Solve $$\frac{1}{x}+\frac{1}{y}=\frac{1}{2}+\frac{1}{z},$$ where $x, y, z$ are positive integers.
So far I have:
If $x=y=z \Rightarrow(x, y, z) = (2, 2, 2)$.
If $x=z \Rightarrow (x, y, z) = (k, 2, k)$, where $k$ is natural.
If $y=z \Rightarrow (x, y, z) = (2, k, k)$, where $k$ is natural.
If $x=y \Rightarrow (x-4)(z+2) = -8 \Rightarrow x = 1, 2,3$. The first two cases yield already known solutions. The third yields $(x, y, z) = (3, 3, 6)$.
Now I tried to look at the case where $a \ne b \ne c$ and try to establish some bounds (e.g. $z > x, y$) but this has not yielded anything useful.
By symmetry, we assume $x\leq y$ throughout this discussion.
Cases:
If $x=1$, then the equation simplifies to $\frac{1}{2}+\frac{1}{y}=\frac{1}{z}$. Then $z=1$ since otherwise the LHS is larger than the RHS. When $z=1$, we have $y=2$.
If $x=2$, then the equation simplifies to $\frac{1}{y}=\frac{1}{z}$, which can be solved for any $y$ and $z$.
If $x=3$, then we have $3\leq y<5$ as possible left-hand-sides (each of which can be checked).
If $x=4$, then $y\geq 4$ and the LHS is less than the RHS and there are no solutions.