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Old 2020-01-27, 21:41   #79
ewmayer
2ω=0
 
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Sep 2002
República de California

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Quote:
Originally Posted by R. Gerbicz View Post
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 very-very strong check, but the two known factors could find a mismatch with probability = 1-(1e-10) which is still quite impressive. Note that we don't get the strength of 1/(q-1) 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 Euler-pseudoprime 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 Pépin-test 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^m-1) = 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 64M-FFT run update from Ryan), and said runs being done on high-end server ssytems with ECC memory.

For F31 I will also be using the now-famous periodic whole-residue check named in honor of the above-quoted forum member. :)

Last fiddled with by ewmayer on 2020-01-27 at 21:43
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