<s>[INST] Positive integers $(a,b, c)$ such that $\frac{1}{a}+ \frac{1}{b} + \frac{1}{c}$ is an integer Are $$(a, b, c) = (\mid 1 \mid, \mid 2 \mid, \mid 2 \mid), (\mid 2 \mid, \mid 4 \mid, \mid 4 \mid), (\mid 2 \mid, \mid 3\mid, \mid 6 \mid), (\mid 1 \mid, \mid 1 \mid, \mid 1 \mid)$$ the only integers such that 
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} $$ is an integer ?
 [/INST] Assuming we only care about positive integers for the time being, notice that if $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ is an integer then it must be at least $1$. If three numbers add up to at least $1$, then at least one of them must be at least $\frac{1}{3}$ - so at least one of $a$, $b$, and $c$ must be no more than $3$.
If none of them is $2$ or less, then $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$ are all no more than a third. But if they add up to at least $1$, then they must all then be exactly $\frac{1}{3}$. So $a = b = c = 3$.
Say the smallest number we have is $2$. Then we can't have two of them (because if $\frac{1}{2} + \frac{1}{2} + \frac{1}{c}$ is a whole number, then $c = 1$ and that's smaller than $2$). But $\frac{1}{2} + \frac{1}{b} + \frac{1}{c}$ can only be at least $1$ if $\frac{1}{b} + \frac{1}{c}$ is at least $\frac{1}{2}$, so we need $\frac{1}{b},\frac{1}{c}$ to be at least $\frac{1}{4}$. Therefore at least one of $b$ and $c$ is at most $4$. So we have either $\frac{1}{2} + \frac{1}{3} + \frac{1}{c}$ (in which case $c = 6$) or $\frac{1}{2} + \frac{1}{4} + \frac{1}{c}$ (in which case $c = 4$).
So far we have $(3, 3, 3)$, $(2, 3, 6)$, and $(2, 4, 4)$, and we've found all of the ones that don't involve a $1$. I'll leave it to you to try to apply this approach to the case when we do have a $1$.
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