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.