CIS 5371 Cryptography
6*. An Introduction to Number Theory
1
Congruence and Residue classes

Arithmetic modulo n, Zn

Solving linear equations

The Chinese Remainder Theorem

Euler’s phi function

The theorems of Fermat and Euler

Quadratic residues

Legendre & Jacobi symbols
2

3
Solving linear equations

4
The Chinese Remainder Theorem
Let m1 ,  , mn be positive integers with gcd( mi , m j )  1,
and let M  m1m 2  mr .
Then the congruence
x  c1 (mod m1 )
x  c2 (mod m2 )
 

x  cr (mod mr )
has a unique solution z  Z M .
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The Chinese Remainder Theorem
Let ( M / mi ). yi  1 (mod mi ). Then for
zi  ( M / mi ). yi
it is easy to see that
z  i 1 zi ci (mod M )
r
is a solution.
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Example
Solve the modular conguence :
x  2 (mod 3)
x  3 (mod 5)
x  4 (mod 7)
We have : M  105,
y1  35 1 (mod 3)  2 (mod 3)
y 2  211 (mod 5)  1 (mod 5)
y3  15 1 (mod 7)  1 (mod 7)
Then z  2  35  2  3  21  1  4  15  1  263 mod 105  53 mod 105
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Example
Solve the modular conguence :
x  1 (mod 2)
x  2 (mod 5)
x  3 (mod 7)
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Euler’s phi function
 (n) is the number of positive integers k for which
gcd( k,n)  1. We have
 (1)  1
 ( p)  p  1, p a prime,
 ( pq)  ( p-1)( q-1), for p,q primes,
( p )  p  p
e
e
e 1
1
 p (1  ), p a prime, e  1.
p
e
It follows that if n  p1 1  p1 2    p1 1 , is the prime
e
e
e
factorizat ion of n, then
1
1
1
 (n)  n (1  )  (1  )    (1  ).
p1
p2
pk
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The theorems of Fermat and Euler

10
Legendre & Jacobi symbols
Let QR p be the set of quadratic residues mod p and let
QNR p  Z p \ QR p be the set of quadratic nonresidue s.
For x  Z p , p prime, the Legendre symbol of x mod p is :
 1 if x  QR p
x
   
 p
 1 if x  QNR p
Let n  p1 p2  pk be the prime factorizat ion of n.
For x  Z n , the Jacobi symbol of n mod p is :
 x
 
n
 x  x   x 

  
   
 p1   p2   pk 
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Legendre & Jacobi symbols
It is easy to see that
x
   x ( p1) / 2 (mod p ), for p prime.
 p
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Legendre & Jacobi symbols
We have :
1
 xy   x   y 
   1, and 
    
n
n
 

 n  n
 x   x   x

    
mn

  m  n
 x  y
If x  y (mod n) then     
 n  n
For m, n odd :
  1
(n -1)/2

  (-1)
 n 
 2
(n 2 -1)/8
   (-1)
 n
 m n 
If gcd( m,n)  1 and m, n  2 then      (-1)(m-1)(n -1)/4
 n  m
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Example
 4 
 7 
Compute 
 and 

 15 
 15 
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