799. Solve the equation in integers

$$
x^{2}-y^{2}=1997
$$
$\triangle$ Let's write the equation in the form:

$$
(x+y)(x-y)=1997
$$

The number 1997 is prime, so there are only four possible cases:

1) $\left\{\begin{array}{l}x+y=1997, \\ x-y=1 ;\end{array}\right.$
2) $\left\{\begin{array}{l}x+y=1, \\ x-y=1997 ;\end{array}\right.$
3) $\left\{\begin{array}{l}x+y=-1997, \\ x-y=-1 ;\end{array}\right.$
4) $\left\{\begin{array}{l}x+y=-1, \\ x-y=-1997 .\end{array}\right.$

It remains to solve each of these systems of equations.

Answer: (999; 998), (999; -998), (-999; -998), (-999; 998).