Proving that a Gaussian integer $a+bi$ with $a,b\neq0$ is composite in $\mathbb{Z}[i]$ if its norm $a^2+b^2$ is composite in $\mathbb{N}$?

Question

It is simple to show for any given composite sum of two squares, that there is at least one Gaussian composite with that norm, but more difficult to show that all Gaussian integers with that norm are composite.

How can we prove this?

Answer 1

If you've established certain facts about the possible composite norms (i.e., you can't have primes $3 \pmod 4$ appear an odd number of times), then you can use Euclid's lemma in the Gaussian primes (which you can establish once you define the gcd): $p$ is prime iff when $p$ divides $ab$, it must divide either $a$ or $b$. So if you know that $p = \alpha\overline{\alpha}$ for some Gaussian prime $\alpha$ divides $x^2 + y^2$, you have that $\alpha$ must divide $p$, which divides $x^2 + y^2 = (x + iy)(x - iy)$. Hence, $\alpha$ divides $x + iy$ or $x - iy$. If the former, you're done. If the latter case, choose $\overline{\alpha}$.

That said, this is a lot of annoying casework and I'd rather just follow the usual path to getting a characterization of the Gaussian primes.

Answer 2

Proof by contradiction.

Suppose $x^2+y^2$ is composite in $\mathbb Z$ with non-zero $x$ and $y$. And suppose $x+yi$ is a Gaussian prime.

Let $x^2+y^2=ab$ with $|a|>1$ and $|b|>1$. Then $x+yi$ divides either $a$ or it divides $b$. Without loss of generality assume that $x+yi$ divides $a$. Therefore $x-yi=\frac{a}{x+yi}b$, which must also be a Gaussian prime. This is only possible if $\frac{a}{x+yi}$ is a unit, implying that either $x=0$ or $y=0$. Contradiction.

So if $x^2+y^2$ is composite in $\mathbb Z$ with non-zero $x$ and $y$, then $x+iy$ is not a Gaussian prime.