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Can anybody help me find all integer solutions of the equation in the title? I have learnt that this kind of equation is solved by writing it in the form $x^4+y^4=z^2,$ but I can´t see how to use it here...

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  • $\begingroup$ Please use mathjax. $\endgroup$ May 29, 2019 at 19:50

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$$(x^2+y^2-13^2)^4=z^2-2z$$ so we add one to both sides: $$(x^2+y^2-13^2)^4+1^4=(z-1)^2$$

But we know that $a^4+b^4=c^2$ has only the trivial solution $(0,0,0)$ (attributed to Fermat by infinite descent, I believe) and thus your original equation has no integer solutions.

Edit Thanks to Mark Bennet for pointing out that I can't read; it also has the trivial solutions $(a,0,a^2)$ and $(0,a,a^2)$. The former is obviously impossible, and the latter gives $a=\pm1$ so $z=0$ or $2$, $x^2+y^2=13^2$ which is a pythagorean triple! Thus $x,y=\pm5,\pm12$ in some order.

Edit 2 as Thomas Andres pointed out in the comments, Fermat's result isn't actually needed to prove that $a^4+1=b^2$ has only trivial solutions.

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  • $\begingroup$ ack whoops i misread the theorem didn't I, thanks for the correction :) really appreciate it $\endgroup$
    – auscrypt
    May 29, 2019 at 19:57
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    $\begingroup$ You don't need Fermat to show that $a^2+1=b^2$ has no other solutions other than $a=0.$ $b^2-a^2=(b-a)(b+a)=1$ means that either $b-a=-1,b+a=-1,$ or $b-a=1,b+a=1.$ In either case, $a=0.$ $\endgroup$ May 29, 2019 at 20:00
  • $\begingroup$ @ThomasAndrews Yup, I agree fully, but OP did specifically mention it. I might just edit that in as an aside $\endgroup$
    – auscrypt
    May 29, 2019 at 20:01

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