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How to solve the following congruence for $x,y$ integers such that $x\neq\pm y$? $$x+y\equiv 2x^2y^2 \pmod {x^4-y^4}.$$

By brute force search in Maple it seems it has only four solutions $(-1,0), (1,0), (0,-1)$ and $(0,1)$, but I couldn't find a proof that there are no more (checked the range $|x|,|y| \leq 1000$ so far). I have noticed that solving pair of congruences $x+y\equiv 2x^2y^2 \pmod {x^2-y^2}$ and $x+y\equiv 2x^2y^2 \pmod {x^2+y^2}$ simultaneously seems to lead to the same four solutions (this at least allows to write this as quadratic equations in $x$ or $y$).

This came from thinking about this question, the above congruence can be seen as a generalization of that equation.

Sil
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  • I checked (1, 1) is also the solution. – sirous Sep 03 '22 at 13:29
  • @sirous I excluded $x=\pm y$ to avoid zero modulus, but yes if we would allow it $(1,1)$ and $(0,0)$ could be considered – Sil Sep 03 '22 at 13:33
  • does this helps to prove those you found are only solutions: We write the congruence like:$x+y=k(x^4-y^4)+2x^2y^2$ we can see that for every k , x=y=1 is a solution. – sirous Sep 03 '22 at 18:31

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