td ari

◦
( )
2
1.2 (Z/11Z)×(Z/31Z)×(Z/341Z)×
341
2. n m = 2n−1
a)n|m−1
b)a, b ∈N∗a|b2a−1|2b−1
c)m
3.
4. Mp= 2p−1p Mp
5. Fn= 22n+ 1
( )
n
a)n
b)a n an−1≡1[n]
1. n =p1×p2. . .×pkp1, p2, . . . , pk
(pi−1) (n−1) i∈J1; kKn
561 10585
2. n
1
a)n2n= 2αα2
(Z/nZ)×a a(2α−1)
1n a 1n
b)n p1p1, p2, . . . , pk
n n n =Qk
i=1 pαi
i
(i)ω pα1
1((Z/pα1 1Z)×,×)
ω
t
t≡ω[pα1
1]i i ∈J2; kKt≡1[pαi
i]
tn−1≡1[n]
(ii)pα1−1
1(p1−1) (n−1) α1= 1 (p1−1)
(n−1)
(iii)n n n =p1×p2×
. . . ×pkp1, p2, . . . , pk(pi−1)
(n−1) i∈Jl;kK
3.
4.85p−16k= 1 (k, p)∈Z2
5 17
5.6k+ 1 12k+ 1 18k+ 1

a a >2
a>2p p -a(a2−1) n= (a2p−1)/(a2−1)
1. n
2. a2p≡1[n]
3.(a2−1)(n−1) p|(n−1)
4.2|(n−1)
5. n a
1.15
2.105(= 3 ×5×7) 13
2
3. n
n
7
17;28 13;10 83;665 97 
p
1. α ∈Z Z/pZx∈Z/pZ
x2≡α[p] 1 + α
p
2. a, b, c ∈Zp-a ax2+bx +c= 0 1 + ∆
p
Z/pZ∆ = b2−4ac
3.Z/83Z
a)x2+ 1 = 0
b)x2+x+ 1 = 0
c)x2−4x+ 13 = 0
d)x2+x+ 21 = 0
p
p+1
2
p!.