For what integer value systems of $p$ and $q$ will one root of the equation $x^{2}+p x+q=0$ be four times the other? How many solutions do we get if we take $p$ and $q$ to be at most two-digit numbers?
Solution. In the equation $x^{2}+p x+q=0$,

$$
\begin{aligned}
& x_{1}=\frac{-p+\sqrt{p^{2}-4 q}}{2} \\
& x_{2}=\frac{-p-\sqrt{p^{2}-4 q}}{2}
\end{aligned}
$$

Suppose that $x_{1}=4 x_{2}$, then

$$
\begin{gathered}
\frac{-p+\sqrt{p^{2}-4 q}}{2}=4 \frac{-p-\sqrt{p^{2}-4 q}}{2} \\
-p+\sqrt{p^{2}-4 q}=-4 p-4 \sqrt{p^{2}-4 q} \\
5 \sqrt{p^{2}-4 q}=-3 p
\end{gathered}
$$

Squaring both sides of the equation, we get

$$
\begin{gathered}
25 p^{2}-100 q=9 p^{2} \\
16 p^{2}=100 q \\
4 p^{2}=25 q \\
p^{2}=\frac{25}{4} q
\end{gathered}
$$

$p^{2}$ and hence $p$ can only be an integer if $q$ is a multiple of 4, i.e., if

$$
q=4 u
$$

Then

$$
p^{2}=25 u
$$

and

$$
p= \pm 5 \sqrt{u}
$$

For $p$ and $q$ to be at most two-digit numbers, it is necessary that

$$
\begin{gathered}
4 u<100 \\
\pm 5 \sqrt{u}<100 \\
25 u<10000
\end{gathered}
$$

or, that

$$
u<25<400
$$

From the first necessary and sufficient inequality, we derive the sought value pairs:

$$
\begin{gathered}
\pm 5,4 \\
\pm 10,16 \\
\pm 25,36 \\
\pm 20,64
\end{gathered}
$$

(Students of the 8th grade of the Debrecen State Real School.)