If $p$ is an odd prime and $\alpha\in\Bbb Z/p\Bbb Z^*$, then $\alpha^2$ is not a primitive root modulo $p$.

Question

Prove true or give a counterexample if false.

If $p$ is an odd prime and $\alpha\in\Bbb Z/p\Bbb Z^*$, then $\alpha^2$ is not a primitive root modulo $p$.

I was trying to prove it to be true, but I am not sure where to start. I was thinking of using Fermat's Little Theorem: if $p$ is a prime and $\alpha\in\Bbb Z/p\Bbb Z^*$, then $\alpha^{(p-1)}=1$ but how does one make the jump from FLT to primitive roots? A primitive root is defined as an element $\gamma=\phi(m)$ but how does that tie into this problem?

Answer 1

$(a^2)^{\frac{p-1}{2}}=a^{p-1}=1 \pmod{p}$.The last step follows from FLT.

Hence, the order of $a^2$ mod $p$ is at most $\frac{p-1}{2}$, so it can't be a primitive root by definition.