From 13.5.1 and 13.5.3, the only case left to consider if \(N\) has a prime of the form \(p=4k+3\), but to an odd power. This seemed to be the bottleneck in our exploration.

By way of contradiction, suppose that it is possible to write \begin{equation*}N=a^2+b^2\; .\end{equation*} First, divide this equation by any factors of \(p\) common to both sides to get \begin{equation*}M=c^2+d^2\end{equation*} The power of \(p\) we divided by (so that \(N=Mp^k\)) must be an even power, since each term on the right-hand side is a perfect square and can only contribute even powers of primes by the Fundamental Theorem of Arithmetic.

Since \(N\) had an odd power of \(p\), we know \(M\) still has an odd power of \(p\) dividing it, yet \(p\nmid c,d\).

Take everything modulo \(p\) to get the congruence \begin{equation*}0\equiv c^2+d^2\text{ (mod }p)\; .\end{equation*} Since \(p \nmid c\), we can multiply this congruence by \(\left(c^{-1}\right)^2\) to get \begin{equation*}0\equiv 1+\left(c^{-1}\right)^2d^2\Rightarrow -1\equiv \left(c^{-1}d\right)^2\text{ (mod }p)\end{equation*}

This is a contradiction, as there is no square root of \(-1\) modulo \(p\) by Fact 13.3.2!