$p\mid a^2+b^2$ Prove that $p\mid a$ and $p\mid b$

Question

Let $p=4k+3$ be a prime number and $a,b\in\mathbb{N}$ so that $p\mid a^2+b^2$

Prove that $p\mid a$ and $p\mid b$

I tried to use Fermat's little theorem, but I don't know what to do next.

Answer 1

Suppose $a^2 \equiv -b^2 \pmod{p}.$ If $a$ or $b$ is divisible by $p$ then it immediately implies both are divisible by $p.$ So suppose they are not. Then $(\frac{a}{b})^2 \equiv -1 \pmod{p}.$ Which implies $-1$ is a square mod $p.$ However, $(\frac{-1}{p}) = (-1)^{\frac{p-1}{2}}.$ Thus, since $p \equiv 3 \pmod{4}$ this is a contradiction. Hence, both are divisible by $p.$

Another solution: Since $p \equiv 3 \pmod{4}, p$ is a Gaussian prime. Hence $p \mid a^2 + b^2$ implies $p \mid a + bi$ or $p \mid a - bi.$ Thus, $p \mid 2a$ which implies $p \mid a.$ Similarly, $p \mid b.$