## Task 1 - 150931

Three consecutive odd natural numbers are to be determined, for which the sum of their squares is a four-digit number consisting of four identical digits.
Let $2 n-1, 2 n+1$, and $2 n+3$ be three consecutive odd natural numbers. Then the sum $S$ of their squares is

\[
\begin{gathered}
S=(2 n-1)^{2}+(2 n+1)^{2}+(2 n+3)^{2}=4 n^{2}-4 n+1+4 n^{2}+4 n+1+4 n^{2}+12 n+9= \\
=12 n^{2}+12 n+11=12\left(n^{2}+n+1\right)-1
\end{gathered}
\]

This sum is supposed to be a four-digit number consisting of four identical digits $z$. Thus, $S=1111 \cdot z$. Since $S$ is odd, $z$ must also be odd. Furthermore, because $S=1110 z+z=3 \cdot 370 z+z$, the sum $S$ leaves the same remainder when divided by 3 as $z$ itself. Since $S$ is 1 less than a multiple of 12 (and thus also of 3), this remainder when divided by 3 must be exactly 2. The only digit $1 \leq z \leq 9$ that satisfies both conditions is $z=5$.

Therefore, $S$ must be 5555, from which we get $n^{2}+n+1=\frac{5556}{12}=463$ and $n=21$ (or $n=-22$, which is excluded because $2 n-1 \in \mathbb{N}$), so the three consecutive odd natural numbers are 41, 43, and 45.