Relation between two gcds

Question

Given two expressions $ \text{gcd}(a^2-d b^2,a-bx)$ and $\text{gcd}(a^2-d b^2,a-by)$ where $a, b$ are coprime, $x$ is an integer-valued variable and the rest all are constant and so $\text{gcd}(a^2-d b^2,a-by)$ is known, what is the relation between $\text{gcd}(a^2-d b^2,a-bx)$ and $\text{gcd}(a^2-d b^2,a-by)$? Is there a way to find $\text{gcd}(a^2-d b^2,a-bx)$ given $x$ and $\text{gcd}(a^2-d b^2,a-by)$ without calculating the actual gcd?

Answer 1

Let $\,e = a^2-db^2.\,$ Then $\,(e,b) = (a^2,b)=1\,$ by $\,(a,b)=1,\,$ so $\,\color{#c00}{b^{-1}}$ exists $\bmod e.\,$ Therefore there is no relation between $\,(e,a-bx)\,$ and $\,(e,a-by)\,$ since as $\,x\,$ varies $\,n = a-bx\,$ can take all values $\bmod e\,$ (so all divisors of $\,e)\,$ since $\bmod e\!:\ a-bx \equiv n\iff x \equiv (a-n)\color{#c00}{b^{-1}}$.