Let's determine $x$ such that

$$
\begin{gathered}
1^{4}+2^{4}+\cdots+x^{4}=1+2 \cdots+x+1^{2}+2^{2}+\cdots+x^{2}+ \\
+1^{3}+2^{3}+\cdots+x^{3}
\end{gathered}
$$
The sum of the second, third, and fourth powers of natural numbers (K. M. L. VIII. 42.):

$$
\begin{gathered}
1^{2}+2^{2}+3^{2} \ldots+x^{2}=\frac{x(x+1)(2 x+1)}{6} \\
1^{3}+2^{3}+3^{3} \ldots+x^{3}=\frac{x^{2}(x+1)^{2}}{4} \\
1^{4}+2^{4}+3^{4} \ldots+x^{4}=\frac{x(x+1)(2 x+1)\left(3 x^{2}+3 x-1\right)}{30}
\end{gathered}
$$

Thus, in the context of our problem:

$$
\frac{x(x+1)(2 x+1)\left(3 x^{2}+3 x-1\right)}{30}=\frac{x(x+1)}{2}+\frac{x(x+1)(2 x+1)}{6}+\frac{x^{2}(x+1)^{2}}{4}
$$

from which:

$$
4 x^{3}+x^{2}-11 x-14=0
$$

One of the roots of this equation is 2, hence:

$$
4 x^{3}+x^{2}-11 x-14=(x-2)\left(4 x^{2}+9 x+7\right)=0
$$

Since the equation \(4 x^{2}+9 x+7\) has no real roots, the only solution to the problem is \(x=2\).

(Ferencz Aczél, Budapest.)

The problem was also solved by: Bartók I., Bayer B, Bogdán G., Hirschfeld Gy., Holzmann J., Kertész F., König D., Lázár L., Messik G., Pilczer P., Póka Gy., Sasvári J., Scharff J., Schlesinger A., Spitzer V., Tóbiás J. L., Wohlstein S.