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Let's attempt to find a primitive root modulo, say, $p=127$. Since $p$ is prime a primitive root exists (more specifically there are $\varphi (\varphi (127))=36$ primitive roots modulo $127$). $\varphi (127) = 2\cdot 3^2\cdot 7$
We know that for $\alpha$ to be a primitive root modulo $127$, it is necessary and sufficient that: $$\begin{cases}\alpha^{126/2}\not\equiv 1\pmod{127}\\ \alpha^{126/3}\not\equiv 1\pmod{127}\\ \alpha^{126/7}\not\equiv 1\pmod{127}\end{cases} $$ in other words $\varphi (127)$ is the least for which $\alpha^{\varphi (127)}\equiv 1\pmod{127}$.
My question is whether there is a further method of eliminating potential candidates? It is very tedious to try them out one by one.
Can we say for certain which $\alpha$ definitely cannot be primitive roots modulo $p$ or any $n$ for that matter, aside from $1$, that is.

EDIT: The linked topic gives a thorough explanation on testing for if a number is a primitive root modulo $p$. My question pertains to, rather, omitting numbers that cannot be primitive roots modulo $p$. It isn't entirely clear from said topic if there is a way or not.

AlvinL
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    $3$ is a generator of $\mathbb{F}_{2^q -1}$ whenever $2^q-1$ is a prime Mersenne number – reuns Apr 16 '16 at 18:20
  • @user1952009 interesting result – AlvinL Apr 16 '16 at 18:26
  • The prime $p=127$ is such that $p-1$ has a rather complicated factorization. Let's instead make it as simple as possible: choose a prime $p$ such that $p-1 = 2q$ for some prime $q$. Then to check that a number is a primitive root modulo $p$, you need only check that it is a non-square modulo $p$. This doesn't generalize well, but is a nice special result related to your line of thinking. – Barry Smith Apr 16 '16 at 23:56

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