tom wins the game.
Solution 2. Once again, we divide the problem into three cases.
(i) a> O. Jim can play in sucha way that in the last move he chooses
one of ao, a2, a4. Since the polynomial has an absolute minimum,
Jim can choose the value of ao such that the polynomial is positive everywhere.
x4P (
!)
Similarly, he can choose a4 for the polynomial
in the same mann er. In his last move, he must set the
a2
a2
value of a2 . If we sets a2 > - 3 + - 0 + 1, then
4a4 4a1
P(x)
> a4x 4 + a3x 3 + ( a~ + a5 +
1) +
~.r
4a4
2
2
X +a4X (x+ 2
4a1
x2
a1x + ao
2
0.
2
+a1
(x+ a:J>
241
Solutions to advanced problems
(ii) a = -1. then all the coeffi.cients are ±1. At first , Jim can choose the
coeffi.cient of x 2 . Then ' he chooses the other coeffi.cients such that
the coeffi.cients of x 3 , x and x 4, x 0 are of opposit e signs and being
the same (respectively), then we get one of polynomials below which
has no real roots.
.
x
4
± x3 - x 2 =i=x + 1,
x
± x3 + x 2 =i=x + 1.
4
(iii) a< Oand a i= -1. lf P (O)and lim P(x) have different signs, then
x➔ oo
the polynomial has real roots and then Tom wins the game. So, if
Jim starts from ao, a4 , then Tom sets different signs for the others
and wins the game. So Jim may start from a3 , a2, a 1 . lf he starts
from a3 or a 1 ( in the polynomial x 4 P ( ¾)), assuming the coeffi.cient
of x 4 is 1 and the coeffi.cient of x 0 is a0 > O, we have the polynomial
g (x) = x 4 + a3x 3 + ao. Let m = max {g(l), g(-1)} . Tom can set
the coeffi.cient of x 2 as a2 < -m. Then, h (x) = g ( x ) + a2x 2, hence
h ( 1) = g ( 1) + a2 :Sm + a2 < O,
h ( -1)
= g ( -1) + a2 :Sm + a2 < O.
Regardless of finał moves of Jim, one of the above values increases
and the other decreases. Hence, one of P(l) and P(-1) remains
negative, so Tom wins. If Jim chooses the coeffi.cient of x 2 , then
assume Tom sets a4 =land
Jim sets ao > O. Now, we have
Let
M = max { T ( -1) , T (
lf M < O, choose -M
R(-1)
R (
1)}.
> a3 > O. Then , R (x) = T (x) + a 3 x 3 and
= T(-1) - a3 < T(-1) :SM< O,
1)= T ( 1)+
<T (
~3
1)-1:S ~ <
7
O.
Regardless of the finał moves of Jim, one of the above values increases and the other decreases, then one of P (1) and P(½) remains
negative, so Tom wins. Finally, let M > O. Set a3 > 4M. Then
R ( -1)
and
= T (-1) - a3 :SM - a3 < O
- +-<M+(1)=T (1)
R -2
2
a3
8 -
a3
8 .
Solutions to advancedproblems
242
Now to win, Jim must choose the coefficient of x to ensure that
M - a3 - a1 > O,
This implies that M - a 3 > a1 > -
a3
a1
M + 8 + 2 > O.
a3
8M
4 - M, then a3 < 3
contrary
to the choice of a3.
41. (A. Golovanov - Tuymada 2013) Prove that for any fourth degree poly-
nomial A(x), there are quadratic polynomials P(x), Q(x), S(x), R(x)
such that
A (x) = P (Q (x)) + R(S (x)).
p (X) = 8 (X) = X2.
We will prove that we can represent any fourth degree polynomial with
a3-:/-O as
2
4
(ax2 + bx + c) 2 + dx + ex + f.
Indeed, fix any non-zero a with a2 i- a4. By comparing the coefficients
of x 3 and x, we find that 2ab = a3 and 2bc = a1. Thus we set
b = a3 and c = a1 = a1a.
2a
2b
a3
Now, by comparing the coefficients of x4, x 2 , x 0 , we find d, e, f. We still
need to resolve the case when a3 = O. For it, consider A(x + 1), whose
coefficient of x 3 is 4a4 i- O. Then, by above proof there are polynomials
P(x), Q(x), S(x), R(x) such that
A (x + 1) = P (Q (x)) + R(S (x)).
Then,
A (x) = P (Q (x - 1)) + R(S (x-' 1)).
Solution 2. It is known that every fourth degree polynomial A(x) can be
written as a product of two quadratic polynomials with real coefficients.
Now, set
A(x) = B(x)C(x),
where B(x) and C(x) are quadratic polynomials with real coefficients.
By moving a constant multiple between them, we may assume the leading
coefficients of Band C are neither the same nor negatives of each other.
Solutions to advanced problems
243
Now,
A(x) = B (x) C (x) = (C (x) + B (x))2 - (C (x) - B (x))2.
4
Hence, set
P (x) =
x2
x2
4 , R (x) = - 4 , Q (x) = C (x)+B (x), S (x) = C (x)-B
(x).
42. (Bµlgarian Mathematical Olympiad 1995) Let
P(x) = xd + ad_1 xd-l + ... + ao
be a polynomial with integer coefficients, where a 0 =/O. Suppose the
roots of P taken with multiplicity are the same as the coefficients ai for
i= O,1, 2, ... , d - 1 also with multiplicity. Find P(x).
Solution. We find that P (x) = (x - ao) · ... · (x - ad-i). Hence
P (O)= ao = (-l)dao · . . . · ad-l·
Since ao =/O, we find that a1 · ... · ad-1 = (-l)d . As ai are integers, we
get lail = 1 for i = 1, ... , d - 1. Now we divide the problem into two
cases.
(i) lail = 1 for i = O,... , d - 1. Then, all the roots of P(x) are ±1,
hence
P(x) = (x- lt(x
+ l)b
for some non-negative integers a, b such that a+ b = d. Then,
P (x) = (xa - axa-l + ...)(xb + bxb-l + ...).
The coefficient of xd-l is ad-l = b - a = ±1. The coefficient of
ad-2 =
hence
a2;a+ b2; ab=
b-
±1,
2
(a - b) - a - b = ±1.
2
Thus, a+b = 3, which gives d = 3 and a= 1, b = 2 or a= 2, b = 1.
So,
P(x) = x 3 + x 2 - x - 1,
or
P(x) = x 3 - x 2 - x + 1.
Solutions to advanced problems
244
(ii) laol> 2. Since P (ao) = O, then the polynomial
P(x)=x
d ±x d-1 ± x d-2 ± •·· ± X + ao
has a root with absolute value greater than or equal to 2. Note that
P (a0 ) = aod ± aod-1 ± aod-2 ± • • · ± ao + a O·
By triangle inequality,
O
IP (ao)I
jaod± aod-1± aod-2± ... ± ao + aoj
2
1
> jaodj- jaodj- jaodj- ... - laol- laol
laol(laol- 2) (laold-l- 1)
laol-1
Since iaol> 2, we find that laol = 2. Moreover, the equality case
for the triangle inequality occurs. Hence aod and ao have opposite
signs. Thus dis even and ao = -2. Thus equality in the triangle
inequality forces ai = (-l)i+ 1 for i = 1, ... , d - 1. If d > 3, then
this gives a2 = -1 and then P (-1) = -d + 2 = O, a contradiction.
Thus d = 2 and
43. (Feng Zhigang - Chinese Western Mathematical Olympiad 2009) Let M
be a subset of JRobtained by deleting finitely many real numbers from
JR. Prove that for any given positive integer n, there exists a polynomial
f(x) with <legf(x) = n such that all its coefficients and its n real roots
are in M.
Solution. Let S = {lxl E JR I x ft.M}. This set is finite by definition.
Let a= max S. Choose any real number k > max {lal, 1}. Then, -k €J.S
and hence -k E M. For any positive integer n, define the polynomial
f (x) = k(x + kr. It follows from k > 1 that degf(x) = n and the
coefficient of xm in the polynomial f(x) is k ·
kn-m > k. Hence,
all the coefficients of f (x) are not in S and thus are in M. As the roots
of J(x) are -k with multiplicity n, they are in M. So, the polynomial
f(x) = k(x + kr satisfies the required condition.
(;:J·
44. (Ye. Malinnikova - Russian Mathematical Olympiad 1996) Does there
exist a finite set M of nonzero real numbers such that for any n E N
there is a polynomial of degree at least n with coefficients in M, all of
whose roots belong to M?
245
Solutions to advanced problems
Solution. The answer is no. Let M = {a 1, ... ,ak} be an arbitrary set
of nonzero real numbers,
r = min {la1I , ... , laki} and R = max{la1I, ... , laki} •
Then R 2::r > O. Consider the polynomial
P (x) = bnxn + bn-IXn-ł + . . . +bo,
where all the coefficients bo, b1, ... , bn and the roots x1, x2, . . . , Xn belong
to the set M. By Vieta's formulas, we have
Moreover,
Hence,
R2
R
nr ~ ~ xf ~ 2 + 2-,
Lr
r
2
n
i=l
i.e.
R2
R
n~ 4 +2 3 =A .
r
r
Therefore, for n> A the required polynomial does not exist.
45. Do there exist 2000 real numbers (not necessarily distinct), not all zero,
such that if we put any 1000 of them as roots of a monie polynomial
of degree 1000, its non-leading coefficients are a permutation of the remaining numbers?
Solution. The answer is no. Assume that none of the 2000 numbers is
zero. We have at least 1000 positive numbers or 1000 negative numbers.
lf we have at least 1000 negative numbers, we put them as roots . Then
by Vieta's formulas the coefficients are all positive. Thus, we have 1000
positive numbers. lf we put them as coefficients, then all the roots are
negative. So, we have exactly 1000 negative and exactly 1000 positive
numbers. lf we put these 1000 positive numbers as roots, we get a
contradiction since among the coefficients of the polynomial there must
be also 500 positive coefficients. Now, assume there are k > O zeroes
among the 2000 numbers. lf k < 1000, put them as zeros and then
by Vieta's formulas the product of the roots is zero and hence there is
another zero coefficient (i.e. the constant term), contradiction. Thus,
k > 1000. Take 1000 of them as zeros. Then, we get the polynomial x1000
and hence the other coefficients are zero. Thus all the 2000 numbers are
zero and this contradicts to the problem's assumption.
Solutions to advanced problems
246
:nare
46. (Russian Mathematical Olympiad) If n > 3 and X 1 < x 2 < · · ·t~
the · roots of an n-th degree polynomial P(x). Further assume a
x2 -
X1
< X3 - X2 < . . · < Xn - Xn-1 ·
Prove that IP(x)I on the interval [x1,xn] attains its maximum value
between the two largest root (i.e., in the interval [xn-1, Xn]).
Solution. Assume the maximal value occurs at a E (xi, Xi+I) and i <
n - 1. Set t = a - Xi, b = Xn - t, note that b E (xn-1, Xn)· By adding
up the inequalities of the problem assumptions, one can easily find that
Xk+m - Xk < Xt+m - Xt,
for any 1 :S k < l :Sn - m. Now,
IP(x)I = Clx - x1I · ... •lx - Xnl•
Then, for i+ 1 :S s < n - 1, we have
jb - Xsl = Xn - Xs - t > Xi+n-s - Xi - t = lxi+n-s - al.
(7)
Now, for 1 :S r < i - 1, we have
lb - Xrl = b - Xr > Xn-1 - Xr > a - Xr = la- Xrl•
(8)
Finally, b was chosen so that
(9)
By multiplying (7), (8), (9), we find IP(b)I > IP(a)I, which contradicts
the choice of a.
47. (Polish Mathematical Olympiad 1998) Let g(k) denote the greatest
prime divisor of an integer k if lkl ~ 2, and g(-1) = g(O) = g(l) = 1.
Find if there exists a non-constant polynomial W (x) with int eger coefficients such that the set {g(W (x)) I x E Z} is finite.
Solution. We prove that there is no polynomial with the given property.
For the sake of proof, let's assume that
W(x) = ao + a1x + ... + anxn
is a polynomial of degree n > 1 with integer coefficients and that the
set of numbers g(W(x)), where x is an integer, is finite. So, there is
a natural number m and there are prime numbers p 1 , ... , Pm with the
following property: if x is an integer and W(x) =/-O, the value W(x)
is not divisible by any prime number other than Pi, ... ,Pm · We will
consider two cases.
247
Solutions to advanced problems
(i) If ao = O, then for each integer x i- O, the value W( x ) is divisibl e by
x. Let b = 1 + P1P2• ... •Pm· We find a natural number k for which
W (kb) -/- O. Such a number exists because the polynomial W has
a finite number of roots . The value of W (kb) is divisi ble by kb and
so it is divisible by b. Hence, the polynomial has at least one prime
divisor different from p 1 , ... ,Pm· This contradicts th e hypothesis .
(ii) If ao i- O, consider the number c = a0p 1p2 • .. . · Pm· We find a
natural number k > 2 for which W(kc) i- a0 . Such a number exists
becaus e the value a0 is taken by the polynomial W only at a finite
number of points. We get the equality
W(kc) = ao + kc[a1 + a2(kc) + ... + an(kcr- 1] = ao + kcq,
where q is an integer representing the number in square brackets
and q i- O because W(kc) i- a 0 . By plugging into this equality the
value of c, we get the equality
W(kc) = aow,
where w = 1 + kąp1p2 · .. . · Pm· Since k > 2 and q i- O, th en
w i- -1, O,1, so w it has a prim e divisor different from PI, ... ,Pm·
Therefor e, also in this case we have a contradiction and th e proof
is complete.
48. (Marian Tetiva) Let f be a non-constant polynomial with int eger coefficients and let k be a positiv e integer. Show that there are infinit ely
many positive integers n such that f (n) can be written in the form
d1d2 · ... · dk+l , where
Solution. Without loss of generality, assume that f(x) has posit ive
leading coeffi.cient. Note that
f (x + f (x)) = f( x) g(x)
for some polynomial g(x) with integer coefficients. Let h (x) = x + f (x),
so that this reads f(h( x )) = f(x)g(x). Then iterating this identity gives
f (h(m) (x)) = f (x)g(x)g(h(x))
· . . . · g(h(m-I) (x)) .
Let <legf (x) = s. Then, the degree of the above polynomials are s,
s 2 - s, s 3 - s 2 , ... , sm - sm - l . If s > 2, we have s < s 2 - s < s 3 - s 2 <
... < sm - sm- l. Hence, for t large enough, we have
g (t) < g (h (t)) < ... < g ( h (m - 1) ( t)) .
Solutions to advanced problems
248
Take m = k + 1, n = h(k+l) (t). Set d1 = g (t), d2 = 9 (h (t)), ... ,
dk = g (h(k) (t)) and dk+l = g (h (k) (t)) · J(t) and we are done.
If s = 1, then f (x) = ax + b. Choose pairwise coprime di such that
gcd(di, a)= 1. By the Chinese Reminder Theorem, the system am+ 1 =
O (mod di) has a solution. Setting n= mb, we find that f(n) = b(l+am)
is a multiple of d 1d2 • • • dk and we are done.
Comm ent. The case s > 2 can also be derived from the Chinese Remainder Theorem, much as the case s = 1.
49. (Titu Andreescu - Mathematical Reflections U450) Let P be a nonconstant polynomial with integer coefficients. Prove that for each
positive integer n there are pairwise relatively prime positive integers
k1, k2, ... , kn such that k1k2 · • • kn = IP(m)I for some positive integer m.
Solution.
By Schur 's theorem , there exist distinct prime numbers
P1,P2, ... ,Pn and positive integers m1, m2, ... , mn such that
P(m1) = O (mod P1)
P(m2) = O (mod P2)
P(mn)
= O (mod Pn).
By the Chinese Remainder theorem, exist positive integer m such that
m = m1
(mod P1)
m = m2
(mod p2)
m = mn
(mod Pn).
We have Vi E {1, 2, ... , n}: m = mi (mod Pi), hence we get
P(m) = P(m i) = O (mod Pi).
Thus P1P2. .. Pn divides P( m). Hence we get
jP(m)I = pf1p~2 . .. p~n. A,
where a1, a2, ... , an > O and A E N. Choosing
k1 - p Q l
k - p02
k
= Pn-1 , kn = p~n • A
then we have i-/- j that gives gcd(ki, kj) = 1 and
-
1 '
2 -
2 ,·•· ,
n-1
On-I
k1k2 · ... · kn = IP(m)J.
Solutions to advanced problems
249
50. (Crux Mathematicorum) Let P(x) be a polynomial with integer coefficients such that for all positive integers n we have P (n) > n. For all positive integers m, there exists a term in the sequence P ( 1), P (P (1)) , ...
which is divisible bym. Prove that P (x) = 1 + x.
Solution.
The leading coefficient of P(x) must be positive.
If
<legP( x) > 2, then there is a positive integer M such that P (n) > 2n
for all n > M . Since 1 < P (1) < P (P (1)) < ... , there exists a positive integer k such that p(k) (1) > M. Set r = p(k) (1) and Take
m = p(k+l) (1) - p(k) (1). Since r > M, we get
p(k+l)
(1) = P (p(k) (1)) > 2p(k) (1),
1.e. m > r.
Therefore, for 1 < i < k, we have 1 < p( i) (1) < r < m.
Hence, m ł p(i) (1) for 1 < i < k. Moreover,
p(k+l)
(1) = m + r = r
(mod m).
By induction, one can find that for all i > 1 + k we have p(i) (1) = r
(mod m), i.e.
p( i+ 1) ( 1)
p ( p( i) ( 1))
P(r)
p (p(k _) (1))
_
p(k+l)
(1)
m+r
r (mod m).
But then m does not divide any term in the sequence. Thus degP(x) =
1. Assume that P (x) = x+b for some integer b. Since P (1) = 1+b > 1,
then b > O and this implies that b > 1. If b > 2, then P (1) = 1 (mod b).
Hence, inductively, we find that p( i ) (1) = 1 (mod b) contradicts the
problem assumption. If P (x) = 2x + b, then P (1) = 2 + b > 1, hence
b > O. If b = O, then p(i) (1) = 2i does not satisfy the condition of
problem . If b > 1 or P (x) = ax + b, where a> 3 the same approach as
above works and leads to a contradiction. Hence, P (x) = 1 + x.
51. (Vlad Matei) Find all polynomials P(x) with integer coefficients such
that
(a 2 + b2 + c2 ) I (P(a) + P(b) + P(c)) for all integers a, b, c.
250
Solutions to advanced problems
Solution. Taking the substition a 1-t -a, we see that
(a 2 + b2 + c2) I (P(-a)
+ P(b) + P(c)).
Then, (a 2 + b2 + c2) I (P(a) - P(-a)). Taking b large enough, we get
P(x) = P(-x) for all x. Now, P(x) = Q(x 2 ) for some polynomial Q(x)
with integer coefficients. Then,
·'
We search for integers with the same value of a 2 + b2 + c2. Note that
hence we easily find that
I t follows that
Q((a2 + b2)2)- Q((a 2 - b2)2) = Q((2ab)2).
Since (a 2 + b2)2 = (a 2 - b2) 2 + (2ab)2,we get that
Q(z + t) = Q(z) + Q(t)
for infinitely many positive integers z, t. Thus, Q(z + t) = Q(z) + Q(t)
for all z, t. Hence, Q(x) = Cx for some integer C and so P(x) = Cx 2 .
52. Let P(x) be a polynomial with integer coeffi.cients such that
(r
22011 _ 82 2011)
I (P(r) _ P(s))
for any positiv e integers rand s. Prove there is a polynomial Q(x) with
2017
integer coeffi.cients such that P (x) = Q(x2 ).
Solution. We prove the generał case for an arbitrary positive integer n
(and the conclusion will follows with n = 2017). We prove the assertion
by induction on n . The case n = O is indeed obvious. Assume the
statement holds for all positive integers less than or equal to n . We
prove the problem for n+ l. Note that
(r2n - s2n) I (r2n+1 - s2n+1), and
(r2n+l - s2n+l) I (P(r) - P(s)).
Hence by the inductive hypothesis, P (x) = R(x 2n) for some polynomial
R(x) with integer coefficients. Set a= r 2n and b = -s 2 n. Since
(a - b) I (R(a) - R(b)),
251
Solutions to advanced problems
2n
2n
2n
2n
2n
2n
we have r + s divides R(r )-R(-s
). We also have that r + s
2
divides P(r) - P(s) = R(r n) - R(s2n).
Thus r 2n +s 2n divides R(s 2n)-R(-s 2n). Taking r large enough, we find
that R(s 2n) = R(-s 2n) for all s. Hence, R(x) = R(-x) for infinitely
many x and hence for all x. Then, R(x) = Q(x 2 ) for some polynomial
Q(x ). This implies that P(x) = Q(x 2n+i) and we are done.
Solution 2. More generally, we will show that if rN -sN I P( r )-P( s) for
any positive integers rand s, then P(x) = Q(xN) for some polynomial
Q(x) with integer coefficients. We first prove the following lemma.
Lemma. Suppose F(x) and G(x) are polynomials with integer coefficients with G(x) monie and G(n)/F(n) for all positive integers n. Then
G(x) I F(x).
Proof. Write F(x) = Q(x)G(x)+R(x)
where A(x) and R(x) have integer
coefficients and deg(R) < deg(G). Then G(n) I R(n) for all positive
integers n, but since R has lower degree, we have /R(n)/ < IG(n)I for
large enough n. Thus R( n) = O for all large n, and hence R is the zero
polynomial. Thus G(x) divides F(x) .
Applying the lemma to the problem, we see that xN - sN I P( x) - P( s)
for all s. Write
N-1
P(x) =
L xiQi(xN)
i=O
for polynomials Qi (x) with integer coefficients.
Since xN - sN I Qi(xN) - Qi(sN), we see that the remainder when we
divide P( x) by xN - sN is
N-1
L xiQi(sN).
i=O
But since xN -sN divides P(x)-P(s), this remainder must also be P(s).
Thus we see that Qi(s) = O for all i= 1, ... , N-1. Since s is an arbitrary
positive integer, we conclude that Qi(x) = O for i= 1, ... , N - 1. Thus
P(x) = Qo(xN).
53. (Polish · Mathematical Olympiad 2009) The sequence of integers
Jo,fi, h, ... is defined by the conditions: Jo= O, fi = 1 and
fn = fn-1 + fn-2,
n= 2,3, ....
Find all the polynomials W with integer coefficients having the following
property : for each natural number n there is an integer k such that
W(k) = fn•
Solutions to advanced problems
252
Solution. We prove that W(x) = c:x+ c, where c: E {-1, 1} and cis an
integer. Observe that
(i) the numbers fi, h , fa, ... are positive;
(ii) the sequence h, fa,14,... is strictly increasing;
C
> 2;
( iii) f n-1 _> l f n 1or
a 11n _
2
(iv) ln+l < 3ln-l for all n> 4.
Property (i) is obvious and implies property (ii) because for n > 3 we
have ln-2 > O, which gives In = In-I+ ln-2 > In-I· Properties (iii)
and (iv) follow from property (ii) because In-I > In-I + ln-2 = In for
all n> 2 and
ln+l =In+ ln-1 = 2ln-l
+ ln-2 < 2ln-l + ln - l = 3ln-l
't/n > 4.
Now, let W be a polynomial satisfying the conditions of the problem.
Then, there is an integer a for which W(a) = Io = O and an integer b
for which W(b) = fi = 1. Hence, the difference W(b) - W(a) = 1 is
divisible by b- a. We get b- a= ±1, i.e. b =a+ c:, where c: E {1, -1} .
In this situation, the polynomial P defined by
P(x) = W(a + c:x)
satisfies P(O) = W(a) = O and P(l) = W(a + c:) = W(b) = 1. It is also
easy to see that the sets of values of the polynomials W and P taken
at the integers are equal. In other words, for any n = O,1, 2, ... , there
exists an integer k such that P(k) = In- Let d and e be integers such
that P(d) =fa= 2 and P(e) = /4 = 3. Then,
(d - 1) I {P{d) - P{l)) = 1 ===>d E {O,2}.
It cannot be d = O, because P(O) = O =/-2. Thus, d = 2. Similarly,
(e - 2) I (P(e) - P{2)) = 1 ===>e E {1, 3},
which gives e = 3. We carne to the conclusion that
P(k) = k,
k = O, 1,2,3.
We will show by induction that
P(fn) = fn,
n= O, 1, 2, ....
We already know that for n < 4 the last equality holds. Assume now that
the equality is satisfied for n = O, 1, ... , m, where m > 4. We want to
253
Solutions to advanced problems
prove that P(fm+1) = fm+I· By the assumptions of the problem , there
is an integer k for which P(k) = fm+I · Suppose first that k < f m-1· As
m - 1 > 2, by virtue of property (iii), we have
fm - k > fm - fm-1 = fm-2 >
1
2fm-1 ·
Moreover ,
(fm - k) I (P(fm) - P(k))
= fm - fm+l = - fm-1·
The number fm - k is ther efore a divisor of fm-I larger than its half ,
which implies the equality f m - k = fm-I·
However , k = fm - fm-I = fm- 2, impossible, because the induction
hypothesi s says PUm-2) = f m - 2, while P(k) = fm+l·
We have therefore proved that k 2: fm-I· Since m > 4, by property (iv),
we get
(k - O) I (P(k) - P(O)) = fm+l < 3fm-1·
Consequently, the number k is a divisor of fm +l greater than one-third
1
of fm+l· Thus, k = fm+l or k = fm+l· Th e first case lea ds to the
2
conclusion that PUm+1) = fm+I, which is the same as the induction
thesis. However, the second case is impossibl e because it would result in
k = 1 or
k-
1
~fm+l- 1 I P(k) - P(l) = fm +l -
1= (1fm+l - 1)+
2
1
2(k - 1) + 1,
and therefore k - 1 j 1, i.e. k ~ 2. So, we would have fm +l = 2k < 4 < !5,
which contradicts the assumption m > 4. The conclusion follows from
th e principle of mathematical induction .
Now, P(f n) = f n for all natural numbers n together with the property
(ii) allows to conclude that P(x) = x occurs for infinitely many int egers
x . Since Pis a polynomial, the equality is true for each real number x.
Since P(x) = W(a+ cx), we get W(x) = c(x - a) for all real numbers x,
where c E { - 1, 1} and a is an integer . To end the solution, it remains
only to notice that every polynomial W(x) = c(x - a) has the desired
pro perty .
54. (Korean Mathematical Olympiad 2008) Find all polynomials P(x) with
integer coefficients such that there are infin it ely many positive integers
a and b such that gcd (a, b) = 1 and (a+ b) I (P(a) + P(b)) .
Solutions to advanced problems
254
Solution. Let
Q (x) = P(x) +/(-x)
and R(x) = P(x) -/(-x).
It is elear that Q(x) is an even polynomial and R(x) is an odd polynomial.
In particular, P(x) = Q(x)+R(x). Since all the monomials in R(x) are of
the form cx 2l+l for some non-negative integer l and integer c, we obtain
that R(a) +R(b) is divisible by a+b (indeed, a 2l+ 1 +b 2l+l is divisible by
a+ b), thus the problem is reduced to the case (a+ b) I (Q(a) + Q(b)).
We know that all the monomials of the polynomial Q(x) are of the form
dx 28 for some non-negative integer s and integer d. Moreover, since
a28+ b28 = 2b28 (mod a+ b). First, assume that Q(x) = dx 2n. Hence,
(a+ b) I 2db2n.
Since gcd(a + b, b) = gcd(a, b) = 1, we find that a+ b I 2d, but 2d has
only finitely many divisors. Thus, a + b can assume only finitely many
values. Since a, b are positive integers, we have only finitely many a, b,
contradiction.
Now assume P(x) = a2nx2n + P1(x), where degP1(x) > 2n and P1(x)
is an even polynomial. Take b large enough such that gcd(b, a2n) = 1.
Choose a such that
bi _
P1 (b) _ a2nb2n + P1 (b) _ P(b)
Ia + - a2n + b2n b2n
- b2n ·
Then,
P (a)+ P (b) = 2P (b) = 2b2n la+ bi= O (mod a+ b).
It is elear that we have infinitely many pairs (a, b) of positive integers
satisfying the problem statement. Thus, all the polynomials of the form
P(x) = a2nx2n + P1(x), where degP1(x) > 2n and P1(x) 1s an even
polynomial
55. (Fedor Petrov - Saint Petersburg Mathematical Olympiad 2002) Let
P( x) be a polynomial wit h real coefficients such that for all integers
P(n + 1) · .. . · P(n + k)
n, k > O the number
P(l) .... . P(k)
is an integer. Prove that
P(0) = O.
Solution.
.
P(n + 1)
Settmg k = 1, we see that
P(l)
is an integer for all
positive integers n. The polynomial
~\:i
takes on integer values for all
255
Solutions to advanced problems
integer x, therefore by the Lagrange Interpolation formula, we see that
it has rational coefficients. Hence replacing P(x) with
NP(x)
Q(x) = P(l)
for a sui table integer N we get a polynomial Q( x) with integer coefficients
that also satisfies the hypotheses of the problem, that is,
Q(n+l)·
... ·Q(n+k)
Q(l) · ... · Q(k)
is an integer. Fix k and define
R (x) = Q (x + 1) · ... • Q(x + k) .
Then, Q(n + 1) · . . . · Q(n + k) = R(n) and Q(l) · ... · Q(k) = R(O) = M.
R(n).
.
Then, we find that for all positive integers n, we have ~ 1s an 1nteger.
2
Note that R( l~ - l) is also an integer and
R(2 JMJ -1) - R(-1)
= O (mod M).
Hence,
R(-1)
Q(O)· . . . ·Q(k-l)
M
Q(l)·
. .. ·Q(k)
Q (O)
Q (k)
is an integer. Hence, Q(k) I Q(O) for all k and so Q(O) = O, which gives
P(O) = O.
56. Let P (x) = x 3 + 3x 2 + 6x + 1975. Find the number of integers a in the
interval [1, 3 2017 ] such that P( n) is divisi ble by 32017 •
Since P(x)
x 3 + 1 = x + l (mod 3), where the second
congruence is by Fermat's little theorem, we see that 3 divides P(x) if
and only if x = -l (mod 3). Since we compute that
Solution.
P(3x - 1) = 27x
3
+ 9x + 1971 = 9(3x 3 + x + 219),
we see that 27 divides P(x) if and only if x
that
3
P (9x - 1) = 729x
= -1 (mod 9). Now note
+ 27x + 1971 = 27Q(x ),
3
where Q (x) = 27x + x + 73. Consider the set A = {1, 2, . . . , 3n} and
assume there are 1 < a < b < 3n such that Q (a)
Q (b) (mod 3n).
Note that
·
=
Q (a) - Q ( b) = (a - b) (27 (a 2 + ab + b2 ) + 1)
= O (mod 3n).
Solutions to advanced problems
256
Since 27 (a2 +ab+ b2 ) +1 is not divisible by 3, we get a-b = O (mod 3n)
which is impossible since a - b < 3n_ Then, Q(l), ... , Q(3n) forms a
complete residue system modulo 3n. Hence, there is an integer Cn such
that Q(en) = O (mod 3n).Let an= 9Cn - 1. Then,
Let n= 2014. Then, 32017 divides P(a 201 4). For any interval of the form
[32014 i + 1, 32014 (1 + i)], where i is a positive integer, there is an integer
Xi such that Q(xi) is divisible by 32014 . Then, Yi = 9xi - 1 and P(yi)
is divisi ble by 32017 . Hence, there are three numbers Yi in the interval
[1,32017 ] such that P(yi) is divisible by 32017 •
57. Let Q (x) = (p - 1) xP - x - 1, where pis an odd prime number. Prove
that there are infinitely many positive integers a such that Q( a) is divisible by pP.
Solution.
At first we prove the following lemma.
Lemma. Let a, b E {1, 2, ... ,pP}. Then, Q (a) "t=Q (b) (mod pP).
Proof. Assume that Q (a)= Q (b) (mod pP). We have
(p - 1) aP - a - 1 = (p - 1) bP- b - 1 (mod pP),
hence
(p - 1) (aP - bP)- (a - b) = O (mod pP).
Since aP = a (mod p), we find that
O= (p - l)(aP - bP)- (a - b) = -2(a - b) (mod p).
Thus, a= b (mod p). Furthermore,
O
(p - 1) (aP - bP)- (a - b)
(a - b) ((p - 1) (ap- l + aP- 2b + ... + bP-1 ) - 1) (mod pP).
Note that
(p - 1) (ap-I+ aP- 2b + ... + bP-1 )-1
= (p - 1)paP- 1-1 "t=O (mod p).
Hence, we get a - b = O (mod pP), contradiction,
{1, 2, . .. ,pP}. Our proof is complete.
since a, b E
No:',we find that the values of Q(l), Q(2), ... , Q(pP) form a complete
res1due system modulo pP, thus there is a unique a E {1, 2, ... ,pP} such
that Q (~) = (mod pP). Now, all the numbers of the form a+ spP,
where s 1s an mteger, satisfy the problem conditions.
?
257
Solutions to advanced problems
Comment. An alternate proof is to consider the sequence
p-l
ao = -
2
-,
.
an = an-l + Q(an-d 1f n 2: l.
Then one can verify that Q(an) is divisible by pn. Now, taking n 2: P
solves the problem.
58. (Oleksiy Klurman - Ukrainian Mathematical Olympiad 2016) Let
Xn = 2016P(n) + Q(n) ,
where P(x) and Q(x) are non-constant polynomials with positive integer
coefficients. Prove that there are infinitely many primes p for which there
exists a square-free integer m such that p I Xm .
Solution. Assume the contrary. Then, in this sequence there are only
finitely many prime divisors. Let p 1 , .. . , Pk be the primes other than
2, 3, 7 (since 2016 = 25 · 32 · 7) which occur. Let no = pf1 • p~ 2 • ••• • pft,
where Qi 2: x1. Consider the sequence an= l +n · no<p(no).Then
Q (an) = Q (1)
(mod no) and P (an)= P (l)
(mod cp(no)).
Since gcd (no, 2016) = 1, we find that 2016P(an)_ 2016P(l) (mod no).
Hence, Xan = x1 (mod no). Since the prime Pi divides no with multiplicity Qi 2: x 1 , which is clearly larger than the multiplicity with which
Pi divides x1, we see that Pi divides Xan with the same multiplicity that
it divides x1. Since Xan has no other prime divisors except 2, 3, 7, we
conclude that Xan = 2016P(an) + Q (an)= 2a · 3b · 7c · X1 . Since
3
XI· ( max { 2a,3b, 7c} ) > X1 · 2a · 3b · 7c = 2016P(an) + Q (an)> 2016n,
for all n sufficiently large, there is A E {2, 3, 7} such that A ~-Co divides 2016P(an) (where Co = log20l6 -v'Xi)-Moreover, A~-Co divides
2016P(an) + Q (an), thus A ~-Co divides Q (an), But for all n large
enough, we have A~-Co > Q (an), contradiction. It remains to show that
in the sequence an there are infinitely many square-free terms , which we
prove at the end of the second solution.
Solution 2. Note that
Xn = 2016P(n) + Q (n) = 2016P(n) - 2016P(l) + Q (n)+ 2016P(l).
Let R(n) = Q(n) + 2016P(l). Since R(0) > O, we can prove that set
of primes dividing the sequence Yn = R( n) is indeed infinite. Write
R(n) = cdnd + ... +co. Then,
R (con) = co(cdcod-Ind + ... + l).
Solutions to advanced problems
258
If we have only finitely many prime divisors_, _say ~1, · · · , Pt, th en set
n = Pt ..... Pt and we have another prime d1v1sor 1~ R (c?p.1· •••·Pt) · p > max (R(O), 2016) , from the pnme d1v1sorsof. the
Now, choose a pnme
.
·nr
ove
that
that
there
is
a
square-free
pos1t1ve
sequence Yn = R (n ) . vve pr
integer a such that Xa is divisible by p.
.
. . .
If p I n, then p I n I (R(n) - R(O)). Since I:,(~) 1s d1v1s1bleby P, then
R(O) is divisible by p, but P > R(O), contrad1ct1on. Thus, gcd(n,p) - 1.
Now, consider the system of congruences below
X
X
-
l
(mod p - l)
n
(mod p).
This system has solutions ak, where
ak = l -
(n - l)(p - 1) + (k + c)p(p - 1).
Choose c such that ak is a positive integer. Now, R (ak) - R (n) = O
(mod p) and P (ak) = P (l) (mod p - l). Finally, since ak - l > O
divides P(ak) - P(l), we can find that P(ak) > P(l). Now,
20l6P(ak)
= 2016P(l)
(mod p).
Hence, Xak = 2015P(n) - 2015P(l) + R (ak) is divisible by p. Now, we
prove that sequence ak contains infinitely many square-free terms.
Lemma. Let Xn = a+ nb, where gcd(a, b) = l and a, b are positive
integers. Then, the sequence contains infinitely many square-free terms.
Proof. In the set {x 1 , ...
, x N}
the multi pies of ą2 are at most
l:,j
+ 1.
Let Pt < ... < Pt < JxN be all the primes smaller than -Jxi,.i. Then,
the multiples of p? in the above set are at most
N
t+-2+•
PI
N
( 1
1
1
)
3N
2 ++
·
·
·
+
--< v'XN+-.
· -+2 < ..fxii+N
Pt
2
2 ·3
t (t - l)
4
We prove that
N
JxN <
82
for all large N.
Indeed, this reduces to N > 64(aN + b) which obviously holds for large
N since the left hand side has higher degree. Hence the number of
.
non-square-free terms 1s at most
7N
8
'
. Our proof is complete .
Remark . The lemma is also an immediate corollary of Dirichlet 's Theorem on primes in arithmetic progressions.
Alphabetical
algebraic number, 29
discriminant, 61, 62, 94, 95
factorization, 11
function
decreasing, 55
increasing, 55
identities, 2
parabola, 67
polynomial
coefficient, 1, 6, 18, 115
cubic, 85
definition, 1
degree , 1
division , 26, 27
double root, 42
elementary symmetric
polynomials , 111
even, 37
greatest common divisor, 28
irreducibility , 14, 16
Lagrange Interpolation
Formula, 31
least common multiple, 28
Index
minimal, 30
monie, 1
odd, 37
quadratic, 59
quartic, 101
root, 39, 115
value, 18
Vieta's formulas, 73, 90, 101,
111
theorem
Bezout's identity for
polynomials, 29
Eisenstein 's Criterion, 16
Fundamental Th eorem of
Algebra, 39
Int ermediat e Value Theor em
'
49
on polyn omia l with integ er
coefficients, 20
polynomial congru ences, 127
polynomia l division, 26
property of the degree, 2
Rational Root Theorem, 45
Schur's Theorem, 132
261
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and Circles, 2016.
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262
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29. Andreescu, T., Dospinescu, G., Straight from the Book, 2012.
30. Andreescu, T ., Mathematical Reflections - The First Two Years, 2011.
31. Andreescu, T., Dospinescu, G., Problems from the Book, 2nd edition,
2010.
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