How many solutions does the equation

$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{1988}
$$

have in the set of natural numbers?
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{x y}=\frac{1}{1988}
$$

Solve the equation for one of the unknowns, for example $y$,

$$
y=1988+\frac{1988 \cdot 1989}{x-1988}
$$

Since $y$ (and $x$) are integers, the denominator on the right side of (2) must divide the numerator. We claim that this divisor cannot be negative.

If $(x-1988)$ is negative, then for $x \geq 0$, $x-1988 \leq 1988$, and thus the absolute value of the fraction in (2) is at least 1989. Furthermore, the fraction is negative, so $y<0$; however, this is excluded by the condition. Therefore, if a pair of natural numbers $(x, y)$ is a solution to (1), then $y$ and of course $x$ must be greater than 1988.

It is easy to see, on the other hand, that for any positive divisor $d$ of $A=1988 \cdot 1989$, the pair $\left(1988+d, 1988+\frac{A}{d}\right)$ is a solution to (1), and different divisors yield different solutions. Consequently, the number of solutions to equation (1) in the set of natural numbers is equal to the number of positive divisors of

$$
A=1988 \cdot 1989=2^{2} \cdot 3^{2} \cdot 7 \cdot 13 \cdot 17 \cdot 71
$$

It is known that if a number $B$ has a prime factorization of $p_{1}^{a_{1}} \cdot p_{2}^{a_{2}} \ldots \cdot p_{k}^{a_{k}}$, then $B$ has

$$
d(B)=\left(1+\alpha_{1}\right)\left(1+\alpha_{2}\right) \cdot \ldots \cdot\left(1+\alpha_{k}\right)
$$

positive divisors. In our case, this value is

$$
d(A)=(1+2)(1+2)(1+1)(1+1)(1+1)(1+1)=144
$$

Thus, equation (1) has 144 solutions in the set of natural numbers.