Wanted: Parametric Solution to the Diophantine Equation $x^2+y^2\equiv 0$ mod $m$ (*)

Question

Given $m=p^{a}q^{b}$, $m>1$, $p\neq q$, primes $\equiv 1$ mod 4, and $a,b\in\mathbb{N}$, then there is $u\in\mathbb{Z}$ with $u^{2}\equiv -1$ mod $m$. Hence $(1,u)$ is a solution to (*), and also $(x,ux)$ for each $x\in\mathbb{Z}$. Now the converse: Let $(x,y)\not\equiv(0,0)$ be any solution to $(*)$. If $gcd(x,y)=1$, then $gcd(x,m)=gcd(y,m)=1$ and $x^{-1}$ mod m exists. Putting $u=yx^{-1}$ mod m we have $u^{2}\equiv -1$ mod $m$. \ There is a strong guess, that this also holds for each non trivial solution $(x,y)$ to $(*)$ with $gcd(x,m)>1$. Proof?

Answer 1

COMMENT.-Because $p$ and $q$ are primes $\equiv 1\pmod4$ and since the set of sums of two squares is closed by multiplication you do have finally the equation $$x^2+y^2=t(X^2+Y^2)$$. Consequently if you take the parameter $t\in\{\text{sums of two squares } r^2+s^2\}$ you get at the end the equation $$x^2+y^2=R^2+S^2$$ and you can do $$(x,y)=(R,S)$$ (Use the formula $(r^2+s^2)(X^2+Y^2)=(rX+sY)^2+(rY-sX)^2$)

Answer 2

The heading asks for a parametric solution but your question asks for a proof of the conjecture that $(yx^{-1})^2$ is congruent to $-1$ modulo $m$.

A counterexample with $m=65$ is given by $x=125,y=25$. Since $(yx^{-1})^2=\frac{1}{25}$ and this is not congruent to $-1$ modulo $65$.