Number of solutions in $\Bbb N^2$ of the equation $\frac1x+\frac1y=\frac1{1995}$ 
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.
 A: 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
$x = y = 2n$.
A: 
$\ 1995x+1995y=xy$. 

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.
