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Show that no non-zero integers $a, b, x, y$ satisfy: \begin{cases} ax-by=16. \\ ay+bx=1. \end{cases}

From Baltic-Way 2021.

\begin{align} &(a+bi)(x+yi)=(ax-by)+i(ay+bx)=16+i. \\ &|(a+bi)(x+yi)|=|16+i|. \\ &\sqrt{a^2+b^2}\sqrt{x^2+y^2}=\sqrt{257}. \\ &(a^2+b^2)(x^2+y^2)=257. \\ &\therefore a^2+b^2=257, x^2+y^2=1 \text{ or } a^2+b^2=1, x^2+y^2=257.(\because 257: \text{ prime.)} \\ \Rightarrow & \text{No solution.} \end{align}

Is my solution right?

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    $\begingroup$ Your solution is correct, and very elegant too +1 $\endgroup$
    – FShrike
    Jun 27 at 16:18
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    $\begingroup$ Very nice solution. I can't say I would've immediately thought to apply complex numbers here. $\endgroup$
    – Deepak
    Jun 27 at 16:49
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    $\begingroup$ I edited your title to be clearer and more specific. $\endgroup$
    – Deepak
    Jun 27 at 16:53
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    $\begingroup$ From comments on your previous post: arithmetic of quadratic rings, usual items $x + y \sqrt d$ or $x + y \omega$ where $\omega = \frac{1 + \sqrt d}{2},$ is the first generalization of Gauss composition; the explicit formula is Dirichlet's method. You might enjoy a book on binary quadratic forms; my favorite is Buell. There is plenty in Dickson's Intro (1929) and other book that willbe fairly easy to find. $\endgroup$
    – Will Jagy
    Jun 27 at 17:49
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    $\begingroup$ Not as elegant, but you could also square the two equalities, then add them up: $$\require{cancel} a^2x^2 + b^2y^2 - \cancel{2abxy} + a^2y^2+b^2x^2 + \cancel{2abxy} = 257 \;\;\iff\;\; (a^2+b^2)(x^2+y^2)=257 $$ $\endgroup$
    – dxiv
    Jun 28 at 1:01

1 Answer 1

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Your solution is indeed valid, and as FShrike writes in the comments, quite elegant. I'm putting this confirmation in a community wiki answer to mark this question as answered. (I think this is how that works!)

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