Let's determine the positive primes $p, q$ for which the equation $x^{4}+p^{2} x+q=0$ has a multiple root.
Let the multiple (at least double) root of the equation be denoted by $a$ (which must necessarily be non-positive). Then

\[
\begin{gathered}
x^{4}+p^{2} x+q=(x+a)^{2}\left(x^{2}+b x+c\right)=\left(x^{2}+2 a x+a^{2}\right)\left(x^{2}+b x+c\right)= \\
=x^{4}+(2 a+b) x^{3}+\left(2 a b+a^{2}+c\right) x^{2}+a(a b+2 c) x+a^{2} c
\end{gathered}
\]

By comparing the corresponding coefficients, we get

(1) $2 a+b=0$, hence $b=-2 a$,

(2) $2 a b+a^{2}+c=0$, thus $c=3 a^{2}$

(3) $a(a b+2 c)=p^{2}$, so $4 a^{3}=p^{2}$,

(4) $a^{2} c=q$, hence $3 a^{4}=q$.

From (3) and (4),

\[
3^{3} p^{8}=3^{3}\left(p^{2}\right)^{4}=3^{3} \cdot 4^{4} \cdot a^{12}=2^{8}\left(3 a^{4}\right)^{3}=2^{8} q^{3}
\]

This is only possible if $p=2$ and $q=3$. Then $a=1, b=-2, c=3$, and indeed $x^{4}+4 x+3=(x+1)^{2}\left(x^{2}-2 x+3\right)$. Gergely Ambrus (Szeged, Radnóti M. Gymnasium, 12th grade)