Finding all integer solutions of $\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12$ If we have integers $x,y,z$ such that $x,y,z \ge 3,$ find all solutions to $$\frac{1}{x} + \frac{1}{y} - \frac{1}{z} = \frac12.$$

I was thinking of first expanding out this, and then simplifying from there, but the equation got very messy. Is there a better method?

As an alternative to Joshua Wang's excellent answer:
By symmetry we may assume, without loss of generality, that $x\leq y$. Clearing denominators yields
$$2xz+2yz-2xy=xyz,\tag{1}$$
and a bit of rearranging then shows that
$$xy(z+2)=2(x+y)z.$$
Of course $z+2>z$ and so $xy<2(x+y)$, or equivalently
$$(x-2)(y-2)<4.$$
Because $x\leq y$ we see that $x<4$, so $x=3$ and then $y<6$. Plugging into $(1)$ and rearranging shows that
$$yz+6y-6z=0,$$
or equivalently $z=\frac{6y}{6-y}$, yielding the following three solutions for $(x,y,z)$:
$$(3,3,6),\qquad(3,4,12),\qquad(3,5,30).$$
