Devoir Maison 1

Z/pZ
p
φ: (Z/pZ)×−→ (Z/pZ)×
x7−→ x2
ker φ={±1}
Im φ(Z/pZ)×(p−1)/2
y∈Im φ x2=y
y
y(Z/pZ)×yp−1
2= 1
y∈(Z/pZ)×
g(Z/pZ)×gp−1
2y g
yp−1
2= 1 y(Z/pZ)×
p≡3 (mod 4) y(Z/pZ)×
y y p+1
4−yp+1 4
Z/nZ
n=pq p q
(Z/nZ)×
n, p, q y ∈ {0,1, . . . , n −1}
n=pq y ∈ {0,1, . . . , n −1}
n
y∈(Z/nZ)×x2=y
n, p, q y ∈ {0,1, . . . , n −1}
n=pq y n
n=pq p q
α∈(Z/nZ)×n p q
m∈ {0,1}
(n, α)αmx2x(Z/nZ)×

m p m = 0
l
α
αp(Z/pZ)×αq(Z/qZ)×
α
n∈N∗(1 + n)n2
(1 + n)m(mod n2) 0 ≤m≤n2−1
(1 + n) (Z/n2Z)×
G(Z/n2Z)×(1 + n)
G
(1 + n)
n=p×q p q
(Z/n2Z)×n ϕ(n)
n ϕ(n)
p q
s: (Z/nZ)×−→ Z/n2Z×
rmod n−→ rnmod n2
a≡b(mod n)s(a)≡s(b) (mod n2)
s
sker s={1}s(x)≡1
(mod n2)s(x) (mod n)n
ϕ(n)

n=pq
n ϕ(n)
Z/nZ×(Z/nZ)×(a, b)(a0, b0) = (a+a0
mod n, bb0mod n)
E:Z/nZ×(Z/nZ)×−→ Z/n2Z×
(m, r)−→ (1 + n)mrn
E E
s
EsE
n
m∈Z/nZr∈(Z/nZ)×
c=E(m, r)
n= 35
p q m = 3 n= 35 r= 8
c m r cϕ(n)(mod n2)
n= 35
A
1A
A0
vAvBvcA
AvAvBvC

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