Let's determine $p$ and $q$ such that the function $x^{2}+p x+q$ takes its minimum value when $x=q$ and that this minimum is $(p+q)^{2}$.

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Determine $p$ and $q$ such that the function $x^{2}+p x+q$ takes its minimum value when $x=q$ and that this minimum is $(p+q)^{2}$.
The function $y=x^{2}+p x+q$ attains its minimum value when $x=-\frac{p}{2}$, at which point the minimum is $\frac{4 q-p^{2}}{4}$. According to the problem, we have

$$
q=-\frac{p}{2} \quad \text { and } \quad(p+q)^{2}=\frac{4 q-p^{2}}{4}
$$

From this system of equations, it follows that

$$
p_{1}=0, q_{1}=0 \quad \text { and } \quad p_{2}=-1, q_{2}=\frac{1}{2}
$$

Thus, the functions in question are:

$$
y=x^{2} \quad \text { and } \quad y=x^{2}-x+\frac{1}{2}
$$

(Rézó Földes, Budapest.)

The problem was also solved by: L. Bánó, E. Bauer, J. Blum, I. Csada, I. Dömény, K. Epstein, V. Erdős, M. Fekete, H. Fodor, J. Friedrich, A. Haar, P. Heimlich, Gy. Jánosy, J. Kiss, Gy. Krampera, R. Kürth, P. Merse, O. Morvai, S. Pichler, S. Ruvald, Gy. Schuster, O. Schwarz, A. Stagl.