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Find the positive integral values $n$ can take for which :${(x+y+z)}^2=nxyz$ is solvable for positive integers.

My try: Due to symmetry WLOG let $x\ge y\ge z$.

Now ,making it in a quadratic in $x+y$,we have $$nxyz={(x+y)}^2+2(x+y)z+z^2$$ or $z| {(x+y)}^2$. But I don't know what to do with this. Also I tried playing with inequalities to get $$nyz\le 9x$$ and the equality condition $x=y=z$, yields $$nx=9$$ or $n$ can take the values $(1,3,9)$ but I can't find more.

I know this question has been asked before but I have not learned that much.

I also tried using a little bit of calculus but nothing promising came up.

V.G
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Albus Dumbledore
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  • @RobertZ the solutions are too complex and i have not learned about pells equation yet. – Albus Dumbledore Sep 05 '20 at 10:43
  • @Robert, there is a recent answer at the same question by https://math.stackexchange.com/users/54563/calvin-lin which uses only contest-level techniques, and considers just finding the appropriate coefficients $n$ – Will Jagy Sep 05 '20 at 12:23
  • @RobertZ Calvin: https://math.stackexchange.com/questions/1930438/equation-with-vieta-jumping-xyz2-nxyz/3599276#3599276 – Will Jagy Sep 05 '20 at 12:27
  • Quantum, Calvin Lin put an answer at that question using methods appropriate for you https://math.stackexchange.com/questions/1930438/equation-with-vieta-jumping-xyz2-nxyz/3599276#3599276 – Will Jagy Sep 05 '20 at 12:29
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    @WillJagy Thanks for pointing out! – Robert Z Sep 05 '20 at 12:33

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