Find the number of solutions of $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{1995}$ in $\mathbb N^2$.
My Attempt:
This can be simplified to $1995(x+y)=xy$ and then further to $$y={1995x\over x-1995}$$
Since $1995=3\cdot5\cdot7\cdot19$, it has $16$ divisors
Now since $x>0$, This yields total of $16$ solutions (the answer according to book). however, Mathematica found $81$ solution
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.