3.25 Given that the roots of the equation $x^{2}+p x+q=0$ are integers, and $p+q=198$, find the roots.
(15th All-Russian Mathematical Olympiad, 1989)
[Solution] Let the two integer roots of the given equation be $x_{1}, x_{2}$, then
$$
\begin{aligned}
x_{1}+x_{2} & =-p, \\
x_{1} x_{2} & =q .
\end{aligned}
$$

Thus,
$$
\begin{aligned}
198 & =p+q=-\left(x_{1}+x_{2}\right)+x_{1} x_{2} \\
& =\left(x_{1}-1\right)\left(x_{2}-1\right)-1 .
\end{aligned}
$$

Therefore,
$$
\left(x_{1}-1\right)\left(x_{2}-1\right)=199 \text {. }
$$

Since 199 is a prime number, we can only have
$$
\left(x_{1}-1, \quad x_{2}-1\right)=(1,199) \text { or }(-1,-199) \text {. }
$$

Thus, $x_{1}=2, x_{2}=200$ or $x_{1}=0, x_{2}=-198$.