Théorie des nombres I

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Olympiades Suisses de Mathématiques

a b k ∈Za=kb a
b b b a b |a n
±1±n
a > 0
p∈N p p
a|b b |c=⇒a|c
a|b1, . . . , a |bnc1, . . . , cn
a|
n
X
i=1
bici.
a|b c |=⇒ac |bd
p p |ab =⇒p|a p |b
a∈Nb∈Za|b=⇒b= 0 a≤ |b|
x, y
x2−y! = 2001.
y≥3y!x
x2y≥6y!
y= 1,2,3,4,5
(x, y) = (45,4)
a, b b > 0
q r 0≤r < b
a=qb +r,
r r = 0 b|a

a
p1, p2, . . . , prn1, n2, . . . , nr
a=pn1
1pn2 2· · · pnr
r.
pinia
a= 1
r= 0
a=pn1
1pn2 2· · · pnr
ra
a(n1+ 1)(n2+ 1) · · · (nr+ 1)
a m
nkm
p1, p2, . . . , pn
N=p1p2· · · pn+ 1 N > 1
q N pkN pk|1
q p1, p2, . . . , pn
a, b pgdc(a, b)a
b a
bppmc(a, b)
a b
(a1, a2, . . . , an) [a1, a2, . . . , an]
c= pgdc(a, b)
c > 0a b x
x|a, x |b=⇒x|c.
pgdc(a, b)=1 a b

pgdc(a, b) = pgdc(b, a)
pgdc(a, b, c) = pgdc(pgdc(a, b), c)
c|ab pgdc(a, c) = 1 =⇒c|b
a|c b |cpgdc(a, b) = 1 =⇒ab |c
d= pgdc(a, b)x y
a=xd b =yd ppmc(a, b) = xyd
a, b ab m
a b m
a=pα1
1pα2 2· · · pαr
rb=pβ1
1pβ2 2· · · pβr
r
a b pkαk, βk≥0
pgdc(a, b) = pmin{α1,β1}
1pmin{α2,β2} 2· · · pmin{αr,βr}
r
ppmc(a, b) = pmax{α1,β1}
1pmax{α2,β2} 2· · · pmax{αr,βr}
r
min{x, y}+max{x, y}=x+y
pgdc(a, b)·ppmc(a, b) = ab.
95 m n
pgdc(m, n) + ppmc(m, n) = m+n.
d m n m =ad, n =
bd ppmc(m, n) = abd
d+abd =ad +bd d(ab −a−b+ 1) = 0
d(a−1)(b−1) = 0 a= 1 b= 1
m=d m |n n |m
a, b n
(a, b)=(a, b +na).
n=±1
c a b c b ±a(a, b)|(a, b ±a)
c a b +a b −a
c(b+a)−a=b(b−a) + a=b
(a, b ±a)|(a, b)

(2541,1092)
(2541,1092) = (2541 −2·1092,1092) = (357,1092)
= (1092 −3·357,357) = (21,357)
= (357 −17 ·21,21) = (0,21) = 21.
(a, b)a, b ≥0
a1= max{a, b}a2= min{a, b}n= 2
an−1=qnan+an+1 0≤an+1 < an
an+1 = 0 (a, b) = ann1
2541 = 2 ·1092 + 357
1092 = 3 ·357 + 21
357 = 17 ·21 + 0.
0 (2541,1092) = 21
a, b x, y
xa +yb = 1.
d= pgdc(a, b)x, y
xa +yb =d.
pgdc(a, b) = anan
(n−1) ak
pgdc(a, b) = xa +yb
21 = 1 ·1092 −3·357
= 1 ·1092 −3(2541 −2·1092)
= (−3) ·2541 + 7 ·1092.