4
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In this paper (Proposition 4) you can find statement :

If $p$ is a prime of the form : $p = 2q + 1$ for some odd prime $q$,

then $2$ is a primitive root modulo $p$ if and only if : $q \equiv 1, 5 \pmod 8$.

Is it true that :

If $p$ is a prime of the form : $p =3\cdot q \cdot (q+1)-1$ for some odd prime $q$,

then $3$ is a primitive root modulo $p$ if and only if : $q \equiv 1, 5 \pmod {12}$.

Legendre symbol for $3$ is :

$\left(\frac{3}{p} \right) = \begin{cases} ~~~1, & \text{if } :p \equiv 1,11 \pmod {12} \\ -1, & \text{if } : p \equiv 5,7 \pmod {12} \end{cases}$

It is obvious that necessary condition is : $q \equiv 1, 5 \pmod {12}$ since :

$3 \cdot(12t+7)^2+3(12t+7)-1 \equiv 11 \pmod {12}$

$3 \cdot(12t+11)^2+3(12t+11)-1 \equiv 11 \pmod {12}$

There is a theorem that states :

$a$ is a primitive root modulo $~p~$ iff $~~ord_p(a) = \phi(p)$ .

So , how to prove that : $~~ord_p(3) =3\cdot q \cdot (q+1)-2 $ ?

EDIT :

Conjecture is false . Smallest counterexample is : $35862917$

  • 0
    @GerryMyerson,Consecutive primes of the form $3\cdot q \cdot (q+1)-1$ , where : $q \equiv 1,5 \pmod {12}$ :$89 , 2609 , 4217 , 24029 , 28517 ,35969 , 67049 , 117017 , ....$2012-02-04

0 Answers 0