Let $R:=\mathbb{Z}[i]$. Prove that every nonzero prime ideal $\mathfrak{P}$ of $R$ belongs to one of the following families:
$\mathfrak{P}=(1+i)R$
$\mathfrak{P}=(a+bi)R$ where $a,b\in\mathbb{Z}$ and $a^2+b^2$ is an odd prime $p$ which is congruent to $1$ modulo $4$
$\mathfrak{P}=pR$ where $p$ is an odd prime which is congruent to $3$ modulo $4$.
Hint: in case 3), let $\alpha\in R$ be written as $c+id$ with $c,d\in\mathbb{Z}$ and suppose $\alpha\notin\mathfrak{P}$. Consider $\alpha\overline{\alpha}=c^2+d^2$; prove that $p$ does not divide $c^2+d^2$, so that there exists an integer $e$ such that $(c^2+d^2)e=1\bmod p$. Conclude that $\alpha\cdot\overline{\alpha}e=1\bmod\mathfrak{P}$.
I can't understand the hints that i'm given. I've proved that $p$ doesn't divide $c^2+d^2$, in fact $p=3\bmod 4$ implies that $p$ is also a Gaussian prime, so if it divides $c^2+d^2=(c+di)(c-di)$ then it should divide one of the two factors, which is impossible.
Hence, being $p$ a rational prime, not dividing $c^2+d^2$, it must be coprime to $c^2+d^2$ so that there exists $e$ such that etc. etc.
And now? I have proven that $\alpha$ is invertible modulo $\mathfrak{P}$. How can i use this?