Reducing primes in $\mathbb Z[i]$ Let $4k+1=p$ be a prime.

Assume you do not know $p=a^2+b^2$(this is what we intend to prove).

Over $\mathbb Z[i]$, how does one prove that $p$ splits into conjugates?

That is, if $p = (a+bi)(c+di)$, then $a+bi=c-di$?

We know that $a-bi=t(c+di)$ where $t\in \mathbb R$, could give: $p=t(a^2+b^2)$. It could be that $t=p/m^2$ where $m^2=a^2+b^2$. How do you show $t=\pm 1$?

Does this work?

$a-bi=t(c+di)\implies (a-bi)(a+bi)=t(c+di)t(c-di)\implies(a^2+b^2)=t^2(c^2+d^2)\implies t^2=1(\mbox{since }a^2+b^2=c^2+d^2=p)\implies t=\pm 1.$
The ring $\mathbf{Z} / p \mathbf{Z}$ is a field with $p-1$ nonzero elements. Since $4 | p-1$, there is a multiplicative subgroup of order $4$.

A field can only have two square roots of unity, therefore $\mathbf{Z} / p \mathbf{Z}$ already has a fourth root of unity $m$.

Pick your favorite reason why this implies $p\mathbf{Z}[i]$ cannot be a prime ideal of $\mathbf{Z}[i]$: e.g. some simple things things you can use are


$\mathbf{Z}[i]/p\mathbf{Z}[i] \cong (\mathbf{Z}/p\mathbf{Z})[i]$ is not a field, because it has too many fourth roots of unity
$i-m \not\in p\mathbf{Z}[i]$
$p | (m-i)(m+i)$ but neither $p | (m-i)$ nor $p | (m+i)$


The ring $\mathbf{Z}[i]$ is a Euclidean domain, which implies it is a unique factorization domain.

Since the prime factorization of $p$ cannot simply be $p$ (nor can it be $(-1) \cdot (-p)$) and cannot be any other product of integers, $p$ must have a prime factor of the form $a+bi$. By conjugation (which is an automorphism of $\mathbf{Z}[i]$ that is the identity on $\mathbf{Z}$), $a-bi$ is also a prime factor of $p$.

Thus $(a+bi)(a-bi) = a^2 + b^2$ is a divisor of $p$. Since it's an integer bigger than $1$, we must actually have $a^2 + b^2 = p$.
Inspired by Will Jagy's comment.

We know that $(2k)!^2\equiv(4k)!\equiv(p-1)!\equiv-1\mod p$ and so $x^2+1\equiv 0 \mod p$ has a solution for $x$. Hence $p|(x^2+1)\implies \frac{(x+i)(x-i)}{p}\in\mathbb Z\implies p=(a+bi)(c+di)$.

We know that $a-bi=t(c+di)$ where $t\in \mathbb R$.

We know that $a^2+b^2=c^2+d^2=p$ as $p^2=p\cdot \bar{p} = (a+bi)(a-bi)(c+di)(c-di)=(a^2+b^2)(c^2+d^2)\implies p=a^2+b^2$ and $c^2+d^2.$

$\mbox{So }a-bi=t(c+di)\implies (a-bi)(a+bi)=t(c+di)t(c-di)\implies(a^2+b^2)=t^2(c^2+d^2)\implies pt^2=p\implies t^2=1\implies t=\pm 1\implies p=\pm(a+bi)(a-bi).$ 

Hence $p=a^2+b^2$ since $p&gt;0$.