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

For $x=2$ we have: $$\frac{2}{y}-\frac{4}{z}=\frac{1}{2}\iff \frac{2}{y}=\frac{8+z}{2z}\iff y=\frac{4z}{z+8}\iff y=\frac{4(z+8)-32}{z+8}\iff y=4-\frac{32}{z+8}.$$ 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 $(x,y,z)=(2,3,24)$.

$$\text{For }y=1\text{ we have: }\frac{1}{x}-\frac{4}{z}=-1\iff \frac{4}{z}=\frac{1+x}{x}\iff z=\frac{4x}{1+x}=4-\frac{4}{1+x}.$$ 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: $\frac{1}{x}-\frac{4}{z}=0\iff z=4x\iff x=\ell ,z=4\ell$, 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 $$\begin{gathered} \text{form: }(x,y,z)=(1,k,2k),\text{ k is a positive integer,} \\ (x,y,z)=(\ell ,2,4\ell ),\ell \text{ is a positive integer,} \\ (x,y,z)=(3,1,3)\text{ and }(x,y,z)=(2,3,24). \end{gathered}$$