If $a^2 + b^2$ is a prime number $p$, with $p \equiv 1$ (mod $4$), then $a + bi$ is prime in the Gaussian integers $\mathbb{Z}[i]$

Question

I know that there are a lot of questions similar to this, but the reason I'm confused is that it seems that you can prove this just knowing that $a^2 + b^2 = p$ and not use the fact that $p \equiv 1$ (mod $4$) by the following:

Suppose $a + bi = xy$ ($x, y\in\mathbb{Z}[i]$). Then $N(x)N(y) = N(a + bi) = p$. This implies that $N(x) = 1$ or $N(y) = 1$ which tells us that $a + bi$ is irreducible and therefore a prime in the ring of Gaussian integers. Do I need to use the fact that $p$ is congruent to $1$? Help would be appreciated!

Answer 1

Your argument is correct, but note that it's only possible to write $p=a^2+b^2$ for integers $a,b$ if $p=2$ or $p\equiv 1$ (mod $4$). This is because the squares mod $4$ are $0$ and $1$.

Answer 2

The bit about $p \equiv 1 \pmod 4$ seems to have tripped you up a little bit, since numbers with a norm of $2$ (specifically $1 + i$, $1 - i$, $-1 - i$ and $-1 + i$) are also prime.

You can go further and apply your argument to other domains because $N(u) = \pm 1$ means $u$ is a unit and that's the only $xy$ possibility if $N(xy)$ is prime.

Answer 3

A few clarifications are in order. First, that $p \in \textbf Z$ is a positive odd prime number. For example, $5 \equiv 1 \pmod 4$ but $-5 \equiv 3 \pmod 4$ yet $(-2 + i)(2 + i) = -5$.

Second, $\textbf Z[i]$ is a unique factorization domain in which all non-unit irreducible numbers are also prime.

Third, the norm function maps the algebraic integers of $\textbf Z[i]$ to the nonnegative integers of $\textbf Z$ (which is to say, $0$ and the positive integers).

With those preliminaries taken care of, your proof is perfectly sound. Since the norm is multiplicative, $N(x) N(y) = p$ does indeed mean that either $x$ or $y$ must be a unit and therefore $a + bi$ is prime.