Solve over integers $\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} = 3$ I am trying to solve $\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} = 3$ over integers, but I have no idea what's the best way to do it. I have tried multiplying both sides by $xyz$ and then figuring out that $z$ divides $xy$, $x$ divides $yz$, etc, but with no effect.

It appears there are no non-trivial (i.e. not $x,y,z = \pm 1$) solutions: sorry, no number theory to do here. (Just some analysis.) Dividing everything by $xyz$ puts the equation in a nice form:
$\frac{1}{x^2} + \frac{1}{y^2}+\frac{1}{z^2} = \frac{3}{xyz}$. 
Staring at that for a moment, it seems like the LHS is always larger than the RHS, and this is easy to show. Suppose WLOG $x \leq y \leq z$. 
Case 1: $x < z^{2/3}$. Then 
$\frac{1}{x^2} + \frac{1}{y^2}+\frac{1}{z^2} \geq \frac{3}{z^2} >\frac{3}{x^3} \geq \frac{3}{xyz}.$
So there are no solutions with $x < z^{2/3}$. 
Case 2: $x \geq z^{2/3}$. Then 
$\frac{1}{x^2} + \frac{1}{y^2}+\frac{1}{z^2} \geq \frac{3}{z^2} > \frac{3}{z^{7/3}} \geq \frac{3}{xyz}$,
since the strict inequality $z^{-2} > z^{-7/3}$ holds for $z > 1.$ So there are no solutions in this case either. 
