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I came across this question while studying primitive roots. I know it has something to do with the fact that if the order of $a$ is $m$ then for every $k \in \mathbb{Z}$, the order of $a^k$ is $m/(m,k)$. The question is as follows:

Let $p$ be an odd prime. Prove that $a^2$ is never a primitive root $\pmod{p}$.

I would appreciate any help. Thank you.

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    Okay. So, what would $m/\gcd(m,k)$ have to be for $a^2$ to be a primitive root?2012-03-19

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