For which values of integer $k$, does the equation $x^2+y^2+z^2=kxyz$ have positive integer solutions $(x, y, z)$

I immediately thought of saying that from symmetry we have that $x\le y \le z$.

Also, $y^2+z^2 \equiv 0 mod x$, $x^2+z^2\equiv 0mody$ and $x^2+y^2\equiv 0modz$.

Moreover through trial and error I worked out that the solutions for $k$ must be $k=1$ or $k=3$ but I have not managed to prove it. I attempted to use inequalities, but that didn't work out either. Could you please explain to me how to solve this question?

  • $\begingroup$ For $k=2$ there a re no solutions, see this post. The case $k=3$ is called "Markoff's equation". Obviously then $(x,y,z)=(1,1,1)$ is a solution. $\endgroup$ Jan 9, 2021 at 14:24
  • $\begingroup$ The triple $(5,29,433)$ is a solution for $k=3$, and there are many more. There are no solutions $x\leq y\leq z\leq1000$ with $k\geq4$. Maybe you should try to prove that $k\geq4$ is impossible. $\endgroup$ Jan 9, 2021 at 14:34
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    $\begingroup$ @ChristianBlatter complete proof in zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf $\endgroup$
    – Will Jagy
    Jan 9, 2021 at 15:51
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    $\begingroup$ @DietrichBurde complete proof in zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf I have been able to use the main idea, that of looking for fundamental solutions, for many problems on this site that amount to "Vieta Jumping." $\endgroup$
    – Will Jagy
    Jan 9, 2021 at 15:56
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    $\begingroup$ @WillJagy Very nice, thank you! Even in German! $\endgroup$ Jan 9, 2021 at 17:51

1 Answer 1


this is called a CW answer; recommend beginning with

Equation with Vieta Jumping: $(x+y+z)^2=nxyz$.

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https://artofproblemsolving.com/community/c6h1726220_if_fraction_integer_then_equal_to_5 Poland 1991 training


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