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legender-congruence.md

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Legender's Congruence

  • Every Subsequent method for integer factorization is based on Simpler Observation based on Legender's Congruence introduced By Adrien-Marle legender (1752 - 1833)

proof :

  • if we want factorization number n which is composed of factor p,q ϵ P ( set of all prime number ) There exist congruence of form ff mod n and y ≠ x mod n Where x,y ϵ ( 2 , 3 , ..... n - 1 )

  • This Congruence can Be Written as f ≡ ( X - Y ) ( X + Y ) ≡ 0 mod n which lead us to that p | (x - y) (x + y) and q | (x-y)(x+y) because of condition and x ≠ y mod n .

  • There are only two option for each Condition .

  • First Condition consider if p | (x - y)(x + y) is chosen .

  • Then We have that p | (x - y) and p * x (x + y) or p * x (x - y) and p | (x + y)

  • but if p | (x - y) and p | (x + y)

  • ∴ n | (x - y) or n | (x + y)

  • ∴ (x - y = k1 * n) or (x + y = k2 * n)

  • ∴ x = y + k1 * n or x = k2 * n - y

  • ∴ x ≡ y mod n or x ≡ -y mod n which is contradict with our condition

  • ∴ p | (x - y) and px (x + y) or px (x - y) and p | (x + y)

  • at the first case if p | (x - y) and px (x + y) we must have that qx (x - y) if q | (x - y)

  • we will have n | (x - y) → x ≡ y mod n ( contradiction )

  • ∴ q*x (x - y) but q | (x + y)(x - y) , q ϵ P

  • ∴ q | (x - y) ∴ gcd( x - y , n) = p , gcd( x + y , n) = q

  • at the second case if p*x (x - y) and p | (x + y)

  • we must have that q*x (x + y) or if q | (x + y) → n | (x + y) → x ≡ -y mod n (contradiction)

  • ∴ qx (x - y) but q | (x + y)(x - y) , q ϵ P

  • ∴ q | (x - y) ∴ gcd(x-y , n) = p , gcd(x+y , n) = q

  • In order word if we can get x and y such that ff mod n and X ≠±Y mod n

  • then we can calculate easily the value of gcd( x+ y , n) By The Euclidean Algorithm Which is usable for this Purpose , so the principle of factorization based on legender's congruence is behind all modern method of factorization .

End of proof :

Example of Factoring By Legender's Congruence

  • Let N = 119 and m = 144 = f

  • f mod 119 = f mod 119 .

  • (d1,d2) = (gcd(12+5,119) , gcd(12-5,119)) = (17,7)

  • where n = 119 = 17 * 7