Find all $x,y$ such that $\frac{1}{x} - \frac{1}{y} \in \mathbb{Z}$ Let $C=D=\mathbb{Z} \backslash \{ 0 \}$. Define a relation $S$ from $C$ to $D$ as follows:
For all $(x,y) \in C \times D$, 
$$(x,y) \in S \Leftrightarrow \frac{1}{x} - \frac{1}{y} \in \mathbb{Z}$$
Question: Write $S$ as a set of ordered pairs. 
Attempt: Clearly the ordered pairs of the form $(x,x)$ and $(-x,-x)$ satisfy the relation. Other than this two trivial pairs, what other pair can we obtain? 
 A: You can start with the fact that $\frac1x, \frac1y \in [-1,1]$, meaning that $\frac1x-\frac1y\in[-2,2]$
This means you have three options:


*

*$\left|\frac1x -\frac1y\right| = 0$. You already covered that.

*$\left|\frac1x -\frac1y\right| = 1$. It is obvious that taking $x=\pm2, y=\mp 2$. These are also the only solutions, since if $|x|\geq 3$, having $\frac1x+\frac1y$ equal $1$ or $-1$ would mean that $\frac1y$ must have an absolute value of more than one half, which is impossible unless $y=1$, but $y=1$ does not make the sum equal to $1$.

*$\left|\frac1x -\frac1y\right| = 2$. This is obviously only possible with $x=\pm1, y=\mp1$

A: Suppose $\frac{1}x - \frac{1}y = n$ for some $n \in \mathbb{Z}$. Then we have $y-x = nyx$. We see that taking $\pmod y$ gives $x \equiv 0 \pmod y$ and taking $\pmod x$ gives $y \equiv 0 \pmod x$. Thus $x|y, y|x \implies y = \pm x$. $y = x$ is a valid solution. $y = -x$ gives $\frac{2}x$ so the only working solutions in this case are $(-1, 1), (-2, 2)$. 
A: Another answer, not as slick as the others perhaps but I don't see the harm in posting it anyway:
Since $\frac{1}{x}, \frac{1}{y} \in [-1,0)\cup(0,1]$, the difference $\frac{1}{x}-\frac{1}{y}$ can only take on one of $5$ possible values: $0,\pm1,\pm2$.
Case-by-case, we have
$$\begin{align} \frac{1}{x}-\frac{1}{y} = 0  &\Longleftrightarrow x = y \\
\frac{1}{x}-\frac{1}{y} = \pm1 &\Longleftrightarrow x = \frac{y}{1 \pm y} \\ &\Longleftrightarrow y = \mp 2 \hspace{5mm} \text{(since $x,y$ must be integers)} \\ \frac{1}{x}-\frac{1}{y} = \pm 2 &\Longleftrightarrow x = \frac{y}{1 \pm 2y} \Longleftrightarrow y = \mp1 \hspace{5mm} \text{(since $x,y$ must be integers)}\end{align} $$
So our full set of solutions for distinct $x,y$ is $(-1,1),(1,-1),(-2,2),(2,-2)$. If we allow $x=y$, then $(x,x)$ is a solution for any $x \in \mathbb{Z}\backslash\{0\}$. 
