Egyptian Fractions?

Note that if $a,b,c>3$ the given sum is impossible. Similarly, $a,b,c>1$ . Thus, at least one of our integers must be $2$ or $3$ . WLOG, say it is $a$ . Then we have the following cases.

• $a=2$ . Then $\frac{1}{b} + \frac{1}{c} = \frac{1}{2}$ . We know that $b>1$ . If $b=2$ then $\frac{1}{c}=0$ , a contradiction. If $b=3$ then $c=6$ (first solution). If $b=4$ then $c=4$ (second solution). If $b\geq 5$ then $\frac{1}{c} \geq \frac{3}{10} \implies c \leq 3$ . The only integers meeting this criteria are $3$ and $2$ , which we have already considered.

• $a=3$ . Then $\frac{1}{b} + \frac{1}{c} = \frac{2}{3}$ . We now consider possible $b>2$ (if $b=2$ then we are in the first case). If $b=3$ then $c=3$ (third solution). If $b\geq 4$ then $\frac{1}{c} \geq \frac{5}{12} \implies c \leq 2$ . Again, we have already considered these cases.

Thus there are $\boxed{3}$ solutions.