Shown $p \in \mathbb{Z} [i]$ is a prime given $p\in \mathbb{Z}$ is a prime and $p$ does not equal $x^2 + y^2$

Question

Suppose $p \in \mathbb{Z}$ is a prime number for which there are no integers, $x$, $y$ such that $p = x^2 +y^2$. How can I go about showing that $p$ is a prime element of $\mathbb{Z} [i]$. Assuming that $\mathbb{Z} [i]$ is a Euclidean Domain, and that the only units are $\pm 1$,$\pm i$.

Answer 1

$$\mathbf Z[\mathrm i]/p\mathbf Z[\mathrm i]\simeq \mathbf Z[x]/(p,x^2+1)\simeq\mathbf Z/p\mathbf Z[x]/(x^2+1) .$$ Thus $p$ is a prime in $\mathbf Z[\mathrm i]$ if and only if $-1$ is not a square modulo $p$. By the 1st supplementary law of quadratic reciprocity, this happens for an odd prime if $p\equiv 3\mod 4$, which is the case when $p$ is not the sum of two squares.

Answer 2

The prime p must be odd because $2=1+1$ This prime $p\in \mathbb Z$ must be prime in $\mathbb Z[i]$ which means neither of the form $p=(a+bi)(c+di)$ (decomposed) nor of the form $p=(a+bi)^2$ (ramified). Hence it must be inert. By elementary algebraic number theory we must have $-1\ne x^2\space (mod\space p)$ i.e. $-1$ cannot be a quadratic residu modulo $p$.

For instance $p=7$ is inert because the squares modulo $7$ are $0,1,2,4$ but $p=5$ is not inert because $2^2=-1\space (mod\space 5)$. It can be proved that all prime $p\equiv 3$ $(mod\space 4)$ is inert in $\mathbb Z[i]$ but this is exactly the given condition for p cannot be a sum of two squares.

Answer 3

We want to prove:

If $p$ is not a sum of two squares in $\mathbb Z$, then $p$ is prime in $\mathbb{Z}[i]$.

Let's prove the contrapositive:

If $p$ is not prime in $\mathbb{Z}[i]$, then $p$ is a sum of two squares in $\mathbb Z$.

Indeed, if $p$ is not prime in $\mathbb{Z}[i]$, then we can write $p=\alpha \beta$, with $\alpha, \beta \in \mathbb{Z}[i]$, not units. Taking norms, we get $p^2=N(\alpha) N(\beta)$. Since $\alpha, \beta$ are not units, we must have $N(\alpha)=N(\beta)=p$. This gives an expression of $p$ as the sum of two squares in $\mathbb Z$.