Is there any positive integer solutions of $x$, $y$, $z$ such that both $$\frac{x^2+y^2+z^2}{x+y+z}$$ and $$\frac{x^2+y^2+z^2}{xyz}$$ are integers?
Edit: Forgot to mention that $x$, $y$, $z$ are distinct positive integers.
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Where did you get this from? (Also x = y = z = 1.) – Parcly Taxel Aug 29 '16 at 11:35
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I can find two solutions right away: $(1,1,1)$,$(3,3,3)$. – Sarvesh Ravichandran Iyer Aug 29 '16 at 11:39
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darn I forgot to write x, y, z are distinct positive integers – user3137471 Aug 29 '16 at 11:50
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5What I wonder is the speed at which people down vote when they can surely understand that they can write something besides trivial solutions. Does this look like a homework question? It is just that I am curious about these set of equations and I can't find any non-trivial solution to it so would love to know analytical take. – user3137471 Aug 29 '16 at 11:54
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A remark, how to start: The second condition can be written as Diophantine equation $$ x^2+y^2+z^2=nxyz, $$ and it is easy to see that we must have $n=1$ or $n=3$. This is already very helpful. Compare also with this question.
For $n=3$ the triples $(x,y,z)$ with $\frac{x^2+y^2+z^2}{xyz}=3$ are known as Markov numbers, and we can produce a Markov tree. Now it gets easier if we also require that $(x,y,z)$ is a solution of the Diophantine equation $x^2+y^2+z^2=m(x+y+z)$.
Dietrich Burde
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Thanks Dietrich, it is really helpful and pushed me to the right direction. I will try to see if along the Markov Tree is there any solution possible for Diophantine equation $$x^2+y^2+z^2=m(x+y+z)$$ Also thanks for the Reference Will Jagy. – user3137471 Aug 29 '16 at 18:04
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1@WillJagy Yes, definitely! Also, I like it to read in German.... – Dietrich Burde Aug 29 '16 at 18:10
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Thanks, I think this is the closest we can get. Tried first million members of markov chain without any success. – user3137471 Aug 31 '16 at 11:23