How many solutions does the equation

$$
x^{2}+y^{2}+2 x y-1988 x-1988 y=1989
$$

have in the set of positive integers?
The left side of the equation can be factored as follows:

$$
\begin{gathered}
x^{2}+y^{2}+2 x y-1988 x-1988 y=(x+y)^{2}-1988(x+y)= \\
=(x+y)(x+y-1988)
\end{gathered}
$$

Since $x+y>0$, it follows that $x+y-1988$ is also positive, hence $x+y \geq 1989$. However, $x+y$ cannot be greater than 1989, as it must be a divisor of 1989. Therefore, the only possibility is $x+y=1989$, and in this case,

$$
x^{2}+y^{2}+2 x y-1988 x-1988 y=1989(1989-1988)=1989
$$

Thus, in the set of positive integers, the solutions to the equation are precisely the pairs $(1,1988),(2,1987), \ldots,(1987,2),(1988,1)$, and the number of these pairs is 1988.

Based on the work of Péter Bilics (Bp., Fazekas M. Gymnasium II. o. t.)