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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?

Kenny Wong
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John Bull
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  • Could you please tell us about how you became interested in this question? Also, what research have you done so far? – Kenny Wong May 15 '23 at 12:11
  • Well, it all stemmed from a question from my friend that I posted on SE a few days back link which@BillDubuque was gracious enough to answer (and I learnt simplifying gcds thanks to him and also about grobner basis function et.al. ) and from there a rabbit hole of exploring new stuff in relation to this. – John Bull May 15 '23 at 12:19
  • Given that $\gcd(a,b)=1$ it follows that $\gcd(a-bx,b)=1$. Then also $$\gcd(a^2-db^2,a-bx)=\gcd((bx)^2-db^2,a-bx)=\gcd((x^2-d)b^2,a-bx)=\gcd(x^2-d,a-bx).$$ I'm not sure much more can be said given just $x$ and $\gcd(a^2-db^2,a-by)$. Is $y$ also given? – Servaes May 15 '23 at 15:42
  • @Servaes Yes, y is also given and both x and y are integers... So, is there a way to relate the two GCDs? By the way, how does one go from $gcd(x^2−d)b^2,a−bx)$ to $gcd(x^2−d,a−bx)$? Could you please explain? – John Bull May 15 '23 at 17:56
  • Logically I get it: Since $gcd(a,b)=1$, it means $gcd(a,b^2)=gcd(a-bx,b^2)=1$. So, the $gcd(a-bx,b^2(x^2-d))$ will not depend on $b^2$ at all. But how does one derive it mathematically? – John Bull May 15 '23 at 18:01

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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 $\bmod e\!:\ a-bx \equiv n\iff x \equiv (a-n)\color{#c00}{b^{-1}}$.

Bill Dubuque
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