I asked this question yesterday, and I think I have come up with an (albeit incomplete) proof of my own that I want to complete. The proof proceeds thus:
Given $x + y = xy$, we infer that $x, y$ are even. Hence, we can take $x = 2x_1$ and $y = 2y_1$ where $x_1, y_1 > 1$ [1]. Substituting this, we get $2x_1 + 2y_1 = 4x_1y_1 \implies x_1 + y_1 = 2x_1y_1$. Similarly, we see that $x_1, y_1$ are even [2] as well, which means $x_1 = 2x_2$ and $y_1 = 2y_2$. Substituting again, we get $x_2 + y_2 = 4x_2y_2$. This repeats infinitely, implying that $x, y$ are divisible by all powers of two. We know this is true only for $(0, 0)$. Hence, our only solutions are $(2, 2)$ and $(0,0)$.
I have the following problems with this proof, the list index refers to parts of the proof:
- Am I allowed to make such an assumption?
- What can I do about the case where $x_n, y_n$ are both odd? I've tried taking $x_1 = 2x_2 + 1$ and $y_1 = 2y_2 + 1$, but I get a contradiction ($x_2 + y_2 = -4x_2y_2$). So do I conclude they can't both be odd?
I am aware this is a triviality to solve using algebra, but I really want to understand the concept of infinite descent through this question.
The question admits of no solution via descent, as there is a single solution for $x,y>0$, not a series of hypothetical solutions that descend to one actual solution.
The starting equation is symmetrical in $x,y$. The trivial solution $(0,0)$ is available by inspection. For $x,y>0$, simple division by either $y$ or $x$ gives $$x=1+\frac{x}{y}\\ y=1+\frac{y}{x}$$ Since $x,y \in \mathbb Z$, it follows $\frac{x}{y},\frac{y}{x}\in \mathbb Z$
Together these imply both $x\ge y$ and $y\ge x$, whence $x=y$
Ergo $\frac{x}{y}=1,\frac{y}{x}=1$, and $x=2,y=2$