It's well known that a positive integer $n$ is a sum of two squares if and only if every prime of the form $4m + 3$ that divides $n$ appears with even multiplicity in the prime factorization of $n$. My question is:
For which positive integers $n$ do there exist $a, b \in \mathbf{Z}$ such that $(a, b) = 1$ and $n = a^2 + b^2$?
I have done some work on this. We can use the arithmetic of $\mathbf{Z}[i]$ to show that no prime of the form $4n + 3$ can divide such an $n$. Indeed, if $p \equiv 3 \bmod{4}$ is a prime that divides $n$, then $p|(a + bi)(a - bi)$ in $\mathbf{Z}[i]$. Because $p$ stays prime in $\mathbf{Z}[i]$, either $p|(a + bi)$ or $p|(a - bi)$, and each of these situations implies $p|a$ and $p|b$, contradicting our assumption that $(a, b) = 1$.
Furthermore, $n$ cannot be divisible by four. If it were then $a^2 + b^2 \equiv 0 \bmod{4}$, and because the only quadratic residues mod four are zero and one, we must have $a^2 \equiv b^2 \equiv 0 \bmod{4}$, meaning $a$ and $b$ are both even, which again contradicts $(a, b) = 1$.
Therefore, any integer $n$ satisfying the desired properties must be of the form $$n = 2^e p_1 \cdots p_m$$ where $e = 0, 1$ and each $p_i \equiv 1 \bmod{4}$. Some numerical exploration suggests that any number of this form can be expressed as a sum of two relatively prime squares. It suffices to prove the case $e = 0$, because if $$p_1 \cdots p_m = c^2 + d^2 \text{ where } (c, d) = 1,$$ then $$2p_1 \cdots p_m = (c + d)^2 + (c - d)^2,$$ and the coprimality of $c$ and $d$ implies $(c + d, c - d) = 1$.
Therefore, the problem is reduced to proving the following statement, which I believe to be true.
Any product of primes of the form $4n + 1$ is a sum of two relatively prime squares.
Can someone provide a hint towards a proof of this?
