Find all integer solutions to $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$

Question

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 $y$?

Answer 1

Another answer that does not follow your approach is like this:

Given integers $x,y$ such that $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$:

  $3x+3y = 2xy$.

  Thus $4xy - 6x - 6y = 0$ and hence $(2x-3)(2y-3) = 9$.

Now it remains to find all factorizations of $9$ as a product of two integers.

Answer 2

Your approach can be continued to a full solution. Let's see it in detail:

Given integers $x,y$ such that $\frac{1}{x} + \frac{1}{y} = \frac{2}{3}$:

  At least one of $x,y$ is positive, and by symmetry we can assume that $y$ is positive.

  $\frac{1}{x} = \frac{2y-3}{3y}$.

  Thus $x = \frac{3y}{2y-3} = 1 + \frac{y+3}{2y-3}$.

  If $y > 6$ then $y+3 < 2y-3$ and so $0 < \frac{y+3}{2y-3} < 1$, which makes $x$ not an integer.

  Thus $y \le 6$.

It only remains to check all $y$ from $1$ to $6$.

Answer 3

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 $(x,y) \in \{ (3,3) , (-3,1) , (1,-3), (2,6),(6,2)\}$.