Let's determine $p$ such that in the equation

$$
x^{2}+(2-p) x-p-3=0
$$

the sum of the squares of the roots is as small as possible.
If the roots of the equation are \(x_{1}\) and \(x_{2}\), then

$$
x_{1}^{2}+x_{2}^{2}=\left(x_{1}+x_{2}\right)^{2}-2 x_{1} x_{2}=(p-2)^{2}+2 p+6
$$

or, rearranged,

$$
x_{1}^{2}+x_{2}^{2}=p-2 p+10
$$

But the function

$$
a x^{2}+b x+c
$$

attains its minimum value when \(x=-\frac{b}{2 a}\), so the sum of the squares of the roots is minimal when

$$
p=1
$$

In this case,

$$
x_{1}^{2}+x_{2}^{2}=9
$$

(László Bánó, Budapest.)

The problem was also solved by. E. Ádámffy, I. Dömény, B. Égető, G. Égető, M. Fekete, B. Gunszt, A. Haar, Z. Harsányi, P. Heimlich, Gy. Jánosy, G. Kertész, Gy. Krampera, I. Kürti, F. Magyari, P. Messer, A. Nagy, M. Pám, P. Rássy, M. Riesz, J. Rosenberg, I. Schöffer, Gy. Schuster, Gy. Schwarz, J. Sonnenfeld, Gy. Söpkéz, J. Stróbl, B. Tóth, the mathematics circle of the V. district secondary school.