Gauss prime divides exactly one integer prime in $\mathbb{Z}[i]$ I am asked to show that a Gauss prime $\pi$ divides exactly one integer prime in $\mathbb{Z}[i]$. 
To show existence, I have tried to use the fact that the product $\pi \overline{\pi}$ is equal to either an integer prime $p$ or the square of integer prime $p^2$. If $\pi$ satisfies the first case, then the statement is immediate. How about when $\pi \overline{\pi}=p^2$?
Also, how do we show that $\pi$ divides no other integer primes (i.e. uniqueness)?

This works in any ring of integers $\mathcal{O}_K$ of a number field $K$ :
Take a proper ideal $I$ of $\mathcal{O}_K$ (here $I = \pi \mathcal{O}_K$).
Note that $J =I \cap \mathbb{Z}$ is a proper ideal of $\mathbb{Z}$, thus $J = n\mathbb{Z}$ for some $n \in \mathbb{N}_{> 1}$. 
If $p \in I$ then $p \in  I \cap \mathbb{Z}= n \mathbb{Z}$ so $n | p$. If $p$ is a prime number, it means that $n = p$.
Finally, if $I$ was a prime ideal then $\mathcal{O}_K/I$ is a finite integral domain (and hence a finite field). Its sub-integral domain (subfield) generated by $1$ is $\mathbb{Z}/n\mathbb{Z}$, therefore $n$ is prime.
