How many integral ordered pairs satisfy this equation

Question

Suppose $n, x, y$ are positive integers

How many ordered pairs (x, y) are there with $\dfrac{xy}{x+y} = n$

Answer 1

$$\frac{xy}{x+y}=n\\\Rightarrow xy=n(x+y)\\\Rightarrow xy-n(x+y)=0$$

As an attempt to factorise, and noting symmetry, we add $\pm\ n^2$ to the LHS, which gives

$$xy-n(x+y)+n^2=n^2\\\Rightarrow (x-n)(y-n)=n^2$$

All we need to do now is count the number of such pairs. But this is simply $$2\tau(n^2),$$

since we are counting integral pairs.

Answer 2

Taking reciprocals, $$ \frac1x+\frac1y=\frac 1n$$ from which we see that necessarily $x,y>n$. So write $x=n+u$, $y=n+v$. Transform the original equation to $$ xy=n(x+y)$$ Now substituting gives $$ (n+u)(n+v)=n(2n+u+v).$$ Expand this and arrive at a surprising simple equation that helps us relate the number of solutions with the number of positive divisors of $n^2$.