4
$\begingroup$

No integer solution of this Diophantine equation $$x^4+y^4+1=z^2$$ is known other than the trivial ones.

While I was reading a paper of Don Zagier, I realized that his idea on the Euler's sum of powers conjecture might be helpful to solving the equation. He observed that if $$z^4-x^4-y^4=R^2-QS$$where $Q,R,S$ are quadratic homogeneous equation over $\mathbb{Q}$, then $Q=0$ is tangent to $z^4=x^4+y^4$ at four points. Then the parameterization of $Q=0$ proves that $z^4(p,q)-y^4(p,q)-x^4(p,q)$ is a square, where $z,y,x$ are quadratic homogeneous equation of $p,q$.

Question: Can the same trick be applied to $x^4+y^4+z^4$? I tried to find a suitable $Q$ but I failed.

$\endgroup$
3
  • $\begingroup$ I found $Q$ genus $1$ (didn't check if it has infinitely many rational points but some might have). Probably you don't want this since it doesn't allow parametrization. $\endgroup$
    – joro
    Nov 11, 2014 at 13:39
  • $\begingroup$ @joro: Yes. If a suitable quadratic $Q$ can be found, then it is possible to solve the equation by solving some pell equation. But I failed to find a suitable $Q$ tangent to $x^4+y^4+z^4=0$ at four points. $\endgroup$
    – Y. Zhao
    Nov 11, 2014 at 14:02
  • $\begingroup$ I suspect it exists. $\endgroup$
    – joro
    Nov 11, 2014 at 14:35

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.