Find all integer solutions to $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$ 
Find all integer solutions $(x, y)$ of the equation 
  $$\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$$

What have done is that: 
$$\frac{1}{x}= \frac{2y-3}{3y}$$
so, 
$$x=\frac{3y}{2y-3}$$
If $2y-3 = +1 \text{ or } {-1}$, $x$ will be an integer, so we choose integer $y$ 
to make $2y-3=1 \text{ or } {-1}$. $y = 2$ or $y = 1$ is such a solution.
Also, $2y - 3$ can be deleted by numerator $3$, so $2y - 3$ can be $3$ or $-3$ too. This gives $y = 3$ or $y = 0$, but $y$ can not be $0$. So far, we have $y=1,2,3$. Finally, $2y-3$ can be deleted by numerator $y$, but how can we find such a $y$? 

If $0<x\le y$ then $\frac23 = \frac1x+\frac1y \le \frac1x+\frac1x$ and so $x\le3$.
If $x<0$ then $\frac23 < \frac1{-x}+\frac23=\frac1y$ and so $y=1$.
Therefore $(x,y) \in \{ (3,3) , (-3,1) , (1,-3), (2,6),(6,2)\}$.
