What is known about the number of integer solutions of $\frac{x+y}{x-y}=n$?

Question

I don't know much anything about Diophantine equations, so forgive me if this is trivial.

I'm looking at the equation $$\frac{x+y}{x-y}=n \quad\quad \text{for} \quad x,y\in\mathbb{Z}\setminus\{0\} \quad \text{and} \quad n\in\mathbb{Z}.$$

I haven't been able to find out whether or not a general solution is know, please do let me know if it is.

I'm counting the number of solutions for a given $x$ or $y$. For instance, if $x=1$ we have $$y=\frac{n-1}{n+1},$$ which has solutions $\{y,n\}=\{\{2,-3\},\{3,-2\},\{-1,0\}\},$ so three in total. (Note that the change of sign on $n$ maps $x$ to $y$ and $y$ to $x$, so the solutions for $y=1$ are $\{x,n\}=\{\{2,3\},\{3,2\},\{-1,0\}\}.$)

The sequence giving the number of solutions for $x=1,2,3,\cdots$ is $$3, 5, 7, 7, 7, 11, 7, 9, 11, 11, 7, 15, 7, 11, 15, 11, 7, 17, 7, 15,\cdots$$

and it is plotted below for the first thousand $x$:

The function giving the solutions is even around $0$, so the sequence for $x=-1,-2,-3,\cdots$ is also $3, 5, 7, 7,\cdots$.

OEIS doesn't find anything.

Q: What is known about this equation? What is known about the sequence? Can we find the number of solutions for an arbitrary $x$?

Answer 1

The fact that $x-y$ divides $x+y$ is a pretty strong one, one of the sort we can often do something with.

In this case, the simplest thing to do is to add them, so we know $(x-y) \mid 2x$. Or rather than expressed in terms of divisibility, we can rewrite the original equation into

$$ \frac{2x}{x-y} = n + 1$$

Now, it's easy. We can make a change of variable $z = x-y$ and $m = n+1$ to turn the equation into

$$ 2x = mz $$

which has, for a fixed $x$, a number of solutions precisely equal to the number of divisors of $2x$.

However, your problem has additional constraints: you required $y \neq x$ and $y \neq 0$, or equivalently $z \neq 0$ and $z \neq x$. $z = 0$ is never a solution, but $z = x$ always is.

Thus, the number of solutions to the problem you give is precisely equal to one less than the number of divisors of $x$.

An explicit formula is to write down the prime factorization of $x$

$$ x = (-1)^s 2^d \prod_i p_i^{e_i} $$

and then the number of solutions will be

$$ (-1) + 2 (d+2) \prod_i (e_i + 1) $$

Answer 2

Note that we can rephrase the question as: for a given $x\neq 0$, how many $y\neq 0$ are there such that $n=\frac{x+y}{x-y}$ is an integer?

To answer this, note that $\frac{x+y}{x-y}=\frac{2x-(x-y)}{x-y}=\frac{2x}{x-y}-1$, so we want to know for how many $y\neq 0$ $\frac{2x}{x-y}$ is an integer, or equivalently, $x-y$ is a divisor of $2x$.

This being said, since $x-y$ takes every integer value other than $x$ exactly once for various $y\neq 0$, we are looking at the number of integer divisors of $2x$ other than $x$. With a little thought, you will find this is, using standard notation (see here), $2\tau(|2x|)-1$ (the factor $2$ appears because you also count negative divisors, and $-1$ appears because we ignore divisor $x$.