Quote:
Originally Posted by R. Gerbicz
Have you checked that:
(3^(2^(2^m)) mod F_n) mod q = 3^(2^(2^m)) mod q
is true for the stored interim residues?
Since you have 3^(2^(2^m)) mod F_n then to get the left side is trivial and you can get quickly the right side, becuase q is small , the check is cheap. Though it is not that veryvery strong check, but the two known factors could find a mismatch with probability = 1(1e10) which is still quite impressive. Note that we don't get the strength of 1/(q1) or 1/znorder(Mod(3,q)) because the order is divisible by a "large" power of two, and that will be lost in the iterated squarings of a wrong residue.

ITYM (3^(2^(2^m  1)) mod Fm) mod q = 3^(2^(2^m  1)) mod q, since the Pépin test is just an Eulerpseudoprime test (which happens to prove rigorous in the special case of the Fm), i.e. computes 3^(2^((Fm  1)/2)) mod Fm = 3^(2^((2^m)/2)) mod Fm. But in any event, that's a good suggestion  fiddled my code to load the final Pépintest residue R from the savefile and for the 2 known small prime factors p=640126220763137 and q=1095981164658689 computed
R == 367500396407933 (mod p) and
R == 95971534699540 (mod q).
I then separately computed 3^(2^(2^m  1)) mod p and q via (2^m1) = 1073741823 iterated squarings modulo p and q, separately, and the results match, which anyone can confirm on even quite modest hardware. So we have quite good confidence in the result, on top of the 2 runs agreeing through ~740 Miter (as of the latest 64MFFT run update from Ryan), and said runs being done on highend server ssytems with ECC memory.
For F31 I will also be using the nowfamous periodic wholeresidue check named in honor of the abovequoted forum member. :)