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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.

Thanks in advance!

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Well, you got the solution for $x^4+y^4=41z^2$? –  user63477 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. –  Hecke 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 $x^2+y^2=d*z^2$. –  Hecke Apr 21 '13 at 17:58
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