Positive Integer Solutions to $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{4}{5}$

Question

I'd like to ask how to generate all positive integer solutions to the equality $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{4}{5}$. How about all integer solutions?

Some solutions include $x=2, y=5, z=10$ and $x=2, y=4, z=20$.

Answer 1

Hint: If $k= \min \{x,y,z \}$ then $$x,y,z \geq k \\ \frac{1}{x}, \frac{1}{y} , \frac{1}{z} \leq \frac{1}{k}\\ \frac{4}{5}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \leq \frac{3}{k} \\ 4k \leq 15 \\ k \leq 3 $$

Answer 2

Hint: if $x,y,z$ are all $\geq 4$ you get $\frac 4 5 \leq \frac 1 4 +\frac 1 4 +\frac 1 4 =\frac 3 4$ which is false. $x=1$ is also not possible, so $x=2$ or $x=3$. That should make things simpler.

Answer 3

This problem is most easily solved case by case. Let's first just for simplicity look for positive solutions where $x\leq y\leq z$. Then if $x\geq 4$, the whole sum cannot possibly become large enough, so $x$ is either $2$ or $3$.

If $x = 2$, then $\frac1y + \frac1z = \cdots$ (and continue from there).

If $x = 3$, then $\frac1y + \frac1z = \cdots$

If we allow negative numbers as well, then assuming $|x|\leq |y|\leq |z|$ we solve it the exact same way, except $x$ is also allowed to be $1$ in addition to $2$ and $3$.