A prime is a Gaussian prime if and only if it is 3 modulo 4

Statement

Let $p$ be a prime number. Then $p$ is a Gaussian prime if and only if $p \equiv 3 \pmod 4$.

Remarks

If $p = 2$, then $2 = (1 + i)(1 - i)$, and $1 + i$ is a Gaussian prime. If $p \equiv 1 \pmod 4$, then $p = a^2 + b^2 = (a + bi)(a - bi)$ by Fermat's theorem on sums of two squares, and $a \pm bi$ are Gaussian primes. If $p \equiv 3 \pmod 4$, then $p$ is not a sum of two squares. So no Gaussian integer has norm $p$, and $p$ cannot split. For example, $3$ and $7$ are Gaussian primes, but $13 = (3 + 2i)(3 - 2i)$.

Explanations

L1 human

An ordinary prime stays prime among the Gaussian integers exactly when it leaves remainder $3$ on division by $4$. The other primes factor, because they are sums of two squares.