Write $x(a^2+b^2)+(2ab)y$ as a product of factors.

Question

Let $a,b,c,x,y \in\mathbb{Z}>1$ and $\gcd(a,b)=\gcd(x,y)=\gcd(a,b,x,y)=1$,

Can $$x(a^2+b^2)+(2ab)y$$be factorized ?

Answer 1

Since $x$ and $y$ are coprime, you can reduce the problem into proving that the squared sum of coprime integers $a$ and $b$ is coprime to $a$ and $b$, that is $$\gcd(a^2 + b^2, a) = \gcd(a^2 + b^2, b) = 1.$$ Which can be done by considering evenness and oddness of $a$ and $b$.

But when both $a$ and $b$ are odd, $\gcd(a^2 + b^2, 2ab) = 2$. Therefore, the expression cannot be written generally as a product of factors because $a$ and $b$ aren't always both odd.

Answer 2

One possible interpretation is this: if $y = r^2 + s^2$ and $x = r^2 - s^2,$ with $\gcd(r,s) = 1$ but $r,s$ not both odd, then the whole thing factors. Same if $y = r^2 + s^2$ and $x = 2rs.$ The point is to make $y^2 - x^2$ a square, which causes the quadratic form in $a,b$ to factor