Egyptian fractions

Re-arranging, you get that

$\frac{1}{n}=\frac{x+y}{xy}$

so $n(x+y)=xy$ . We know $x>n$ and $y>n$ , so let $x-n=a$ and $y-n=b$ . $n(a+b+2n)=(a+n)(b+n)$

We may cancel to get $ab=n^2$ . Since there are the same number of solutions of $(a,b)$ as $(x,y)$ we really are interested in how many factors $n^2$ has. You already know its prime factorization:

$n^2={P_1}^2{P_2}^2...{P_{10}}^2$

You can choose 0, 1, or 2 of each prime factor to find $a$ and then $b=\frac{n^2}{a}$ . So there are $3^{10}$ ordered pairs!

It will be proven that for arbitrary primes $p_{1}, p_{2}, ..., p_{10}$ , the number of ordered solution $(x, y)$ is exactly $3^{10}$ with $x, y \in N$ .

Obviously the equality $\frac{1}{n} = \frac{1}{x} + \frac{1}{y}$ is coungruent to the equality $(x-n)(y-n) = n^{2}$ .

For $n = p_{1} p_{2} p_{3} ...p_{10}$ . Therefore, $n^{2} = p_{1}^{2} p_{2}^{2} ...p_{10}^{2}$ .

Because $p_{k} \in prime \forall k \in N$ , therefore $n^{2}$ have exactly $3^{10}$ positive factors. It implies that there exist exactly $3^{10}$ different possible values of $(x-n)$ . Therefore, so does $n$ .