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Forum: Viliam Furik 2021-03-23, 12:39
Replies: 14
Views: 2,136
Posted By Dr Sardonicus
The fact that S(P-2) mod p doesn't tell you...

The fact that S(P-2) mod p doesn't tell you anything about whether 2^P - 1 may be prime when P = 2^p - 1 is prime, disproves the existence of a general "S(p-2) mod p" test for Mersenne primes.
...
Forum: Viliam Furik 2021-03-22, 15:24
Replies: 14
Views: 2,136
Posted By Dr Sardonicus
I mentioned the pattern merely because it gives...

I mentioned the pattern merely because it gives examples of primes P for which S(P-2) mod P is constant (2), but S(P-2) (mod 2^P - 1) is not.

You determined that S(P-2) == 2 (mod P) for P = 3 =...
Forum: Viliam Furik 2021-03-22, 13:04
Replies: 14
Views: 2,136
Posted By Dr Sardonicus
I did notice one pattern: if p > 2 and P = 2^p -...

I did notice one pattern: if p > 2 and P = 2^p - 1 is prime, then m = P - kronecker(3, P) = P + 1 = 2^p.

Now P - 2 = 2^p - 3 > p for p > 2, so Mod(2, m)^(P-2) = 0. So in this case, S(P-2) == 2...
Forum: Viliam Furik 2021-03-22, 01:17
Replies: 14
Views: 2,136
Posted By Dr Sardonicus
For the purpose of calculating remainders, the...

For the purpose of calculating remainders, the recurrence (mod p) is fine. Integer modulo arithmetic is certainly faster than polynomial modulo arithmetic.

The lift function takes a specific...
Forum: Viliam Furik 2021-03-21, 23:32
Replies: 14
Views: 2,136
Posted By Dr Sardonicus
If all you want is the remainder mod p, you can...

If all you want is the remainder mod p, you can speed things up a bit. The following Pari-GP code exploits the fact that S(p-2) = trace(u^(2^(p-2))) where u = Mod(x+2, x^2 - 3) is a unit or norm...
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