20180724, 21:27  #254 
"Mark"
Apr 2003
Between here and the
16E2_{16} Posts 

20180725, 01:36  #255 
"Gary"
Aug 2015
Texas
2^{4}·3 Posts 

20180725, 06:44  #256 
Banned
"Luigi"
Aug 2002
Team Italia
1001010011011_{2} Posts 

20180725, 16:24  #257 
"Mark"
Apr 2003
Between here and the
2×29×101 Posts 

20180726, 22:50  #258 
Sep 2003
2,579 Posts 
Is there some fast way to verify very big Fermat factors?
For all Mersennes in our typical ranges, modular exponentiation verifies a factor almost instantly. The entire database of millions of known factors can be verified in seconds. For Fermat factors, though, modular exponentiation slows down drastically when n gets into the tens of thousands. Let alone huge ones like n, m, k = 3329780,3329782,193. There must be some other way. 
20180727, 00:33  #259  
"Gary"
Aug 2015
Texas
2^{4}·3 Posts 
Quote:
I often verify all known factors for n < 1,000,000 whenever I make changes to pmfs or get access to a new system, to make sure it's working correctly. It take about 5 hours, with about 47 minutes being required for n = 960901 alone. 

20180807, 16:11  #260  
Sep 2003
5023_{8} Posts 
Quote:
The command was ./factorverify fermat vv FERMAT.txt and the files are attached. 

20180807, 23:29  #261 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{7}×71 Posts 
Try on the same platform and compare times:
1. download pfgw 2. put all factors in a file 3. run pfgw N k l gos2 file 
20180808, 01:01  #262  
Sep 2003
5023_{8} Posts 
Quote:
I just wanted to verify that the known factors of Fermat numbers really are factors, for instance, to check that 193 * 2^3329782 + 1 really does divide 2^(2^3329780) + 1 I used GMP even if it's slower than more specialized software, because it's a very widely used and thoroughly tested library, a precompiled version is a standard component of most Linux distributions, and it's possible to formulate the test in a few simple lines of code, and therefore have extremely high confidence in the result. Maybe you could use pfgw to verify that the known Fermat factors really are prime factors and not composite factors? But maybe not for the goal I wanted. Or am I misunderstanding? 

20180808, 04:40  #263 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
9088_{10} Posts 
Yes

20180808, 05:55  #264 
"Robert Gerbicz"
Oct 2005
Hungary
2545_{8} Posts 
Just interestingly, if m=k*2^(n+2)+1  F_n (where k can be even) and 0<k<2^(n+2) then m is prime! And this holds for the known factors in the list (surely) for say n>100.

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