Quelques problèmes classiques d`arithmétique II Parimaths

A(a1, a2)B(b1, b2)C
(a1+b1, a2+b2)OACB
|a1b2−a2b1|
V V
R2O
A(a1, a2)B(b1, b2)V A +B
(a1+b1, a2+b2)−A(−a1,−a2)s sA
(sa1, sa2)O A B
(O, A, B)
V sA +tB s t
V A B C
m n C =mA +nB R(A, B)
R(A, B)C s
t[0,1[ C=sA +tB R(A, B)
A(1,0) B(0,1)
A(2,−1) B(3,1)
R(A, B) (x, y)x y x ≡3y
mod 5

m n A B
(x, y)x≡my mod n
A B R(A, B)
A1(1,0) B1(0,1)
A2(1,1) B2(0,1)
A3(1,1) B3(1,2)
A4(2013,−2012) B4(−2014,2013)
X V A(X)R=R(A, B)V
(R)
A(X)X
R
A(X)>(R)
C D X C + (−D)∈R
X
X
XA(X)>4 (R)X
R
A(X)>4 (R)
R(A, B) = R(C, D)R(A, B)R(C, D)
R(A, B)
A B
R
R
R C
OC 2q(R)
π
R D
Dp(R)
R E(e1, e2)
|e1|+|e2| ≤ p2 (R)

a m
a m x x2≡amod m
m=pα1
1...pαr
r
m a m
a pαi
ii
a pii
m
m p
a
p=
0p|a
1p-a a p
−1p-a a p
p6= 2
•−1
p= (−1)p−1
2
•aa
p≡ap−1
2mod p
•a b a
pb p=ab p
•2
p= (−1)p2
−1
8
•q p
p
qq p= (−1)p−1
2
·
q−1
2
a b
ab

d≥2
p p2|d x y
x2−dy2= 1.
Sa+b√d a b z =a+b√d∈ S
z=a−b√d N(z) = a2−db2=zz
S → R2, z 7→ (z, z)R
RR2
t > 0r > 0R2Xt,r
t2x2+y2
t2≤r Xt,r
Xt,r
Xt,r t r
M
N(z) = M z ∈ S
(1,0) (−1,0)
(x1, y1)
•x1>0y1>0
•(x, y)x > 0y > 0x1+y1√d≤
x+y√d

z1=x1+y1√d
n xnynxn+yn√d=
(x1+y1√d)n(xn, yn)
(xn, yn) (−xn,−yn)
x2−11y2= 1
x2−11y2= 5
a b
xn=an +b
a
p p
n > 0
aij (i, j)n
f:Rn→R,(x1, ..., xn)7→ Pi,j aij xixj
f(x1, ..., xn)=0
m
m
d N d
N < 0d > 0
m x2−dy2≡Nmod m
x2−dy2=N