Confusion? II

There's a lot of general theory here. Firstly, the Gaussian integers $\mathbb{Z}[i]$ form a Euclidean domain. Since $p = 4m+1$ is prime, we note that $[(2m)!]^2 \; \equiv \; (2m)!\prod_{j=1}^{2m}(p - j) \; \equiv \; (2m)! \prod_{j=2m+1}^{4m}j \; \equiv \; (4m)! \; \equiv \; (p-1)! \; \equiv \; -1 \pmod{p}$ by Wilson's Theorem. Thus we deduce that $p$ divides $\big((2m)! + i\big)\big((2m)! - i\big)$ in $\mathbb{Z}[i]$ , but $p$ divides neither $(2m)! + i$ nor $(2m)! - i$ . Thus $p$ is not prime in $\mathbb{Z}[i]$ and hence (since we are in a Euclidean domain), $p$ is not irreducible in $\mathbb{Z}[i]$ . Thus we can find $u,v \in \mathbb{Z}[i]$ such that neither $u$ nor $v$ is a unit and $p = uv$ . But then $p^2 = |u|^2|v|^2$ and since $|u|^2$ and $|v|^2$ are both integers greater than $1$ ( $u,v$ are not units) we deduce that $|u|^2=|v|^2=p$ . If $u=x+iy$ for integers $x,y$ , we deduce that $p = x^2+y^2$ , as required.