solutions des exercices.

u v b =au d =cv bd = (ac)(uv)ac |bd
n|n+ 7 n|n n |(n+ 7) −n= 7 n1 7
n n + 7
n2+ 1|n n = 0 n2+ 1 ≤n n > 0n2+ 1 > n2≥n
n= 0
n2|n n|n(n+ 1) n(n+ 1) n
2|n+ 1 n+ 1|n(n+ 1) n(n+ 1)
n n + 1 n+ 2 3
n(n+ 1)(n+ 2) 3 n(n+ 1)(n+ 2) = 3u
n(n+ 1) n(n+ 1)(n+ 2) 3u
u
6=3×2|3u

c=−an2−bn n | −an2n| −bn n |c
3 = −n5+ 2n4+ 7n2+ 7n n
n|3n−3−1 1 3
−1 3
a|n(n+ 2) = n2+ 2n a |n2+n+ 5 −(n2+ 2n) = −n+ 5
a|(n+ 2) + (−n+ 5) = 7 a= 1 a= 7
=×2 +
=×2 +
=×1 +
=×3 +
d d 10100 10121 + 10813 + 10 d
P GCD
10121 + 10813 + 10 = 10100 ×(1021 + 10713) +
10100 =×1099 +
10 10 −10 −5
−2−1 1 2 5 10 8

q= 1 b= 109r= 5
P GCD(109+ 5,109) = P GCD(109,5) = 5
5|109
131
221 = 13 ×17
p≥3p−1p+ 1
4p−1
p−1 = 4k p = 4k+ 1 p+ 1 p= 4k−1
4k−1p1< p2<··· < prp1= 3 p2= 7
n= 4(p1. . . pr)−1n≥4p1−1≥11 >2n
4k+1
4k+ 1 n
4k−1pipi4p1. . . prpi
1
P GCD(n, n + 2) = P GCD(n, 2) n n + 2
n2n
n2−1 = (n+ 1)(n−1) (n2−2n+ 1) = (n−1)2n−1
n−1−1 1 n= 0 n= 2
n2−2n+ 1 = 1
=− × 2
=− × 2 = ×5− × 2
=−=×3− × 7

2x+ 1 8 8 1 2 4
8 2x+ 1 |8y2x+ 1 |y
7 9
(u0, v0)=(−3,−4) (u, v)
k∈Zu= 7k−3v= 9k−4
P GCD(16,26) = 2 n n
n= 2n08x+ 13y=n08 13
n0= 1 (x0, y0)=(−8,5)
n0(−8n0,5n0) (x, y)
k∈Zx= 13k−8n0y=−8k+ 5n0
18100 = (2 ×32)100 = 2100 ×3200 v3(18100) = 200
2100 + 2200 = 2100 ×(1 + 2100)
2v2(2100 + 2200) = 100
7b100
7c= 14
14 49 98 49 = 72
73= 243 v7(n) = 14 + 2 = 16
n
k10k|n10k|n2k|n5k|n
k2k5kn
min v2(n), v5(n)
v5(n) 20 5 4
52= 25 v5(n) = 20 + 4 = 24
50 v2(n)≥50 >24
24 24 n

5
7
b= 0 a|b a 6= 0 a0|b2
b2= 0 b= 0 a, b 6= 0
a2|b2p2vp(a)≤2vp(b)
vp(a)≤vp(b)p a |b
vp(a2)≥1vp(a2) = 2vp(a)vp(a2)≥2p2|a2
p vp(ab) = vp(a) + vp(b)vp(P GCD(a, b)) =
min vp(a), vp(b)vp(P P CM(a, b)) = max vp(a), vp(b)
vp(P GCD(a, b)P P CM(a, b)) = vp(a) + vp(b) = vp(ab)p
ab P GCD(a, b)P P CM(a, b)
x= 8x0y= 8y0x+y= 128 x0+y0=128
8= 16
8x0×8y0= 8 ×440 x0y0=440
8= 55 x0y0
1 5 11 55 x0+y0= 16 5 11
(x, y) = (40,88) (88,40)