20110821, 22:21  #1 
"William"
May 2003
New Haven
2^{3}·5·59 Posts 
SNFS Polynomial for 919^871
If it survives ECM preparation, what polynomial would you use for 919^871? I'm thinking
x^6 + 919x^3 + 919^2, x=919^10 That results in SNFS size of 178, which is a little small for a sextic. But for a quintic I only see x^5 + 919^7 * x^2 + 919 x=919^12 and that coefficient of 919^7 looks too large. (Both derived by dividing out 919^291 and figuring an SNFS polynomial for 919^58 + 919^29 + 1 Is there something better I have missed? 
20110822, 01:40  #2  
Nov 2003
2^{2}×5×373 Posts 
Quote:


20110822, 22:57  #3 
"Daniel Jackson"
May 2011
14285714285714285714
7^{2}·13 Posts 
Here's a poly from factordb for 919^{87}1 (166 digit cofactor): http://www.factordb.com/snfs.php?id=1100000000225474717
Last fiddled with by Stargate38 on 20110822 at 22:59 
20110823, 00:00  #4  
Jul 2003
So Cal
2^{3}×257 Posts 
Quote:


20110823, 00:11  #5 
"William"
May 2003
New Haven
2360_{10} Posts 
BAD CHOICE.
This is difficulty 258 digits, MUCH MUCH harder than the 178 digits example I gave. It looks like the factordb's polynomial finder needs to learn some tricks about removing algebraic factors. William 
20110823, 02:39  #6 
"Ben"
Feb 2007
D21_{16} Posts 
Also, something is wrong with the way it picks several of the parameters. Setting mfbr/a to more than twice lpbr/a doesn't make any sense. That will result in many more vain 2LP factorizations. Also, I've never seen r/alambda set so high. I'm less sure that this is bad, but it deviates from the status quo by a couple tenths. Extrapolating from the end of a lookup table, maybe?

20110823, 04:59  #7  
Oct 2004
Austria
4662_{8} Posts 
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