TD - Cryptographie par résidu quadratique

a6= 0 b x
x2≡amod b.
x a b
p q n =p.q
Z/nZ?
x2≡amod b(b−x)2≡amod b
a n a p q
a6= 0 p x y =p−x
aZ/nZa p q
x1n−x1x2n−x2
(Z/pZ?,×)p−1
2p
Z/nZ?
(b−x)2=b2−2bx +x2=x2=amod b
a=x2+kpq a ≡x2mod p p q
x y a =x2=y2mod p x2−y2= (x−y)(x+y) = 0
mod pZ/pZp x −y6= 0 mod p
x+y= 0 mod p y =p−x
amod nmod pmod q a 6= 0 mod p
a u1=u u2=p−u p v1=v v2=q−v
q
a n ui.q.q−1[p]+vj.p.p−1[q]mod n1≤i, j ≤2
x1n−x1x2n−x2
g(Z/pZ?, .)x
g=x2gp−1= 1 mod p g p
gp−1
2=−1xp−1=−16= 1 p6= 2
xZ/pZg p
g2ip−1
2
g2ix=gigi+p−1
2=−x=p−x
g(Z/pZ?, .)
(u, v)∈Z/pZ×Z/qZ Z/nZ
p+1
2p(p+1)(q+1) 4nn+p+q−3 4
n

a < n
x a 6= 0 n
n
x1x2
(n−x1) (n−x2)a n
n
u=x1−x2mod n v =x1+x2mod n u.v ≡0 mod n
1≤u, v < n p q
n u v
n
Z/nZ(x) = x2mod n
u.v =x2
1−x2 2=a2−a2= 0 mod n
1≤x1, x2< n x16=x2u=x1−x26= 0 x16=n−x2
v=x1+x2=x1−(n−x2)6= 0 1 ≤u, v < n
u.v =k.n n =p.q u.v u < p.q p q
p u q u p.q u n, u
n n/pgcd( )
O(t2)O(tlog2tlog log t)
x2mod n O(t1+)
a
mod p a

n=pq
p q
xA< n a =x2
Amod n
n a
xA
n
r
y=r2mod n z =xA.r mod n
y z
a.y −z2= 0 mod n
r
xA
z y =z2/a mod n
r y =r2mod n y
b∈ {0,1}b
z=r y n
z=xA.r mod n y.m mod n
y.ab−z2= 0 mod n
k
p
q
xA
xAamod n
xaa n
xA
p q xAxAa
p
gZ/pZ?y < p
ymod p y/g y
y1y2
1=y y1=g y
y1yk=g
i y12.i 2.i + 1 y
p

r y
r z
r z xAr
xA
y
k2−k