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25466Re: [PrimeNumbers] GFN3 conjecture

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  • djbroadhurst
    Feb 2, 2014
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      Jaroslaw Wroblewski wrote:

      > 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).

      That's great, Jarek! 

      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


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