Find all positive integers for the following question Find all positive integers that makes the result of $$\frac{1}{x}+\frac{1}{y}$$
integers
 A: We have: $n = \dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{xy}\to x+y=nxy\to x = nxy-y=y(nx-1)\to y = \dfrac{x}{nx-1}$. Thus $nx - 1 \leq x \to 1\leq n \leq \dfrac{x+1}{x} \leq 2$. Can you continue?
A: The expression $$\frac1x+\frac1y$$ where $x$ and $y$ are positive integers takes integer values only when $$x=y=1$$ OR $$x=y=2$$.
To prove this, assume without loss of generality that $x\geq y>2$. Notice that $$\frac1x\leq\frac1y<\frac12$$ so that $$\frac1x+\frac1y<2\left(\frac12\right)=1$$
Clearly, no integer values can be obtained. To show that the selected values force an integer outcome, simply substitute them.
EDIT: To handle the case when $x=1$ and $y>1$ and the case $x=2$ and $y>2$.
Observe that if $x=1$ and $y>1$, then the expression becomes $1+\frac1y$. As $0<\frac1y<1$ for this case, the result is not an integer. Similarly, if $x=2$ and $y>2$, then the expression is $\frac12+\frac1y$, but since $0<\frac1y<\frac12$, the result is again not an integer.
A: So to begin with, let's state the obvious. $(1,1)$ 
Then there come halves. We can sum the halves to zero or to a nonzero integer. By this I mean $(2,2)$
Other than that, if x or y > 2, then the maximum sum, which would be $\frac{2}{min(x,y)}$, would be less than one. If x or y was one in this situation, then the expression wouldn't equate to an integer. Plugging zero into the equation would make the sum undefined. 
