Number of integral solutions for an equation

Question

How do we approach this kind of problem of finding number of positive integral solutions to $$\frac{1}{x}+\frac{1}{y} = \frac{1}{n!}$$

Here $n$ is given.

Answer 1

$$n!(x+y)=xy$$ $$xy-n!(x+y)=0$$ $$xy-n!(x+y)+(n!)^2=(n!)^2$$ $$(x-n!)(y-n!)=(n!)^2$$

Now, one can express $(n!)^2$ as a product of two factors and generate solutions for $x$ and $y$.

Answer 2

Let $(x,y)$ be a solution. We isolate $x$ : $\frac{1}{x} = \frac{1}{n!} - \frac{1}{y}$. We need $y>n!$. Also $x = \frac{n! y}{y - n!}$.

Hence $y - n! \mid n! y$. The number of pairs solutions $(x,y)$ is the number of integers $y>n!$ such that $y - n! \mid n! y$. Let us find the number of such integers.

We write $y = n! + \alpha$. Then it becomes $\alpha \mid n!(n!+\alpha)$, so $\alpha \mid (n!)^2$. Denoting $d( (n!)^2)$ the number of divisors of $(n!)^2$, there are exactly $d( (n!)^2 )$ possible values for $\alpha$, therefore for $y$, therefore for the pairs.

Finally there are $d( (n!)^2 )$ solutions (number of divisors of $(n!)^2$)

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(For $n=10$, you find $2295$ non-ordered pairs)