if $p \mid (a^2 + b^2 )$, $p \nmid a$ and $p \nmid b$. Prove that there exists an integer $c$ such that $c^2 \equiv −1 \mod p$.

Question

Given $p$ is prime, I'm not really sure what im supposed to do with the information that $p \nmid a$ and $p \nmid b$ in order to conclude $c^2 \equiv −1 \mod p$.

Answer 1

If $p \nmid a$, there is such $c$ that $ca\equiv b \mod{p}$. Then $a^2(c^2+1)=(ac)^2+a^2\equiv b^2+a^2 \mod{p}$. So $p\mid a^2(c^2+1)$ what implies $p\mid (c^2+1)$