20150527, 20:14  #1 
"Daniel Jackson"
May 2011
14285714285714285714
2·5^{2}·13 Posts 
How To Calculate SNFS Poly?
How do I calculate a SNFS poly? Is there an easy formula based on N?
Last fiddled with by Stargate38 on 20150527 at 20:17 
20150527, 20:21  #2 
"Curtis"
Feb 2005
Riverside, CA
4781_{10} Posts 
Give an example of the form of the number you wish to factor. There are many cases.

20150527, 20:35  #3 
"Daniel Jackson"
May 2011
14285714285714285714
2·5^{2}·13 Posts 
(n^{2p}+1)/(n^{2}+1) with prime p
n^{2[SUP]x}[/SUP]+1 a^{n}±b^{n} (take out algebraic factors first) i.e. 4607450010708798^{8}+1 (known factor before I factored it: 449). I wasted 23 CPUdays doing GNFS on it, before I realized that I could have done it in a matter of hours with SNFS. Last fiddled with by Stargate38 on 20150527 at 20:42 
20150527, 21:30  #4 
"William"
May 2003
New Haven
3×787 Posts 
Have you read the Mersenne Wiki article on polynomial selection?

20150527, 21:47  #5 
Nov 2003
2^{2}×5×373 Posts 

20150527, 21:52  #6  
Nov 2003
2^{2}·5·373 Posts 
Quote:
For your first question (n^2p + 1)/(n^2+1) may I suggest performing the division and looking at the result? One gets a dense polynomial of degree 2p2........ n^(2^x) + 1 is algebraically prime. for a^n +/ b^n, start by factoring it, completely. Or are you asking about the primitive cofactor? Last fiddled with by R.D. Silverman on 20150527 at 21:54 

20150527, 22:12  #7 
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
Yafu can do automatic snfs poly selection for the example as well as the second two forms (though I don't know if it can take advantage of the power of 2 exponent, or if that's even possible).
Last fiddled with by Dubslow on 20150527 at 22:13 
20150527, 23:09  #8  
Nov 2003
16444_{8} Posts 
Quote:
The OP asked HOW. Suggesting to use a black box does not answer that question. Indeed. Your statement "I don't know........possible." suggests that you should not have even tried to answer this question. (Hint: Once you have decided the degree for the SNFS algebraic polynomial for 2^2^n+1, it is trivial first year algebra to find the representation). 

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