Sums of two squares via $\mathbb{Z}[i]$

$-1$ is a square mod $p$. For $p\equiv1\pmod4$, by Euler's criterion $(-1)^{(p-1)/2}=1$, so there is $a$ with $a^2\equiv-1\pmod p$. Thus $p\mid a^2+1=(a+i)(a-i)$.

$p$ is not irreducible in $\mathbb{Z}[i]$. If $p$ were prime in $\mathbb{Z}[i]$, then $p\mid a+i$ or $p\mid a-i$, which is impossible (the imaginary part is $\pm1$, not divisible by $p$).

Factorization. So $p=zw$ with $z,w$ non-units. Taking norms: $p^2=N(p)=N(z)N(w)$, and since $z,w$ are non-units, $N(z),N(w)>1$. So $N(z)=N(w)=p$.

Conclusion. If $z=a+bi$, then $a^2+b^2=p$. Every prime $p\equiv1\pmod4$ is a sum of two squares.