Let $p$ be a prime so $p\equiv3\pmod4$. If $p|a^2+b^2$, then $p|a,b$ 
Let $p$ be a prime so $p\equiv3\pmod4$. If $p| a^2+b^2$, then $p| a,b$

How do I prove this small theorem? I know that it's quite useful. Are there other small theorems like this one? I am mostly searching elementary proofs, so not involving to complicated stuff...

I hope I didn't miss something and I think it is pretty elementar:
Using Fermats Little Theorem: $a^p\equiv a\mod{(p)}$ and $b^p\equiv b\mod{(p)}$. Now we get that $a^{p+1}+b^{p+1}\equiv a^2+b^2 \equiv 0 \mod{(p)}$. Because $4\mid p+1$ we can write $p+1=4k$ , for some $k\in\mathbb{N}$. Now we get: $0\equiv a^{4k}+b^{4k}\equiv a^{4k}+(-a^2)^{2k}\equiv a^{4k}+a^{4k}\equiv 2a^{4k} \mod{(p)}$. So now that means $p$ divides $2a^{4k}$, but bcs $p>2$ it cant divide the 2 so it has to divide $a^{4k}$, and if it is a factor of it, it has to be also a factor of $a$, in other words $p\mid a\Rightarrow p\mid b$.
