Problem 1217

If $x_1$ and $x_2$ are the roots of the equation $x^2 + p x + q = 0$, then by Viete's formulas we obtain
$$\begin{cases} x_1 + x_2 = -p \\ x_1 x_2 = q \end{cases}$$
Because $198 = p + q = -(x_1 + x_2) + x_1 x_2 = (x_1 - 1)(x_2 - 1) - 1$, it follows that $(x_1 - 1)(x_2 - 1) = 199$. Since $x_1, x_2$ are integers and $x_1 - 1, x_2 - 1$ are integers and $199$ is a prime number
$$\begin{cases} x_1 - 1 = 1 \\ x_2 - 1 = 199 \end{cases} \text{ or } \begin{cases} x_1 - 1 = -1 \\ x_2 - 1 = -199 \end{cases}$$
The solution $(x_1, x_2)$ of the system is $(2, 200)$ and $(0, 198)$. From $(2, 200)$, we have $p = -202$, $q = 400$, $x_1 = 2$, $x_2 = 200$, and from $(0, 198)$, $p = -198$, $q = 0$, $x_1 = 0$, $x_2 = 198$.