- Feb 2, 2014Jaroslaw Wroblewski wrote:

That's great, Jarek!> For prime p=1 (mod 12) we know 3 to be a quadratic residue mod p.> For prime p=3k^3+1 number 3 is a cubic residue mod p, because

> 3=(-1/k)^3 (mod p).> Since k is even, we have also p=1 (mod 12)

> and 3 is a quadratic residue mod p.> Hence 3 is a 6th power residue, i.e 3^((p-1)/6) = 1 (mod p).

Just want I wanted and was not able to prove by myself.

Dziękuję bardzo!

David

2014-02-02 <david.broadhurst@...>:Conjecture: If N = 3*8^n+1 is prime, then 3^((N-1)/6) = 1 mod N.

Comment: We know of 35 primes of this form, namely those with

n = 2,4,6,10,12,22,63,67,92,136,

146,178,736,1056,1063,1304,11450,14098,14895,16050,

18264,18394,19991,26730,71107,101031,127483,236656,267326,305591,

610832,763870,1694102,2344547,3609782

and each has 3^((N-1)/6) = 1 mod N, which is equivalent

to saying that N is a divisor of GF(m,3) = 3^(2^m) + 1

for some m < 3*n - 1.

David - << Previous post in topic Next post in topic >>