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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?
WLOG let $x\leq y$. Then, we have that $x = 3$ because we are given that $x\geq 3$, and if $x\geq 4$, $\frac{1}{x}+\frac{1}{y}\leq \frac{1}{2}$.
Then, we must solve $\frac{1}{y}-\frac{1}{z} = \frac{1}{6}$. Note that $y = 3,4,5$ because they are the only values such that $y\geq 3$ and $\frac{1}{y}>\frac{1}{6}$. Then, we can go into casework:
For $y = 3$, we get $\frac{1}{z} = \frac{1}{6}$ and thus $z = 6$.
For $y = 4$, we get $\frac{1}{z} = \frac{1}{12}$ and thus $z = 12$.
Finally, for $y = 5$ we get $\frac{1}{z} =\frac{1}{30}$ and thus $z = 30$.
We must permute $x$ and $y$ when they are different to get all of the solutions because of the earlier WLOG.
Thus, the only solutions are $\boxed{(3,3,6),\ (3,4,12),\ (3,5,30),\ (4,3,12),\text{ and }(5,3,30).}$
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 $(x,y,z)$: $$(3,3,6),\qquad(3,4,12),\qquad(3,5,30).$$
