Number of solutions in $\Bbb N^2$ of the equation $\frac1x+\frac1y=\frac1{1995}$

Question

How many solutions are there in $\Bbb N^2$ of the equation $\frac1x+\frac1y=\frac1{1995}$?

My working:

$\frac{x+y}{xy} = \frac1{1995} $
$\ 1995x+1995y=xy $
$\ y= \frac{1995x}{x-1995} $
For this to have integral solutions $\ x-1995\ |1995$, i.e. $\ x-1995\ |3\cdot5\cdot7\cdot19\cdot x$

Problem:
I don't know how to proceed further to find the number of solutions that satisfy this. It would be great if someone could help.

Answer 1

This is probably a duplicate, but, anyway.

If $\frac1x +\frac1y = \frac1{n} $ then, multiplying by $nxy$, $n(x+y) = xy $, so $0 =xy-n(x+y) =xy-n(x+y)+n^2-n^2 =(x-n)(y-n)-n^2 $ so $n^2 = (x-n)(y-n) $ (actually, I think I did this a while ago).

So, for each factorization $n^2 = uv$, set $x-n = u, y-n = v$ so $x = u+n, y = v+n$.

There are at least two distinct factorizations, $n^2 = 1\cdot n^2 = n \cdot n$. The solutions for these factorizations are $x = n+1, y=n+n^2$ and $x = y = 2n$.

Answer 2

$\ 1995x+1995y=xy$.

Well at this point I'd note there are a heck of a lot of common factors.

Let $d \gcd(x,y)$ and $x'd =x$ and $y'd = y.$

So $1995(x'+y') = x'y'd$ So $x'$ and $y'$ and both divide $1995(x'+y')$ but $\gcd(x',y'+'x) = \gcd(x',y') = (y',y'+x') = 1$ so $x'|1995$ and $y'|1995$. $1995 = 3*5*7*19.$

So possibilities are $x'=1, y'1, d= 3990$ works.

$x' =35; y'=19, d=132$ works, etc.

To formalize: $x' = 3^i_35^i_57^i_719^i_{19}; y'=3^j_35^j_57^j_719^j_{19}; (i_k, j_k) \in \{(0,0),(1,0),(0,1)\}, d =(x'+y')3^{1-i_3-j_3}5^{1-i_5-j_5}7^{1-i_7-j_7}19^{1-i_{19}-j_{19}}; x = x'd; y = y'd.$

There are 81 solutions, not accounting for symmetry. 41 accounting for symmetry.