If the expression $x^{4}+1$ is to be divided by the trinomial $x^{2}+p x+q$. What should $p$ and $q$ be so that the division can be performed without a remainder?
$$
\begin{aligned}
& x^{4}+1: x^{2}+p x+q=x^{2}-p x+p^{2}-q \\
& \frac{x^{4}+p x^{3}+q x^{2}}{-p x^{3}-q x^{2}}+1 \\
& \frac{-p x^{3}-p^{2} x^{2}-p q x}{\left(p^{2}-q\right) x^{2}+p q x+1} \\
& \quad \frac{\left(p^{2}-q\right) x^{2}+p\left(p^{2}-q\right) x+q\left(p^{2}-q\right)}{p\left(2 q-p^{2}\right) x-q\left(p^{2}-q\right)+1}
\end{aligned}
$$

For the remainder to be zero for any value of \( x \), it is necessary that:

$$
\begin{array}{r}
p\left(2 q-p^{2}\right)=0 \\
q\left(p^{2}-q\right)=1
\end{array}
$$

From (1), \( p' = 0 \) and \( p'' = \sqrt{2 q} \). Substituting these values into (2), we get:

$$
\begin{aligned}
& q^{2}=-1 \\
& q^{2}=+1
\end{aligned}
$$

Thus, the corresponding values of \( p \) and \( q \) are:

$$
\begin{gathered}
p=0, \quad q=\sqrt{-1} \\
p=\sqrt{2}, \quad q=1 \\
p=\sqrt{-2}, \quad q=-1
\end{gathered}
$$

and hence the factors of the expression \( x^{4}+1 \) are:

$$
\begin{gathered}
x^{2}+\sqrt{-1} \\
x^{2}+\sqrt{2 x}+1 \\
x^{2}+\sqrt{-2 x}-1
\end{gathered}
$$

(Tivadar Heymann, realist. VIII. grade student, Győr) The problem was also solved by: József Rosenberg, realist. VIII. grade student, Győr.