I'm trying to show that $a^2 \equiv b^2 \mod p$ implies that $a \equiv b \mod p$ and $a \equiv -b \mod p$, where $p$ is prime.
Since $a^2 - b^2 \equiv 0 \mod p$, I know that $(a + b)(a - b) \equiv 0 \mod p$. Then either $a + b \equiv 0 \mod p$ or $a - b \equiv 0 \mod p$.
But I don't know how to show that both are true. Could someone give me a hint?