Proving that the norm of an irreducible Gaussian integer is equal to $p$ or $p^2$

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.

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.

(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.)