The sequence of natural numbers has 4 consecutive numbers, the sum of which, when multiplied by the sum of their squares, gives three times the number that is the sum of the cubes of the respective numbers; which are these four numbers? (ld)
Let $x-1, x, x+1, x+2$ be the four sought numbers. We denote the sum of the first powers of these four numbers by $S_{1}$, the sum of the second powers by $S_{2}$, and the sum of the third powers by $S_{3}$.

$$
\begin{gathered}
x-1=\left(x+\frac{1}{2}\right)-\frac{3}{2} \\
x+2=\left(x+\frac{1}{2}\right)+\frac{3}{2} \\
x=\left(x+\frac{1}{2}\right)-\frac{1}{2}
\end{gathered}
$$

and

$$
x+1=\left(x+\frac{1}{2}\right)+\frac{1}{2}
$$

thus

$$
\begin{gathered}
S_{1}=4\left(x+\frac{1}{2}\right) \\
S_{2}=4\left(x+\frac{1}{2}\right)^{2}+\left(\frac{3}{2}\right)^{2}+\left(\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}=4\left(x+\frac{1}{2}\right)^{2}+5
\end{gathered}
$$

and

$$
\begin{aligned}
S_{3}=4\left(x+\frac{1}{2}\right)^{3} & +3\left[\left(\frac{3}{2}\right)^{2}+\left(\frac{3}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}\right]\left(x+\frac{1}{2}\right)= \\
& =\left(x+\frac{1}{2}\right)\left[4\left(x+\frac{1}{2}\right)^{2}+15\right]
\end{aligned}
$$

But according to the problem,

$$
S_{1} S_{2}=3 S_{3}
$$

or

$$
4\left(x+\frac{1}{2}\right)\left[4\left(x+\frac{1}{2}\right)^{2}+5\right]=3\left(x+\frac{1}{2}\right)\left[4\left(x+\frac{1}{2}\right)^{2}+15\right]
$$

or

$$
4\left(x+\frac{1}{2}\right)^{2}=25
$$

from which

$$
x+\frac{1}{2}= \pm \frac{5}{2}, x_{1}=2 \text { and } x_{2}=-3
$$

thus

$1,2,3,4$ and $-4,-3,-2,-1$

are the sought numbers.

The problem was solved by: Dévai E., Eckstein V., Fekete M., Friedländer H., Fuchs A., Füstös P., Harsányi Z., Heimlich P., Jánosy Gy., Kiss J., Koffler B., Miklóssy K., Neumann L., Paunz A., Pichler S., Rajz E., Reich Z., Schlesinger Gy., Schuster Gy., Steiger J., Szécsi I., Tandlich E., Tóth B., Viola R., Wáhl V.