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 ?
 A: 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 $1$.
A: We can consider the positive case, i.e. $a,b,c>0$.
We can categorize into the 3 cases(from simple to copmlex):
1): $a=b=c$. Then $\frac1a+\frac1b+\frac1c=\frac3a\in\mathbb Z$, so $a|3$, hence $a=1,3$. 
2): $0<a<b<c$. Then $1\leq\frac1a+\frac1b+\frac1c<\frac3a$, $1\leq a<3$, So $a=1,2$.
If $a=1$, then $\frac1b+\frac1c<\frac22=1$, impossible, so $a=2.$
Then $b\geq3$(since $b>a$), and
$$1\leq\frac1a+\frac1b+\frac1c<\frac12+\frac2b\leq\frac12+\frac23<2,$$
thus the sum equals to $1$. Moreover, if $b\geq4$, then $\frac1b+\frac1c<\frac24$, which means the sum is not an integer. 
  So $a=2,~b=3~,\Rightarrow c=6.$
3): $a=b<c$. By the method of  2),  we have $1\leq\frac2a+\frac1c<\frac3a,~\Rightarrow a<3$.
If $a=1$, then $\frac1c<1$,  impossible. 
If $a=2$, $\frac1c<1$, still impossible. 
Hence, in this case, there is no result.
4): $a=b>c$. By the above argument, $1\leq\frac2a+\frac1c<\frac3c$. 
$c=1,2$.
If $c=1$, then $a=2$.
If $c=2$, $\frac2a\in \mathbb Z+\frac12$, and $a\geq3$. So $a=4$.
PS: For the non-positive case, there are infinite solutions. 
e.g. $a=b=2n,~c=-n$, where $n\in \mathbb N^+.$
