Test de primalité de Lucas, nombres de Mersenne.

N
N
N
N
N
Z/nZn
Fpp
N2N−1N
1N
1N
a2aN−11
N
N
aN−1≡1N a N
N a
N N −1=2eq q
aq≡1N
i0≤i≤e−1a2iq≡ −1N
N=
3277 a= 2 a∈ {1,...N −1}N
a

N
a a N
N
1/4a
N N =Qppfp
A a
Y
p
#{x:qx ≡0 (p−1)pfp−1}
+
e−1
X
i=0 Y
p
#{x:q2ix≡(p−1)pfp−1/2 (p−1)pfp−1}.
A=1 + 2ωE −1
2ω−1Y
p
(q, p −1),
ω N E =pep
epp ep=v2(p−1) P
A/(N−1) ω= 1 P= 1 fp= 1 P≤
(p−1)/(pfp−1) ≤1/(p+ 1) ≤1/4fp>1ω > 1
(q, p −1) ≤(p−1)/2epQp(p−1) ≤N−1
P≤1 + 2ωE −1
2ω−1/2ωE.
ω= 2
P≤1/2
p−1N−1p
N
N
N
N+ 1 (Un)
U0= 0 U1= 1 Un+1 =SUn−P Un−1,
n≥2S P D
Q(x) = x2−Sx+P p

DP (Un)p r r0
QFp[Un]p= (rn−r0n)(r−r0)−1
(D
p) = 1 rp−1=r0p−1= 1 Up−1≡0p(D p) = −1
rp=r0Up+1 ≡0p e
Ue≡0p Un≡0p e n
(Un)
(N, DP )=1
(U(N+1)/p, N)=1 N+ 1
UN+1 ≡0N.
N
d
N e
Ue= 0 N+ 1|e|d+ 1 N=d
N2n−1N+ 1
N N
n Mn= 2n−1n
M31
p q p
Mqp≡1q p ≡ ±1 8
(Vn)
Vn+1 =SVn−P Vn−1n≥1V0= 2 V1=S U2n=UnVn
N UN+1 =U(N+1)/2V(N+1)/2
V(N+1)/2≡0N V(N+1)/2
V2n=V2
n−2Pnvn=V2n
vn=v2
n−1−2P2n−1P= 1
−1a a0QCaa0= 1
−1
(D
N) = −1D3
a= 2 + √3Q Q(x) = x2−4x+ 1
P= 1 D
N=−1UN+1
2≡
0N
a a0p2 + √3 =
√3+1
√2a√3−1 √2a0x2−x√6 + 1

Z
q
N(3
q) = −1 2
q α 6Fq2r
r0FqX2−αX + 1
rq=−r0r0q=−r Uq+1 ≡0q
(V)V0= 2 V1=√6V2= 4
vk=V2k+1 k≥0
n Mn
vn−2≡0Mnvk+1 =v2
k−2k≥0v0= 4
2n−1MnMn
6
(p, 2p+ 1) p≡3 4
k > 1p= 4k+ 3 2p+ 1
2p≡1 2p+ 1