20060206, 12:39  #1 
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(xy)(x+y) and hopefully gcd((xy),n) yields to a nontrival factor of n. Does the additional prerequisite x =! y (mod n) implies that gcd((xy),n) always yields to an nontrivial factor and n does not divide (xy) nor (x+y), respectively? 
20060206, 13:37  #2  
Tribal Bullet
Oct 2004
6747_{8} Posts 
Quote:
jasonp 

20060206, 21:03  #3  
"Nancy"
Aug 2002
Alexandria
100110100011_{2} Posts 
Quote:
Iff x !≡ ±y (mod n), the gcd will find a nontrivial factor. As Jason points out, the likelyhood of that depends on the number of prime factors in n. Alex Last fiddled with by akruppa on 20060206 at 21:04 

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