Let the roots of the equation $x^{2}+p x+q=0$ be $x_{1}$ and $x_{2}$. Construct the fourth-degree equation whose roots are: $x_{1}, x_{2},-x_{1},-x_{2}$.
First solution. If $x_{1}$ and $x_{2}$ are the roots of the equation $x^{2} + p x + q = 0$, then $-x_{1}$ and $-x_{2}$ are the roots of the equation $x^{2} - p x + q = 0$. The sought equation is therefore

$$
\left(x^{2} + p x + q\right)\left(x^{2} - p x + q\right) = 1
$$

or

$$
x^{4} - \left(p^{2} - 2 q\right) x^{2} + q^{2} = 0
$$

(Pál Sárközy, Pannonhalma.)

Second solution. The sought equation is:

$$
\left(x - x_{1}\right)\left(x - x_{2}\right)\left(x + x_{1}\right)\left(x + x_{1}\right) = 0
$$

or

$$
\begin{gathered}
\left(x^{2} - x_{1}^{2}\right)\left(x^{2} - x_{2}^{2}\right) = 0 \\
x^{4} - \left(x_{1}^{2} + x_{2}^{2}\right) x^{2} + x_{1}^{2} x_{2}^{2} = 0
\end{gathered}
$$

But

$$
x_{1}^{2} + x_{2}^{2} = \left(x_{1} + x_{2}\right)^{2} - 2 x_{1} x_{2} = p^{2} - 2 q
$$

and

$$
x_{1}^{2} x_{2}^{2} = q^{2}
$$

so from (1) we get:

$$
x^{4} - \left(p^{2} - 2 q\right) x^{2} + q^{2} = 0
$$

(Ervin Kirchknopf, Budapest.)

Number of solutions: 40.