If in a certain base the digits of a number $N$ are:

$$
1,2,3, \ldots n, n-1, n-2, \ldots 3,2,1
$$

then $\sqrt{N}$ is an $n$-digit number, all of whose digits are: 1.
If the base number of the number system is $a$, then

$$
\begin{gathered}
N=a^{2 n-2}+2 a^{2 n-3}+3 a^{2 n-4}+\ldots+n a^{n-1}+(n-1) a^{n-2}+(n-2) a^{n-3}+ \\
+\ldots+3 a^{2}+2 a+1
\end{gathered}
$$

and

$$
a \times N=a^{2 n-1}+2 a^{2 n-2}+3 a^{2 n-3}+\ldots+n a^{n}+(n-1) a^{n-1}+
$$

$$
+(n-2) a^{n-2}+\ldots+3 a^{3}+2 a^{2}+a
$$

Subtracting (1) from (2):

$$
\begin{aligned}
(a-1) N=a^{2 n-1}+a^{2 n-2} & +a^{2 n-3}+\ldots+a^{n}-a^{n-1}-a^{n-2}-\ldots-a^{3}-a^{2}-a-1= \\
= & a^{n} \cdot \frac{a^{n}-1}{a-1}-\frac{a^{n}-1}{a-1}=\frac{\left(a^{n}-1\right)^{2}}{a-1}
\end{aligned}
$$

from which

$$
N=\left(\frac{\left(a^{n}-1\right)}{a-1}\right)^{2}
$$

and thus

$$
\sqrt{N}=a^{n-1}+a^{n-2}+a^{n-3}+\ldots+a^{3}+a^{2}+a+1
$$

which is a $n$-digit number where every digit is: 1.

(Jenő Filkorn, Nyitra.)

The problem was also solved by: Bayer B., Burján K., Czank K., Feldmár V., Kerekes T., König D., Krausz B., Gy. Krisztián, A. Kürth, Gy. Lukhaub, Gy. Lupsa, Gy. Póka, Á. Rosenberg, J. Scharff, 1. Schmidl, K. Smodics, H. Szmodics, P. Weisz.