Find the number of positive integer solutions to the equation $$a^2+b^2 = p_1p_2p_3 $$ where the $p_i$ are distinct primes each congruent to $1$ mod $4$.
My take:
We can show each $p_i$ can be written as a sum of two squares since they are congruent to $1$ mod $4$, and also that such a representation is unique. We know that the product of sum of two squares is a sum of two squares since $(x^2+y^2)(z^2+w^2) = (xz+yw)^2 + (xw-yz)^2 = (xz-yw)^2 + (xw+yz)^2$
So $p_1p_2$ can be written as a sum of two squares in two ways and for each such representation $(p_1p_2)p_3$ can be written as a sum of two squares in two ways. So in all, $p_1p_2p_3$ can be written as a sum of two squares in $4$ ways.
So it seems there are $4$ solutions to the given equation (not considering ordered pairs).
Am I missing any other solutions?
Assuming the smallest values of $p1, p2, p3 = 5, 13, 17$, we get that $5*13*17 = 1105$.
For this, we have $a = 23, b = 24$ || $a = 33, b = 4$ || $a = 32, b = 9$ || $a = 31, b = 12$.
That makes 4 for the first set, at least.
I believe there are none for $p1, p2, p3 = 5, 13, 29$. The actual theory, I have no idea.