A number theory problem by Trung Le

We can use Simon's Favorite Factoring Trick

$xy=x+y\\ xy-x-y=0\\ x\left( y-1 \right) -y=0\\ x\left( y-1 \right) -y+1=1\\ x\left( y-1 \right) -\left( y-1 \right) =1\\ \left( x-1 \right) \left( y-1 \right) =1$

Since we are looking for integral solutions and all the terms on LHS are integral, we express 1 as $a\times b$ where $a$ and $b$ are integers and then use it to solve the problem.

$1=1\times 1\\ 1=\left( -1 \right) \times \left( -1 \right)$

For $a = b = 1$ ,

$x-1=1, \quad y-1=1\\ x=2, \quad y=2$

which is a valid solution. For $a = b = -1$ ,

$x-1=-1, \quad y-1=-1\\ x=0, \quad y=0$

which is invalid because $x$ and $y$ are 0, and they don't satisfy the given restrictions.

Hence $x = y = 2$ is the only integral solution.

The solution is Yes. Prove: Let x and y be non-zero integers and x + y = xy. The equation is equivalent with xy - x - y + 1 = 1, or (x-1)(y-1) = 1 This can happen only if x-1 = y-1 = 1, or x-1 = y-1 = -1 For x-1 = y-1 = 1, we have x = 2, y = 2 For x-1 = y-1 = -1, we have x = 0, y = 0. The second solution is rejected because x and y are non-zero integers. The only values of x and y are x=2 and y = 2.

from the equation, x|y and y|x. Therefore x = y, and the equation becomes 2x = x^2, so x = 2, since x .ne. 0. Ed Gray