I have already proven one direction, which is $p\equiv3$ $(mod$ $4)$ implies $p$ irreducible. Now, stating that $p$ is irreducible I can't get to the conclusion that $p\equiv3$ $(mod$ $4)$. I have tried many things but the one that got me closer to the solution was to suppose that $p\not\equiv3$ $(mod$ $4)$, and find a contradiction. I've found a contradiction in the cases where $p\equiv2$ $(mod$ $4)$ and $p\equiv1$ $(mod$ $4)$ for $p$ being an even prime number (i.e. $2$) but it's getting really difficult to prove the remaining statement, that is $p\equiv1$ $(mod$ $4)$ is not possible for $p$ odd prime.
To solve the first implication I've used that $p$ is compound (not irreducible) in $\mathbb{Z}[i]$ iff can be expressed as the sum of two squares, I don't know if it could come in handy again.
