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A friend of mine posed this question: Assume three constants $a,b$ and $d$ and one variable x. $a,b$ & $d$ are all integers and $a$ and $b$ are coprime.$x$ can also be only an integer.

Now, is it possible to calculate $GCD(a*x-d*b,a-b*x)$ without having to do all the multiplications. In other words, is a simplification to this GCD expression possible?

I tried many things and have not managed to find if it is possible or not. Any clues or opinions are welcome.

Thanks

John Bull
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  • I suggest looking at examples, try to spot a useful pattern (if there is one). – lulu May 10 '23 at 12:49
  • It can be simplified to $,(bx-a,a^2-db^2).,$ Is that the type of simplification you seek? – Bill Dubuque May 10 '23 at 19:06
  • Yes.. How do you get there? – John Bull May 10 '23 at 21:38
  • Eliminate $x$ like we did here or here. Yours is special case of the cross multiplication elimination in the latter (let $m$ be your gcd). – Bill Dubuque May 10 '23 at 21:53
  • Thanks a lot.. Can you please write the full solution here so that I can upvote.. Also it might be useful to people in the future. – John Bull May 10 '23 at 21:59
  • Also in that case, can $\frac{a^2−d b^2}{GCD(bx−a,a^2−d b^2)}$ be simplified too apart from the obvious LCM relation? – John Bull May 10 '23 at 22:53
  • @BillDubuque I tried this: $a x=d b (mod) m$ & $-b*x=-a (mod) m$ which leads to $\frac{a}{b}\equiv x \equiv \frac{db}{a}$. Now if we cross multiply, we get $a^2 \equiv x \equiv db^2$. How to proceed after that?..

    On the other hand, if I assume $a x - d b=n_1 m$ & $a-bx=n_2 m$ , then substituting, I get $a^2-db^2=n_1mb+n_2mb$

     – John Bull May 11 '23 at 10:37
  • @BillDubuque On the other hand, if I assume $a x - d b=n_1 m$ & $a-bx=n_2 m$ , then substituting, I get $a^2-db^2=m(n_1b+n_2a)$. What is the next step? – John Bull May 11 '23 at 10:45

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