Number of solutions to $1/x+1/y=1/1995$ in $\mathbb N^2$ [duplicate]

First, note $\frac{1}{1995}>\frac{1}{x}$ and so $x>1995$ and, likewise, $y>1995$. Next, we have: $$ 1995(x+y)=xy\quad\implies\quad (x-1995)(y-1995)=1995^2=3^25^27^219^2. $$ Write $x-1995$ as $3^i 5^j 7^k 19^l$. Then, each quadruple $(i,j,k,l)$ uniquely pins down $x$ and $y$. There are 3 options each for $i,j,k,l$ (i.e. $\{0,1,2\}$), so there are $3^4=81$ solutions in total. Your book appears to be in error.

$81$ looks correct to me. It is the number of divisors of $1995^2$

Take any divisor $d$ of $1995^2$.

Then $d$ also divides $1995^2+1995d = 1995(1995+d)$.

So I can set $x=1995+d$, a positive integer and have $y=\dfrac{1995^2}{d}+1995=\dfrac{1995x}{x-1995}$ also a positive integer giving $\dfrac{1}{x}+\dfrac{1}{y}=1995$. In other words

$$\dfrac{1}{1995+d}+\dfrac{1}{\frac{1995^2}{d}+1995}=1995$$

will be one of the $81$ solutions