Solve the equation

$$
1996 x+1998 y+1=x y
$$

in the set of integers.
Notice that

$$
1996 \cdot 1998=(1997-1)(1997+1)=1997^{2}-1=3998008
$$

After rearranging and factoring the equation of the problem:

$$
(x-1998)(y-1996)=1997^{2} .
$$

Since 1997 is a prime number, $1997^{2}$ can only be expressed as the product of two integers in the following 6 ways:

$x-1998=-1997^{2}, y-1996=-1999^{2}$, so $x=-3990007, x-1998=1, y-1996=1997^{2}$, so $x=1999$, $y=3990005; x-1998=1997, y-1996=1997$, so $x=3995, y=3993 ; x-1998=1997^{2}, y-1996=1$, so $x=$ 3990007, $y=1997 ; x-1998=-1, y-1996=-1997^{2}$, so $x=1997, y=-3986013 ; x-1998=-1997$, $y-1996=-1997$, so $x=1, y=-1 ; x-1998=-1997^{2}, y-1996=-1$, so $x=-3986001, y=1995$.

This solves the problem.

Zoltán Szilasi (Debrecen, KLTE Gyak. Gimn., 9th grade)

Note. Gergely Kiss (Fazekas M. Fóv. Gyak. Gimn., 10th grade) similarly examined equations of the form

$$
k(a-b) x+k(a+b) y+b^{2}=k^{2} x y
$$

The direct generalization of this problem is given by the type where $k=1$, and the parameter $a$ is prime.