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)?

If $\pi$ is a Gaussian prime from $\mathbb{Z}[i]$ that divides exactly one prime from $\mathbb{Z}$, it divides only that one prime.
Actually, it divides two primes.
Now, the product $\pi \overline \pi$ is the norm function. Let's say $\pi = a + bi$, where $a, b \in \mathbb{Z}$. Then $\pi \overline \pi = (a + bi)(a - bi) = a^2 - abi + abi - (-1)b^2 = a^2 + b^2 = N(\pi)$. So if $p$ is a prime from $\mathbb{Z}$, we can set $a = p$ and $b = 0$, and so $N(\pi) = p^2$. If on the other hand, $N(\pi) = p$ rather than $p^2$, then $\pi$ must be a complex number with a nonzero real part and nonzero imaginary part.
The really important result we need here is that the norm function is multiplicative. That is, $N(\alpha \beta) = N(\alpha) N(\beta)$. For example, $N(4 - 5i) = N(4 + 5i) = 41$. And $(4 - 5i)(4 + 5i) = 41$, and indeed $N(41) = 41^2 = 1681$.
So if $N(\alpha \beta) = n$, that must mean that $N(\alpha)$ is an integer from $\mathbb{Z}$ that is a divisor of $n$. For example, if $N(\alpha \beta) = 10$, we might have $N(\alpha) = 2$ or $5$. And we can do that because $10$ is composite. But if $n = p$ is prime, then the only possibilities for $N(\alpha)$ are $1$ and $p$. The same goes for $N(\beta)$. Remember that if $p$ is negative, its norm is nevertheless positive (and this is also true in other imaginary quadratic rings).
Therefore, $\pi$ can only divide two primes from $\mathbb{Z}$. In the previous example of $4 + 5i$, we see that $$\frac{41}{4 + 5i} = 4 - 5i$$ and $$\frac{-41}{4 + 5i} = -4 + 5i.$$
