How many integer solutions does the following equation have?

$$
\frac{1}{x}+\frac{1}{y}=\frac{3}{1981}
$$
How many integer solutions does the following equation have?

$$
\frac{1}{x}+\frac{1}{y}=\frac{3}{1981}
$$

By eliminating the fractions and rearranging, we get

$$
3 x y-1981 x-1981 y=0
$$

The left side can be factored if we multiply by 3 and add $1981^{2}$:

$$
(3 x-1981)(3 y-1981)=1981^{2}
$$

This equation has $x=y=0$ as a solution, which is clearly not a solution to (1), but all other solutions of (2) are solutions of (1), and all solutions of (1) are solutions of (2), since multiplying by $xy$ only introduces the additional solutions $x=0$ or $y=0$, and if one of $x, y$ is 0, then the other is also 0 according to (2).

Therefore, it is sufficient to count the non-zero integer solutions of (2). If $x$ and $y$ are non-zero integers, then according to (2), the number $A=3 x-1981$ is a divisor of $N=1981^{2}=7^{2} \cdot 283^{2}$, different from -1981. Conversely, starting from any divisor $A$ of $N$, we will get some integer solution of (2) if $A+1981$ and $\frac{1981^{2}}{A}+1981$ are both divisible by 3. Since 1981 gives a remainder of 1 when divided by 3, this is necessary and sufficient for $A+1$ to be divisible by 3. Therefore, the suitable values of $A$ are the following:

$-1,-7,-283,-7^{2},-283^{2},-7^{2} \cdot 283,-7 \cdot 283^{2},-7^{2} \cdot 283^{2}$, and the number of integer solutions of (1) is 8.