Hellenic Mathematical Olympiad

If $x \ge 3$ and $y \ge 3$, then we have:
$$\frac{1}{x} + \frac{2}{y} - \frac{4}{z} \le \frac{1}{3} + \frac{2}{3} - \frac{4}{z} = 1 - \frac{4}{z} < 1,$$
and hence the equation is not satisfied. Hence we may have: $x \le 2$ or $y \le 2$.

* For $x=1$ we have: $\frac{2}{y} - \frac{4}{z} = 0 \Leftrightarrow z = 2y \Leftrightarrow y = k, z = 2k$, where $k$ is a positive integer. Hence $(x, y, z) = (1, k, 2k), k \in \mathbb{Z}$ positive.

* For $x=2$ we have:
$$\frac{2}{y} - \frac{4}{z} = \frac{1}{2} \Leftrightarrow \frac{2}{y} = \frac{8+z}{2z} \Leftrightarrow y = \frac{4z}{z+8} \Leftrightarrow y = \frac{4(z+8)-32}{z+8} \Leftrightarrow y = 4 - \frac{32}{z+8}.$$
Since $y$ is positive integer, it follows that $z+8$ must be a positive divisor of $32$ greater than $8$. Therefore $z=8$ or $z=24$, and finally we find the solutions: $(x, y, z) = (2, 2, 8)$ and $(x, y, z) = (2, 3, 24)$.

$$\text{For } y=1 \text{ we have: } \frac{1}{x} - \frac{4}{z} = -1 \Leftrightarrow \frac{4}{z} = \frac{1+x}{x} \Leftrightarrow z = \frac{4x}{1+x} = 4 - \frac{4}{1+x}.$$
Since $z$ must be positive, $1+x$ must be positive divisor of $4$ greater than $1$. Therefore we have $x=1$ or $x=3$, and finally we find the solutions
$$(x, y, z) = (1, 1, 2) \text{ or } (x, y, z) = (3, 1, 3).$$

* For $y=2$ we have: $\frac{1}{x} - \frac{4}{z} = 0 \Leftrightarrow z = 4x \Leftrightarrow x = \ell, z = 4\ell$, where $\ell$ is a positive integer. In this case we find the solutions: $(x, y, z) = (\ell, 2, 4\ell)$, where $\ell$ is a positive integer.

Hence, taking in mind overlapping of solutions we can write the solutions in the
$$\text{form: } (x, y, z) = (1, k, 2k), \text{ k is a positive integer,} \\ (x, y, z) = (\ell, 2, 4\ell), \ell \text{ is a positive integer,} \\ (x, y, z) = (3, 1, 3) \text{ and } (x, y, z) = (2, 3, 24).$$