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Continuing for Brahmagupta-Fibonacci Identity...

Find non-zero integers $a, b, x, y$ satisfy: \begin{cases} ax+by=\alpha \\ ay-bx=\beta \\ \end{cases}

I also want various solutions to this kind of question. Again, I want many solutions, including solutions that use Brahmagupta-Fibonacci and other solutions.

I want to try with $\alpha=24, \beta=7.$ For my method, the answer to the question is:

$(a, b, x, y)=(\pm1, \pm2, \pm2, \pm11), (\mp2, \pm1, \mp11, \pm2), (\mp2, \pm11, \mp1, \pm2), (\pm11, \pm2, \pm2, \pm1), (\mp3, \pm4, \mp4, \pm3), (\pm4, \pm3, \pm3, \pm4)$

Xander Henderson
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1 Answers1

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\begin{cases} ax+by=24 \\ ay-bx=7 \end{cases}

Solution with Brahmagupta-Fibonacci Identity:

\begin{align}&(ax+by)^2+(ay-bx)^2=(a^2+b^2)(x^2+y^2)=24^2+7^2=625. \\&(a^2+b^2, x^2+y^2)=(5, 125), (25, 25), (125, 5). \\\ \\& \Rightarrow (a, b, x, y)=(\pm1, \pm2, \pm2, \pm11), (\mp2, \pm1, \mp11, \pm2), (\mp2, \pm11, \mp1, \pm2), (\pm11, \pm2, \pm2, \pm1), (\mp3, \pm4, \mp4, \pm3), (\pm4, \pm3, \pm3, \pm4)\end{align}

Xander Henderson
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    What is the point to repeat as an answer something you have already given in your question ? Moreover, this is a particular case that has only an interest if you explain the method you have used to obtain these solutions. – Jean Marie Jun 28 '22 at 09:28
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    Answers should be answers. There is no reason to use a "spoiler box" to hide parts of the answer. – Xander Henderson Jun 28 '22 at 14:32