Proving that the norm of an irreducible Gaussian integer is equal to $p$ or $p^2$ Question: Suppose $\pi \in \mathbb{Z}[i]$ is irreducible. Prove that there exists a prime number $p$ such that $N(\pi) = p$ or $N(\pi) = p^2$.
My attempt:
I genuinely do not know where to begin. There is a hint which says that I must take an ordinary prime number $p$ such that it is a factor of $N(\pi)$.
 A: Have you learned about the difference between unique factorization domains (UFDs) and non-UFDs yet? $\mathbb Z[i]$ is a UFD, which makes certain things easier.
Suppose $N(\pi) = pq$, where $p$ and $q$ are "ordinary" primes, and $p \neq q$. Since we're working in a UFD, this means that we can find Gaussian integers $a$ and $b$ such that $N(a) = p$, $N(b) = q$ and $ab = \pi$. But since neither $a$ nor $b$ are units, this contradicts $\pi$ being irreducible. Therefore $N(\pi) = pq$ is impossible if $\pi$ is irreducible. This also takes care of ruling out the possibility that $q$ is an "ordinary" composite number.
Quick little exercise: of $1 + i$ and $1 + 3i$, one is irreducible and one is reducible. Determine which is which.
But it doesn't rule out the possibility that $N(\pi) = p^2$ but $\pi$ is the square of some Gaussian integer $z$ such that $N(z) = p$. However, please take this on faith for the time being, but for an "ordinary" positive prime $p \equiv 3 \pmod 4$ there is no solution in integers to $x^2 + y^2 = p$ and therefore $N(z) = p$ is impossible. Those primes are irreducible in this domain. 
If "ordinary" positive prime $p \equiv 3 \pmod 4$, then $p, -p, pi, -pi$ are all irreducible, and their norms are all $p^2$.
One more quick little exercise: of 17 and $-19i$, one is irreducible and one is reducible. Determine which is which.
A: Remember that the norm is multiplicative. If $\pi = \alpha \beta$, then $N(\pi) = N(\alpha) N(\beta)$. If $\pi$ is indeed irreducible and prime, that means either $N(\alpha) = 1$ or $N(\beta) = 1$. But if $N(\alpha) \neq 1$ and $N(\beta) \neq 1$, then $\pi$ is not irreducible as originally asserted.
Then consider the three possibilities for $\pi$: either it is purely real, purely imaginary, or complex. For example: $-3, 3i, 2 + i$. The norms of the first and second examples are both $9$, the square of a prime. The norm of the third example is $5$, itself a prime.
A: (fixed a typo: $\gamma\mid p \mid (x+yi)(x-yi),$, was $\gamma\mid p \mid (x+yi)(x+yi),$)
Suppose $\pi=x+yi$ (where $x$ and $y$ are integers) is irreducible, then $x-yi$ is also irreducible. $N(\pi)=x^2+y^2>1$. Let $p$ be a prime integer divisor of $N(\pi)$, and let $\gamma$ be an Gaussian integer and an irreducible divisor of $p$. Then since $$\gamma\mid p \mid N(\pi)=(x+yi)(x-yi),$$
we have either $$\gamma \mid (x+yi),$$ or $$\gamma\mid (x-yi),$$ 
and since both $x+yi$ and $x-yi$ are irreducible, we have $N(\gamma)=N(x+yi)$ or $N(\gamma)=N(x-yi)$, but $N(x+yi)=x^2+y^2=N(x-yi)$, so $$1<N(x+yi)=N(x-yi)=N(\gamma) \mid N(p)=p^2,$$
therefore, $N(x+yi)=p$ or $p^2$. 
(Note that we avoid using the classification of Gaussian primes and only use the definition of irreducible Gaussian integers.)
A: Since $\mathbf Z[i]$ is a UFD, an element $\pi$ is irreducible if and only if the principal ideal $\langle \pi \rangle$ is a prime ideal, and you can directly apply the theory of ramification of primes in the ring of integers of a number field. But I guess that you ask for a direct proof using only the factoriality of $\mathbf Z[i]$. Recall that uniqueness of the decomposition into irreducible elements is only up to units (that is, invertible elements) which in the case of $\mathbf Z[i]$ are $\pm 1$ and $\pm i$.
Let $p$ be a prime of $\mathbf Z$ dividing $N(\pi) = \pi \cdot \pi'$, where $\pi'$ is the conjugate of $\pi$ is also irreducible. Because of the UFD property, this implies that (up to units) the decomposition of $p$ in $\mathbf Z[i]$ is $\pi$, or $\pi'$, or $\pi \cdot \pi'$. Taking norms then gives $N(\pi) = N(\pi') = p^2$ or $p$ . Note that the second case occurs if and only if $p \equiv 1 \bmod 4$.
