## Task A-1.1.

Determine all triples of consecutive odd natural numbers whose sum of squares is equal to a four-digit number with all digits being the same.
## Solution.

We seek a natural number $k$ and a digit $x \in\{1,2, \ldots, 9\}$ such that:

$$
\begin{aligned}
(2 k-1)^{2}+(2 k+1)^{2}+(2 k+3)^{2} & =\overline{x x x x} \\
12 k^{2}+12 k+11 & =1111 x \\
12\left(k^{2}+k\right)+11 & =12 \cdot 92 x+7 x
\end{aligned}
$$

It follows that $7 x-11$ is divisible by 12.

Therefore, $x$ must be odd.

For $x \in\{1,3,7,9\}, \quad 7 x-11$ is not divisible by 12.

For $x=5$ we have $7 \cdot 5-11=24=2 \cdot 12$. Then:

$$
\begin{aligned}
12 k^{2}+12 k+11 & =5555 \\
12 k^{2}+12 k & =5544 /: 12 \\
k^{2}+k=k(k+1) & =462=21 \cdot 22
\end{aligned}
$$

Thus, $k=21$, and the sought numbers are 41, 43, and 45.