Arithmétique dans l`ensemble des entiers relatifs

n∈Nn= 10q+r q ∈Nr∈J0,9K
7|n7|q−2r
7|11228 7 |15637
11 2123 + 3121
n∈N
(i) 6 |5n3+n, (ii) 7 |32n+1 + 2n+2,(iii) 11 |38n×54+ 56n×73,
(iv) 5 |22n+1 + 32n+1,(v) 9 |4n−1−3n, (vi) 152|16n−1−15n.
n>2 10 |22n−6
n∈N2
52n−1 2n+2
(a, b)∈Z2n∈N∗
a≡b[n]⇒an≡bn[n2].
(x, y)∈N27|x7|y7|x2+y2
(x, y)∈Z2
(i)xy = 3x+ 2y, (ii)1
x+1 y=1 5,(iii)x2−y2−4x−2y= 5.
n∈N40nn!|(5n)!
x2+y2= 3z2(x, y, z)∈Z3
x2+y2+z2=x2y2(x, y, z)∈Z3
(x, y, z)x, y z
4542
n∈N∗√n∈Q√n∈N
n∈N∗
n n
n∈N\ {0,1}n=
N
Y
k=1
pαk
k
d(n)n
n d(n)
Y
d|n
dX
d|n
d.
n∈N∗d(n)n
1
n
n
X
k=1
d(k).

(3123 −5) ∧25
(a, b)∈Q2
a+b∈Zab ∈Z⇒(a, b)∈Z2.
(a0, a1, . . . , an−1)∈Znx∈Q
xn+an−1xn−1+··· +a1x+a0= 0,
x∈Z
(a, b, n)∈(N∗)3an∧bn= (a∧b)n
(a, b)∈(N∗)2(a+b)∧(a∨b) = a∧b
(a, b)∈(N∗)2(a2+ab +b2)∧ab =a2∧b2
n∈N
(i) (n2+n)∧(2n+ 1),(ii) (15n2+ 8n+ 6) ∧(30n2+ 21n+ 13).
(x, y)∈Z2
(i) 95x+ 25y= 45,(ii) 20x−53y= 3.
(x, y)∈N2
(i)x∧y+x∨y=x+y, (ii)x∨y+ 11(x∧y) = 203.
(N∗)2
(i)x∧y= 18
x∨y= 540 (ii)x∧y= 10 x+y= 100
(a, b, c)∈(N∗)3
a∨b= 42, a ∧c= 3 a+b+c= 29.
(d, m)∈(N∗)2
(x, y)∈(N∗)2
x∧y=d x ∨y=m.
(un)u0= 14 un+1 = 5un−6
n∈Ndn=un+1 ∧un
(m, n)∈(N∗)2
2m−1 2n−1
pgcd(2m−1,2n−1)
(a, b)∈(N∗)2
∃n∈N, ab =n2⇔ ∃(m, n)∈N2, a =m2b=n2.
∀n∈N, n + 1 |2n
n.

n∈N
ak=k·n!+1 k∈J1, n + 1K
n∈NFn= 22n+ 1 (m, n)∈N2
m6=n Fm∧Fn= 1
(ϕn)n∈N
ϕ0= 0, ϕ1= 1 ∀n∈N, ϕn+2 =ϕn+1 +ϕn.
ϕn+1ϕn−1−ϕ2
n= (−1)nn∈N∗
ϕn∧ϕn+1 = 1 n∈N∗
ϕn+m=ϕmϕn+1 +ϕm−1ϕn(m, n)∈N∗×N
ϕm∧ϕn(m, n)∈N2
∀n∈N\ {0,1}, n
X
k=1
1
k!/∈N.
(a, b)∈N∗×N∗pgcd(a, b)=1
f: div(a)×div(b)→div(ab),(k, `)7→ k`
div(n)n∈N∗
n∈N∗
n n!+2
a p ap−1
a= 2 p
n∈N∗2n+ 1 n
2
p>5p2−1 24
n∈N\ {0,1}
n
∀k∈J1, n −1K, n |n
k.
p∈N
∀n∈Z, np≡n[p].
n∈N∗p∈N
p n!
30!
3 4
p∈Np>5
p⇔p|(p−1)!.

6 5n3+n=−(n−1)n(n+ 1) 6
7 32n+1 + 2n+2 = 9n·3+2n·4=2n·7 = 0
11 38n·54+ 56n·73= 5n·9+5n·2 = 0
5 22n+1 + 3n+1 = 4n·2+9n·3=4n·5 = 0
9 4n+1 −1−3n= 3(4n−1) = 0 3 |4n−1
15216n+1 −1−15n= 15(16n−1) = 0
x2+y27
3 (x, y, z) = (0,0,0)
4x, y z
x, y ∈Z∗2
4 (x, y, z) = (0,0,0)
n= 3α·5βα, β > 1
(3α/2·5β/2)(α+1)(β+1) = 4542 α= 6 β= 3 n= 36·53
d(n)=(α1+ 1) ···(αN)
Y
d|n
d=√nd(n)X
d|n
d=
N
Y
i=1
pαi+1
i−1
pi−1.
ln(n)
a, n 6= 0 a=p/q b =m/n p, m ∈Z
q, n ∈N∗p∧q= 1 m∧n= 1 a+b=pn+mq
qn ∈Z
q|n n |q n =q
ab = (pm)/q2∈Zq∧pm = 1 q=n= 1
pgcd(a, b)=1
p p |a2+ab +b2ab p |a−b p |a+b
p|a p |b p |a p |b
(x, y) = (−9−5k, 36 + 19k) (x, y) =
(24 + 53k, 9 + 20k)
(21,2,6) (3,14,12) (6,14,9)
3|a3|c b = 29 −a−cpgcd(b, 3) = 1
b|42 = 3 ×14 b|14 b∈ {1,2,7,14}29 −b=a+c
3b2 14
m>n2m−1 2n−1
2m−1=2r(2nq −1) + 2r−1=2r(2n−1)(2n(q−1) +··· + 1) + 2r−1
pgcd(2m−1,2n−1) = 2pgcd(m,n)−1
p aka`p
p6n
ϕm∧ϕn=ϕm∧n
p n! + 1 p6n
p|(n! + 1) −n! = 1
n k =p
p n!Pkbn/pkc
p1, . . . , pk
m= 4p1. . . pk−1m>2m= 3[4] m
p p = 3[4] p|m p |p1···pkp|1