HLMA304 - Corrigé 2ème session - Juin 2015 Fichier

p x Z/pZxp=x x
xp−1= 1
(77)72
Z/2Z(77)7= (7)49 = (1)49 = 1 1
(77)75
Z/5Z(77)7= (7)49 = (2)4.12+1 = ((2)4)12.2 = 2
2
(77)710 (77)7
(x≡1 [2]
x≡2 [5] , x ∈Z.
x= 7
pgcd(2,5) = 1 {7 + k.2.5, k ∈Z}
0 9 7 7
(77)7=q.10 + 7 q7
n∈N2(n2+ 1) ±1 3
Z/3Z{0 = 02,1 = 12= 22}
n2= 0 2(n2+ 1) = 2.1 = −1n2= 1 2(n2+ 1) = 2.2 = 1
p, n ∈N∗p, p + 2(n2+ 1) p+ 4(n2+ 1)
3
Z/3Z= 2(n2+ 1) = ±1
{p, p + 2(n2+ 1), p + 4(n2+ 1)}={p, p +, p −}={0,1,2},
p, p +, p −0p, p + 2(n2+
1), p + 4(n2+ 1) 3
p= 3
p p ≥2p+ 2(n2+ 1) ≥4p+ 4(n2+ 1) ≥6
3p= 3
(x≡2 [35]
x≡ −3 [99] , x ∈Z.

35 99
99 = 2.35 + 29,35 = 1.29 + 6
29 = 4.6+5,6 = 1.5+1
1=6−5=6−(29 −4.6) = −29 + 5.6 = −29 + 5.(35 −29) = 5.35 + (−6).29
= 5.35 + (−6).(99 −2.35) = (−6).99 + (17).35.
etl0ensembledessolutionsest :
∈Z
(Fn)n∈N
F0= 0, F1= 1,∀n≥2, Fn=Fn−1+Fn−2
∀(m, n)∈N∗2,pgcd(Fm, Fn) = Fpgcd(m,n)
n≥1
Fn+1Fn−1−F2
n= (−1)n.
n
pgcd(Fn, Fn+1)=1
m, n ≥1
Fm+n=FmFn+1 +Fm−1Fn.
m
m, n ≥1d≥1Fm+nFnFm
Fnpgcd(Fm+n, Fn) = pgcd(Fm, Fn)
∀(m, n)∈N∗2,pgcd(Fm, Fn) = Fpgcd(m,n)