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GFN3 conjecture

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  • djbroadhurst
    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,
    Message 1 of 5 , Feb 2, 2014
      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
    • Jaroslaw Wroblewski
      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
      Message 2 of 5 , Feb 2, 2014
        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).

        Jarek


        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


      • djbroadhurst
        ... That s great, Jarek! Just want I wanted and was not able to prove by myself. Dziękuję bardzo! David 2014-02-02
        Message 3 of 5 , Feb 2, 2014
          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


        • thamermm15
          Hi every body . My theory to finding primes will be the best theory in number theory . I can find any prime number by TMSM theory . Thamer M . Masarweh ‏من
          Message 4 of 5 , Feb 3, 2014
            Hi every body . My theory to finding primes will be the best theory in number theory . 


            I can find any prime number by TMSM theory . 


            Thamer M . Masarweh 

            ‏من جهاز الـ iPad الخاص بي

            في ٠٣‏/٠٢‏/٢٠١٤، الساعة ٥:٣٦ ص، كتب Jaroslaw Wroblewski <jaroslaw.wroblewski@...>:

             

            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).

            Jarek


            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


          • djbroadhurst
            ... It is now some time since he and his brother, Ragheb, had their profiles in the Jordan Times. Until Thamer has something of substance to communicate,
            Message 5 of 5 , Feb 3, 2014

              Thamer Masarweh spammed:


              > I can find any prime number by TMSM theory

              It is now some time since he and his brother, Ragheb,
              had their profiles in the Jordan Times.

              Until Thamer has something of substance to communicate,
              regarding primality, he might refrain from posting to this list?

              David (with condolences to Jarek, whose proof was spammed)
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