How to show no other elements besides $\pm 1$ will be in the kernel of $h: \mathbb Z_p^* \rightarrow \mathbb Z_p^*$; $h(\bar{a}) = \bar{a}^2$.

Question

Let $p$ be a prime and let $h: \mathbb Z_p^* \rightarrow \mathbb Z_p^*$ be defined by $h(\overline{a}) = \overline{a}^2$.

Since $h(\overline{xy}) = \overline{xy}^2 = \overline{x}^2 \overline{y}^2 = h(\overline{x}) h(\overline{y})$, $h$ is a homomorphism. Now, we can see that the kernel of $h$ will include $\pm \overline{1}$.

Now, I am having trouble trying to formally show that no other elements will be in the kernal. I am trying to show that if $a$ is any other element besides $\pm \overline{1}$, then $a \not\equiv 1 \pmod{p}$. We can see this through examples but I'm not sure how to show in general.

Answer 1

$\mathbf Z_p$ is a field, and in a field, a polynomial of degree $d$ has at most $d$ roots.

Answer 2

If $a^2\equiv 1$ then $p|a^2-1$. So by Euclid $p|a+1$ or $p|a-1$. So $a \equiv \pm1\bmod p$

Answer 3

$\mathbb Z_p^*$ is isomorphic to the cyclic group on $p-1$ elements. Since $p-1$ is even and there are clearly at most two solutions to $$x+x\equiv 0\pmod{2n}$$ it follows that there are at most two solutions to $$x\cdot x\equiv 1\pmod p$$