Positive pairs, anyone?

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

For $1 = 1 \times 1$ , $x = 2$ and $y = 2$ .

For $1 = -1 \times -1$ , $x = 0$ and $y = 0$ .

Since $x$ and $y$ are positive, the only pair is $x = 2$ and $y = 2$ .

$x+y=xy$ y=\(\frac{x}{x-1} =1+ $\frac{1}{x-1}$ ) For y to be positive, $\frac{1}{x-1}$ must also be positive, which is only true when $x=2$ (for positive values of $x$ )

Let y = cx. Then x + cx = cx^2. Either x = 0, or dividing by x, 1 + c = cx. Then c|1. so c = 1; Then 2 = x, and y = cx = 2. Ed Gray