If p is an odd prme and (t,p) = 1, then $t^2$ is not a primitive root $\pmod p$ If p is an odd prime and (t,p) = 1, then $t^2$ is not a primitive root $\pmod p$.
proof:  If g is a primitive root $\pmod p$ and if t is an integer such that (t,p) = 1, then there exist an integer k such that $0 \le k \le \varphi(p) -1$ and $t \equiv g^k\pmod p$.
Then suppose $t^2$ is a primitive root, then $g = t^2$. S.
Maybe if p is an odd prime, then g is a primitive root $\pmod p$, then $g^{\frac{p-1}{2}} \equiv -1 \pmod p$?
I don't know if this helps.
I don't really understand. Can someone please help me start? Thank you.

A start: We have by Fermat's Theorem that $t^{p-1}\equiv 1\pmod{p}$.
It follows that $(t^2)^{(p-1)/2}\equiv 1\pmod{p}$.
