<s>[INST] Find all integer solutions to the equation $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$ I was going through one of my Mathematics books and I came to this problem:
Find all integer solutions to the equation: $\frac{xy}{z}+\frac{xz}{y}+\frac{zy}{x}=15$
I tried a few things to begin with but none went in the right direction, any suggestions? 
Another thing, if we have the left side of the equation to be equal to 9, there are supposed to be 16 solutions in integers. What way can they be found? 
 [/INST] Since obviously $xyz\ne 0$, you have $$(xy)^2+(xz)^2+(yz)^2=15xyz\iff (y^2+z^2)x^2-(15yz)x+(yz)^2=0$$ hence the discriminant must be non-negative.
$$(15yz)^2-4(y^2+z^2)(yz)^2\ge 0$$ It follows $$y^2+z^2\le 56$$ This involves only the set $\mathcal S$ of  the first seven squares
$$\mathcal S=\{1,4,9,16,25,36,49\}$$ Searching for $x=y=z$ one has $3x^4=15x^3$ whose only solution is $x=5$ which can be done with $y=\pm 5$ and $z=\pm 5$.
We make then a table in which we discard the cases $x=y=z$ and also $y=z$ because in those cases the equation $2X^2-15X+1=0$ has no rational solution.
$$\begin{array}{|c|c|}\hline x & y^2+z^2 & yz\\\hline 1&5&4\\\hline1&10&9\\\hline 1&17&16\\\hline1&26&25\\\hline1&37&36\\\hline1&50&49\\\hline2&13&36\\\hline2&20&64\\\hline2&29&100\\\hline2&40&144\\\hline2&53&196\\\hline 3&25&144\\\hline3&34&225\\\hline 3&45&324\\\hline4&41&400\\\hline4&52&576\\\hline\end{array}$$
Because of the congruence $$x^2(y^2+z^2)+(yz)^2\equiv 0\pmod 5$$ the discarding of this sixteen it is straightforward. Thus there are only four solutions basically refered to the $(|x|,|y|,|z|)=(5,5,5)$  changing1 signes   in two of the variables. 
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