If $a,p\in\mathbb{N}$ with $p$ prime, have to show that if $a²\equiv1\pmod p $, then $a\equiv1\pmod p$ or $a\equiv p-1\pmod p$

Question

If $a,p\in\mathbb{N}$ with $p$ prime, have to show that if $a²\equiv1\pmod p $, then $a\equiv1\pmod p$ or $a\equiv p-1\pmod p$

I'm studying congruence, and I have no idea where to start this demonstration, if anyone can do it, preferably detailed, or go giving ways, and I doing, I will thank enough.

Answer 1

Hint: The ring $\mathbb{Z}/p\mathbb{Z}$ is a field, so the polynomial $x^2-1$ has at most $2$ solutions over $\mathbb{Z}/p\mathbb{Z}$.

Answer 2

$a^2\equiv1\bmod p$ iff $p$ divides $a^2-1=(a-1)(a+1)$ iff $p$ divides $a-1$ or $p$ divides $a+1$ iff $a\equiv \pm 1 \bmod p$. Finally, note that $-1 \equiv p-1 \bmod p$.

Answer 3

If $a^{2}\equiv 1\pmod p\Longrightarrow p\mid a^{2}-1\Longrightarrow p\mid (a+1)(a-1)$ then $p\mid (a+1)$ or $p\mid a-1$

If $p\mid (a-1)\Longrightarrow a\equiv1\pmod p$.

If $p\mid (a+1)\Longrightarrow a\equiv-1\pmod p$.

As $-1\equiv p-1\pmod p$ and $a\equiv-1\pmod m$ then $$a\equiv-1\equiv p-1\pmod m\Longrightarrow a\equiv p-1\pmod p.\;\;\;\;\;\Box$$