Is there any solution such that..

Question

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.

darn I forgot to write x, y, z are distinct positive integers — user3137471, Aug 29 '16 at 11:50
What 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

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

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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 $x^2+y^2+z^2=m(x+y+z)$.

edited Apr 13 '17 at 12:20

Community

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answered Aug 29 '16 at 15:56

Dietrich Burde

130,978

reference I like http://zakuski.utsa.edu/~jagy/Hurwitz_A_1907.pdf – Will Jagy Aug 29 '16 at 17:59
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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@WillJagy Yes, definitely! Also, I like it to read in German.... – Dietrich Burde Aug 29 '16 at 18:10
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

score -1 · Answer 2 · answered Aug 29 '16 at 11:41

Special case

Well, if $x = y = z$ then your equations become

$$\frac{3x^2}{3x} = x$$

$$\frac{3x^2}{x^3} = \frac{3}{x}$$

If they are coupled, then you notice that $x = 1$ is the simplest result. There is also $x = 3$: $3/x$ is an integer if and only if $x = 1$ or $x = 3$.

Unless you count negative integers too, in that case you have also $x = y = z = -1$ and $x = y = z = -3$

Is there any solution such that..

2 Answers2