25466Re: [PrimeNumbers] GFN3 conjecture
- 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.
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,
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.
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