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

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?

• 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. Jan 9, 2021 at 14:24
• 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. Jan 9, 2021 at 14:34
• @ChristianBlatter complete proof in zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf Jan 9, 2021 at 15:51
• @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." Jan 9, 2021 at 15:56
• @WillJagy Very nice, thank you! Even in German! Jan 9, 2021 at 17:51