Primes in $\mathbb Z[i]$ I've been trying to proof that if $\pi\in Z[i]$ and $\pi$ is prime, then $\pi\overline \pi$ is a prime in $\mathbb Z$ or is the square of a prime  in $\mathbb Z$.
I took a prime $p$ in $\mathbb Z$ such that $\pi|p$, then $\overline \pi |p$, which is why $\pi\overline\pi |p^2$. I am stuck in that, so I need that $\pi\overline \pi=p^2,p$.
Thanks in advance :).
 A: You have shown that $\pi\overline{\pi}$ divides $p^2$ and that $\pi\overline{\pi}\in\Bbb{Z}$, so it remains to show that $\pi\overline{\pi}\neq1$. But this is clear because $\pi$ is prime, hence by definition it is not a unit.
A: Let  $\pi=a+bi$, $a,b$ integers.Suppose that it is a prime. First check  that $\bar\pi=a-bi$ is also a prime.
Let $n=|a+bi|=a^2+b^2$. Let $r$ be a prime dividing integer $n$. So $a^2\equiv -b^2 \mod r$. If $r\equiv 3\mod 4$, then the only possibility is that $a\equiv b\equiv 0\mod r$, so $n$ is divisible by $r^2$.
If $r\equiv 1 \mod 4$, then $r=c^2+d^2$ for some integers $c,d$, so $c+di | (a+bi)(a-bi)$ which contradicts the fact that $a+bi$ (hence $a-bi$) is prime.
So we need to consider only the case  that $n=m^2$ where $m$ is a product of primes $p_1...p_k$, each $p_i\equiv 3\mod 4$. Again, if $k>1$, then $p_1$ is a prime dividing the  product of two primes $(a+bi)(a-bi)$ which is impossible, so $k=1$.
Thus $n$ is either a prime or a square of a prime.
The opposite implication is obvious.
