20070722, 10:22  #1 
"Michael"
Aug 2006
Usually at home
2^{4}·5 Posts 
Integer Factorization 2
By Fermat's factorization method we have
X^{2}  Y^{2} = pq = n. Y = (p  q) / 2 Consider the fraction p/q. Since gcd(p, q) = 1 integers x and y exist such that, py  qx = 1. If we can find x and y such that x/y ~ p/q we have, by the theory of linear congruences, (py  qx)/2 = e, a small number. We can arbitrairly choose e and find x and y. So, instead of factorizing pq we can more easily factor pqxy. Example; pq = 2003 x 4001. y = 999. but (761 x 2003)(381 x 4001) = 1524383 x 1524381 in this case y = (1524383  1524381) / 2 = 1, which is easily found by trial and error. so a multiple of n can be easier to factor. The problem is obvious, we don't know x and y. But we can produce P = p_{1}p_{2}...p_{k} where these p's are odd primes. Let x_{j}y_{j} = P. There will be many such factorizations but one will more closely approximate p/q than the others. Call this pair x_{i} and y_{i} such that x_{i}/y_{i} ~ p/q. So, we try to factor Ppq instead. We now have Y = (qx_{i}  py_{i}) / 2. The larger P is the more choices of x_{i} and y_{i} we have. We can force the issue if we make P immensly large in which case we are guarenteed to find a suitable x_{i} and y_{i}. The only objection to this is that we must work with immense integers, but we are guarenteed of finding Y very quickly. Any comments? 
20070722, 13:23  #2  
Nov 2003
2^{2}·5·373 Posts 
Quote:
This is well known. Look up Lehman's Algorithm. 

20070722, 17:27  #3 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
3^{3}×389 Posts 

20070722, 20:07  #4  
Jul 2007
17_{10} Posts 
Quote:
Lehman optimized the idea as a "booster" to Fermat. His algorithm starts a Fermat factorization, but then after enough (failed) tests, it adds a multiplier to keep the effective rate as fast as possible. This reduces the factoring complexity of factoring from Fermat's O(n^(1/2)) to O(n^(1/3)). See R. Lehman, "Factoring Large Integers", Mathematics of Computation, 28:637646, 1974. 

20070723, 08:07  #5 
"Michael"
Aug 2006
Usually at home
2^{4}·5 Posts 
Thanks for your lucid reply fenderbender. My question was which devil is easiest to contend with; Trial and Error or Magnitude? I did not know this had been explored before. Perhaps it could be developed?
Last fiddled with by mgb on 20070723 at 08:08 
20070723, 12:55  #6  
Nov 2003
2^{2}·5·373 Posts 
Quote:
It has *already* been developed. It is a purely exponential time method and is not competitive with either ECM or indexcalculus methods. 

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