TITLE: Natural number solutions to $\frac{xy}{x+y}=n$ (equivalent to $\frac 1x+\frac 1y=\frac 1n$)
QUESTION [6 upvotes]: I have a question about the following problem from a Putnam review:
Let $n\in \mathbb{N}$. Find how many pairs of natural numbers $(x, y)\in \mathbb{N}\times \mathbb{N}$ solve
$$
\frac{xy}{x+y}=n.
$$
I have found some solutions for all $n$, such as $(2n, 2n)$ and $(n+1, n(n+1))$, but I feel as though this is the wrong approach, as the question is only asking for the number of solutions to the equation.
I don't really want a complete solution, but any hints would be greatly appreciated.

REPLY [5 votes]: This can be rewritten as 
$$\frac 1x+\frac 1y=\frac 1n$$
$$\frac 1y=\frac 1n-\frac 1x$$
$$y=\frac{nx}{x-n}$$
Changing variables to $k=x-n$:
$$y=\frac{n(n+k)}{k}$$
$$y=\frac{n^2}k+n$$
The left side is an integer. What does that tell you about the allowed values of $k$, and the corresponding number of valid $x$ values?