TD1 - IECL - Université de Lorraine

◦
1.(a, b)∈N×N∗
a b
2.(a, b)∈N×N∗
u←0
v←a
v>b
n←0
c←1
10nb6v
n←n+ 1
n←n−1
10nbc 6v
c←c+ 1
c←c−1
u←u+ 10nc
v←v−10nbc
(u, v)
v
a=bu +v
•
•a b
( )
1. m ∀x∈R
xm+ 1 = (x+ 1)(xm−1−xm−2+. . . + 1).
2. n ∈N∗2n+ 1 n2
3. n ∈NFn= 22n+ 1 {Fn}n
•Fnn∈J0; 3KFn
n>5F5= 641 ×6700417
Fnn>5
•
n
n= 2m2

( )
b>2n
n=akbk+ak−1bk−1+. . . +a1b+a0=akak−1. . . a1a0b
∀i∈J0; kKai∈J0; b−1Kaib
1. a0a1, . . . , ak
P db(b, n)n b
2. n ∈Nb
n=akak−1. . . a1a0b
P bd(b, L)L= [a0, a1, . . . , ak−1, ak]b
n
3. n = 146554 12
4. n = 163567
P=Xn+c1Xn−1+. . . +cn−1X+cn∈Z[X]P
(un)n∈Nu0= 14 un+1 = 5un−6
n
2
n
n
n∈N∗
•P GCD(n2+n, 2n+ 1)
•P GCD(15n2+ 8n+ 6,30n2+ 21n+ 13)
( )
1. a >2n>2an−1a= 2 n
2.
n= 11
3. p 4k+ 3 k∈N∗2(p−1)/2≡(−1)k+1[p]
N= 2(p−1)/2(p−1)/2!N

4. Mp= 2p−1p p 4k+ 3 2p+ 1
Mp
(Ln)L0= 4 Ln+1 =L2
n−2
n>3Mn= 2n−1Mn|Ln−2
0
x2+y2=z2(x, y, z)∈(N∗)3
•x y z
•
•x(z−y)/2 (z+y)/2