 mersenneforum.org > Math Kraitchik's factorisation method
 Register FAQ Search Today's Posts Mark Forums Read 2006-02-06, 12:39 #1 Robertcop   Feb 2006 3 Posts Kraitchik's factorisation method Hi, I'm a new member here and I have a question concerning Kraitchik's factorisation method. Let x^2 = y^2 (mod n). It follows that n|(x-y)(x+y) and hopefully gcd((x-y),n) yields to a non-trival factor of n. Does the additional prerequisite x =! y (mod n) implies that gcd((x-y),n) always yields to an non-trivial factor and n does not divide (x-y) nor (x+y), respectively?   2006-02-06, 13:37   #2
jasonp
Tribal Bullet

Oct 2004

67318 Posts Quote:
 Originally Posted by Robertcop Hi, I'm a new member here and I have a question concerning Kraitchik's factorisation method. Let x^2 = y^2 (mod n). It follows that n|(x-y)(x+y) and hopefully gcd((x-y),n) yields to a non-trival factor of n. Does the additional prerequisite x =! y (mod n) implies that gcd((x-y),n) always yields to an non-trivial factor and n does not divide (x-y) nor (x+y), respectively?
See p. 48 of Per Leslie Jensen's thesis 'Integer Factorization', where he reproduces a table I vaguely remember seeing elsewhere. For n the product of two prime factors, there are 8 combinations possible and 6 of them yield a nontrivial factor.

jasonp   2006-02-06, 21:03   #3
akruppa

"Nancy"
Aug 2002
Alexandria

2,467 Posts Quote:
 Originally Posted by Robertcop Does the additional prerequisite x =! y (mod n) implies that gcd((x-y),n) always yields to an non-trivial factor and n does not divide (x-y) nor (x+y), respectively?
Not by itself. If x² ≡ y² (mod n), then x ≡ ±y (mod p) for all p|n. We get the trivial factorisation iff for all such p either (1) xy (mod p) or (2) x ≡ -y (mod p). Your condition rules out case (1), you still need to rule out case (2). Thus:

Iff x !≡ ±y (mod n), the gcd will find a non-trivial factor.

As Jason points out, the likelyhood of that depends on the number of prime factors in n.

Alex

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