Nonnegative Integer solutions of $x+y-xy=0$

Question

I would like to see other methods, besides the one I use here to find all the nonnegative integer solutions of an equation like $$x+y-xy=0$$.

This is the one I used:

First we note that for $x=1$ we get $1+y-y=0$, but $1+y-y=1$ so $x\neq 1$. Then as $x\neq 1$ we have $y=\frac{x}{x-1}=1+\frac{1}{x-1}$. Furthermore $y(0)=0$ and $y(2)=2$ so $x=y=0$ and $x=y=2$ are solutions. Then if $x>2$, $x-1>1>0$ so as $x-1>1$ we have $\frac{1}{x-1}<1$ and as $x-1>0$ we have $\frac{1}{x-1}>0$ so $0<\frac{1}{x-1}<1$ for $x>2$, and so $1<1+\frac{1}{x-1}<2$ for $x>2$. As $y=1+\frac{1}{x-1}$, we get that for $x>2$, $1<y<2$. As there is no integer between $1$ and $2$, for $x>2$, $y$ can't be a integer. So the only nonnegative solutions of the equations are $x=y=0$ and $x=y=2$.

How else can I prove this, using less theory (For example, I think it can be proven using only Peano and a little more)?

Answer 1

$$ xy - x - y = 0 $$ $$ xy - x - y + 1 = 1 $$ $$ (x-1)(y-1) = 1. $$ With integer variables, the choices are $$ x-1 = 1, y-1 = 1, $$ or $$ x-1 = -1, y-1 = -1. $$

Answer 2

We have $$x+y-xy=0$$ $$\implies x+y=xy$$ $$\implies \text{sum of two non-negative integers}=\text{product of two non-negative integers}$$ For non-negative integer solution of the above equation, we have $x=0$, & $y=0$

Notice, $$\frac{x}{xy}+\frac{y}{xy}=\frac{xy}{xy}$$ $$\implies \frac{1}{x}+\frac{1}{y}=1$$ $$\implies \text{sum of inverse of two non-negative integers}=1$$ For non-negative integer solution of the above equation, we have $x=2$, & $y=2$

Answer 3

If $x+y-xy = 0$, then $1 =xy-x-y+1 =(x-1)(y-1) $.

Therefore $x-1$ and $y-1$ must be integers that divide $1$. The only solutions are $x=y=2$ and $x=y=0$.