CORRECTION DES EXERCICES POUR LE MARDI 12 JANVIER

•a+b ab
P GCD(a+b;a) = 1 d(a+b)
a d a +b−a=b d a b a b
d= 1 P GCD(a+b;a)=1 P GCD(a+b;b)=1
P GCD(a+b;ab) = 1
•a+b a2−ab +b2
a2−ab +b2=a2+ 2ab +b2−3ab = (a+b)2−3ab
P GCD(a2−ab+b2;a+b) = P GCD(a+b;−3ab)d
P GCD(a2−ab +b2;a+b)a+b ab
u v (a+b)u+ (ab)v= 1 −3(a+b)u−3abv =−3
d=P GCD(a+b;−3ab)d a+b d −3ab d −3(a+b)u−3abv
d−3d d = 1 a+b a2−ab +b2
d= 3 a+b a2−ab +b2
P GCD(a+b;a2−ab +b2)
P GCD(a+b;a2−ab +b2)=3 a+b P GCD(a+b , 3) = 3
P GCD(a+b;a2−ab +b2) = P GCD(a+b, 3)
P GCD(a+b , a2−ab +b2) = 1 P GCD(a+b , 3)
P GCD a +b3ab
(a+b)2−3ab a2−ab +b2P GCD(a+b , a2−ab +b2) = 1
P GCD(a+b , 3) = 1 P GCD(a+b;a2−ab +b2) = P GCD(a+b, 3)
P GCD(a+b;a2−ab +b2) = P GCD(a+b, 3)
½xy = 5292
P GCD(x , y) = 6 (x;y)∈N2x
y x0=x
6∈Ny0=y 6∈N
xy = 5292 ⇔36x0y0= 5292 ⇔x0y0= 147 P GCD(x , y) = 6 ⇔P GCD(6x0,6y0) = 6 ⇔P GCD(x0, y0) = 1
½x0y0= 147
P GCD(x0, y0) = 1
147 = 2 ×72(1 + 1) ×(2 + 1) = 6
(x0;y0)x0y0= 147 (1 ,147) ; (3 ,49) ; (7 ,21) ; (21 ,7) .
(7,21) (21 ,7)
(x , y)x0y0(6 ,882) (18 ,294) (294 ,18)
(882 ,6)
xy = 5292 x0y0x0y0= 147 P GCD(x , y) = 6
x0y0P GCD(x0, y0) = 1 P GCD(1 ,147) = 1
P GCD(3 ,49) = 1 S={(6 ,882) ; (18 ,294) ; (294 ,18) ; (882 ,6)}
(AB)xA6=xB(AB)
y=mx +p m =yB−yA
xB−xA
=−7−9
7+7 =−16 14 =−8 7
(AB)y=−8
7x+p A ∈(AB)yA=−8 7xA+p9 = −8 7×(−7) + p

p= 9 −8 = 1 (AB) : y=−8
7x+ 1 (AB)∩E
(x;y)y=−8
7x+ 1 y=−8 7x+ 1 ⇔8x+ 7y= 7
8x+7y= 7 x y P GCD(8 ; 7) = 1
(AB)∩(E) = ∅
8x+ 7y= 7 (x0;y0) = (0 ,1)
8x+ 7y= 7
(x , y) 8x+7y= 7 8x+7y= 8x0+7y08x−8x0= 7y0−7y
8(x−x0) = 7(y0−y)y0−y∈Z8(x−x0)P GCD(7 ,8) = 1
x−x0k∈Zx−x0= 7k
x= 7k8×7k= 7(y0−y)
8k=y0−y y = 1 −8k(x , y) 8x+ 7y= 7 k∈Z
(x , y) = (7k , 1−8k)
8x+ 7y= 7 k∈Z8x+ 7y= 8(7k) + 7(1 −8k) = 56k+ 7 −56 = 7
(E) (7k , 1−8k)k∈Z
(AB)∩(F) (x , y)
(AB)∪(E)−76x67−76y69−767k67
k∈Z−16k61k= 0 k= 1 k=−1
−761−8k69⇔ −86−8k68⇔1>k>−1k∈ {−1,0,1}k=−1
(x , y) = (−7,9) k= 0 (x , y) = (0 ,1) k= 1 (x , y) = (7 ; −7)
(AB)∩F={(−7,9) ; (0 ; 1) ; (7 ; −7)}
P GCD(165 ; 98) 165 = 98 + 67 98 = 67 + 31
67 = 31 ×2 + 5 31 = 5 ×6 + 1 5 = 5 ×1+0 P GCD(165 ,98) = 1
P GCD(165 ; 98) = 1 (E) : 165x+ 98y= 1
(E)
1 = 31 −5×6 = 31 −(67 −31 ×2) ×6 = 31 ×13 −6×67 = (98 −67) ×13 −6×67
= 98 ×13 −19 ×67 = 98 ×13 −(165 −98) ×19 = 98 ×32 −19 ×165
(x0;y0) = (−19 ; 32) (E)
(x , y) (E) 165x+ 98y= 165x0+ 98y0
165x−165x0= 98y0−98y165(x−x0) = 98(y0−y)y0−y∈Z
165(x−x0)P GCD(165 ,98) = 1 x−x0
k∈Zx−x0= 98k x =−19 + 98k
165 ×98k= 98(y0−y) 165k=y0−y y = 32 −165k
(x , y) 165x+ 98y= 1 k∈Z(x , y) = (−19 + 98k , 32 −165k)
(E)k∈Z
165x+ 98y= 165(−19 + 98k) + 98(32 −165k) = 165x0+ 98y0= 1
(E)S={(−19 + 98k , 32 −165k) ; k∈Z}
N M
x
x= 17N+ 4
x= 11M+ 9 17N+ 4 = 11M+ 9
17N−11M= 5 (E) : 17N−11M= 5 (N;M)
17 = 11+6 11 = 6+5 5 = 11−6 = 11−(17−11) = 11×2−17
(N0;M0) = (−1 ; −2) (E)
(N;M) (E) 17N−11M= 17N0−11M0
17(N−N0) = 11(M−M0)M−M0∈Z17(N−N0)P GCD(11 ; 17) = 1

N−N0
k∈ZN−N0= 11k N =−1 + 11k17(N−N0) = 11(M−M0)
17 ×11k= 11(M−M0) 17k=M−M0M=−2 + 17k x = 17N+ 4
x= 17(−1 + 11k) + 4 = −13 + 187k k ∈Z06x6200 0 6−13 + 187k6200
13 6187k6213 0 <13
187 6k6213 187 <2k k = 1
x= 187 −13 = 174
k= 1 N=−1 + 11 = 10 M=−2 + 17 = 15
15 ×11 = 165