TD2 - IECL - Université de Lorraine

◦
a b c
a=bq +r06r < b P GCD(a, b) = P GCD(b, r) = P GCD(b, b −r)
P GCD(89,55)
(Fn)n∈N∗F1=F2= 1 n>3
Fn+2 =Fn+1 +Fn
1. k ∈N∗a > b >0a b k
a>Fk+2 b>Fk+1
2.∀k∈N∗∀a, b ∈Nb < Fk+1 a b
k
3. k ∈N∗Fk+2 Fk+1 k
4. Fn+1/Fnϕ
(1 + √5)/2Fn∼ϕn/√5
a b O(ln b)
n
0
m, n ∈N
P GCD(2n−1,2m−1) = 2P GCD(m,n)−1.
(d, x, y) 899 493
Fk+1, Fk

x y y
x5
x y
a1, . . . , ann
P GCD(a1, . . . , an)u1, . . . , un
P GCD(a1, . . . , an) =
n
X
i=1
uiai.
[a1, . . . , an]
[u1, . . . , un]Pn
i=1 uiai=P GCD(a1, . . . , an)
n n >2
n6
6=1+2+3
◦1 2p−1p
n>2σ(n)n n
σ(n)=2n
1. a)n σ(n) = n+ 1
b)r∈N∗σ(2r)=2r+1 −1
c)m n σ(mn) = σ(m)σ(n)
2. q = 2p−1
n=
q
X
j=1
i=q(q+ 1)
2= 2p−1(2p−1)
n
1. k >2n= 2k−1m m
2. m (2k−1) m= (2k−1)d
3. d +m=σ(m)
4. d >2 1 + d+m6σ(m)d= 1
5.2p−1(2p−1) (2p−1)
IIIeme
9XV IIIeme