So this is a question off Facebook (I know). $a, b$ are positive integers. The answer to the problem turns out to be $a-b=0$ as the only two obvious solutions are $a=b=0, 2$ (and we select two as it can't be zero).
My question is how do we prove this? I was trying
$a+b=ab$
$\dfrac{a}{a-1}=b$
But I can't proceed by substituting this in $a-b$ and solving the polynomial as I would have to first assume that $a-b=0$.
How do we prove this? How can we prove that the only value $a$ and $b$ can take is $2$ (and hence $a-b=0$)? Alternatively, how can we prove that if $a-b$ is non-zero, then $a$ and $b$ aren't integers? How can we prove that there are infinitely many solutions if the constraint is shifted to real numbers?