<s>[INST] Solve $ \frac{1}{xy} = \frac{x}{yz} + 1, \:\: \frac{1}{yz} = \frac{y}{zx} + 1, \:\: \frac{1}{zx} = \frac{z}{xy} + 1. $ Solve the following system for positive integers:
$$ \frac{1}{xy} = \frac{x}{yz} + 1,  $$
$$ \frac{1}{yz} = \frac{y}{zx} + 1,  $$
$$ \frac{1}{zx} = \frac{z}{xy} + 1.  $$

Attempt:
Equivalent to
$$ z = x^{2} + xyz,  $$
$$ x = y^{2} + xyz,  $$
$$ y = z^{2} + xyz.  $$
Multiplying all:
$$ xyz = (x^{2}+xyz)(y^{2}+xyz)(z^{2}+xyz) = (xyz)^{2} + (xy)^{3}z + (xz)^{3}y + (yz)^{3}x + x^{4}(yz)^{2} + y^{4}(xz)^{2} + z^{4}(xy)^{2}  $$
 [/INST] We see that the first equation is equivalent to:
$$\frac{1}{xy} = \frac{x}{yz} + 1\ \Rightarrow z = x^{2}+xyz \ \Rightarrow z(1-xy) = x^{2}$$
Since $x^{2}$ and $z$ are positive, we must have $1-xy > 0$. However, since $x$ and $y$ are positive integers, this is impossible, so the system has no solution. $\blacksquare$
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