If p is an odd prme and (t,p) = 1, then $t^2$ is not a primitive root $\pmod p$

Question

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.

Answer 1

A start: We have by Fermat's Theorem that $t^{p-1}\equiv 1\pmod{p}$.

It follows that $(t^2)^{(p-1)/2}\equiv 1\pmod{p}$.

Answer 2

Consider the group homomorphism $q\colon (\mathbb Z/p\mathbb Z)^\times\to (\mathbb Z/p\mathbb Z)^\times$, $x\mapsto x^2$. Since $-1\not\equiv 1\pmod p$ and $q(-1)=g(1)$, it has nontrivial kernel, i.e. its image is a proper subgroup, hence cannot contain any generator of the group.