Three consecutive odd numbers, the sum of the squares of which is a four-digit number whose digits are equal. Which are these numbers?
I. solution. Let the middle odd number be $n$, and the common digit of the sum be $j$, then

$$
(n-2)^{2}+n^{2}+(n+2)^{2}=3 n^{2}+8=\overline{j j j j}=1111 \cdot j
$$

The left side is odd, so $j$ is also odd. Furthermore, when divided by 3, the left side gives a remainder of 2, while the right side gives the same as $j+j+j+j=4 j$, or simply $j$. Among the odd digits, only $j=5$ gives a remainder of 2.

From the resulting equation $3 n^{2}=5547$, we get an integer solution, $n=43$, so the numbers are $41,43,45$. (Naturally, $-45,-43,-41$ also works.)

Cseh Ágnes (Tatabánya, Árpád Gymn., I. o. t.)

II. solution. Let the smallest of the three numbers be $2 k-1$. Thus

$$
\begin{gathered}
(2 k-1)^{2}+(2 k+1)^{2}+(2 k+3)^{2}=12 k^{2}+12 k+11=1111 j \\
12\left(k^{2}+k+1-92 j\right)=7 j+1
\end{gathered}
$$

and the right side is only divisible by 12 when $j=5$.