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Solve the equation: $x^4+y^4=d~z^2,$ where $x,y,z$ are positive integers,and $d>1$ is a given square-free integer.

I know if $p$ is an odd prime and $p|d,$ then $t^4\equiv -1 \pmod p$ is solvable,so $p\equiv 1 \pmod 8$,but this is not sufficient.

PS:I just want to know how to determine whether $x^4+y^4=d~z^2$ has any positive integer solutions,not ask how to get all the solutions.

jjagmath
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lsr314
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  • Well, you got the solution for $x^4+y^4=41z^2$? – Inceptio Apr 21 '13 at 17:31
  • @Inceptio,Not yet,I just want a more general solution,so I removed the equation $x^4+y^4=41z^2$ from the post.But if you have any ideas,you can show it with this special equation. – lsr314 Apr 21 '13 at 17:35
  • Using the methode that I explain here you can find integral solutions of $x^2 + y^2 = 41z^2$, by reducing them to rational solutions of $x^2 + y^2 = 41$ and noting that $5^2 + 4^2 = 41$. Then I would hope that there is an easy way to find out if these solutions can be squares. (For instance, hope that they are $2$ or $3$ mod $4$.) – Myself Apr 21 '13 at 17:46
  • @Myself,thank you,but it's difficult to find a square solution in the solutions of $x^2+y^2=d*z^2$. – lsr314 Apr 21 '13 at 17:58
  • If d is congruent to 3 mod 4 then it has no solutions – Guruprasad Mar 22 '25 at 23:07
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    Somewhat related: https://math.stackexchange.com/questions/367398/on-the-elliptic-curve-x4y4-193z2 also https://math.stackexchange.com/questions/26548/how-to-get-fermats-descent-working-on-the-conic-x4y4-2z2 and https://math.stackexchange.com/questions/339410/x4y4-2z2-has-only-solution-x-y-z-1 – Gerry Myerson Jun 06 '25 at 02:32

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