Diophantine Equation #1

Firstly, $n$ should be factored to help with calculations later on. $n = 12345678987654321 = 3^{4} \times 37^{2} \times 333667 ^{2}$ .

Now for some algebra:

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

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

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

$nx + ny = xy$

Note that $x + y$ can never equal zero for positive integers $x$ and $y$ , so that last step is acceptable.

$nx + ny - xy - n^{2}= -n^{2}$

$(x - n)(n - y)= -n^{2}$

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

For $y$ to be a positive integer, $x$ must be greater than $n$ and $x - n$ must be a factor of $n^{2}$ . Let $k = x - n$ ; now the problem becomes finding the number of factors $k$ of $n^{2}$ . Since $n = 3^{4} \times 37^{2} \times 333667 ^{2}$ , $n^{2} = 3^{8} \times 37^{4} \times 333667 ^{4}$ . The number of factors of $n^{2}$ is thus $(8 + 1) \times (4 + 1) \times (4 + 1) = 225$ . But the problem is asking for distinct solutions, so the number of solutions $(x, y, n)$ is $1 + \frac{225 - 1}{2} = \boxed{113}$ . The number of distinct solutions is equal to the tuple $(2n, 2n, n)$ and half of all the other tuples because of symmetry.