- 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 - 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).Jarek2014-02-02 <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 - Jaroslaw 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 - 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 . Masarwehمن جهاز الـ iPad الخاص بيFor 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).Jarek2014-02-02 <david.broadhurst@...>:

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 Thamer Masarweh spammed:

It is now some time since he and his brother, Ragheb,> I can find any prime number by TMSM theory

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)