1. Find three consecutive odd numbers $a, b$, and $c$ such that $a^{2}+b^{2}+c^{2}$ is a four-digit number composed of the same digit (written with the same digit).
Solution. Let $a=2 n-1, b=2 n+1$ and $c=2 n+3$ be three consecutive odd numbers. The sum of their squares is

$$
a^{2}+b^{2}+c^{2}=(2 n-1)^{2}+(2 n+1)^{2}+(2 n+3)^{2}=12 n(n+1)+11
$$

Thus, the number $12 n(n+1)+11$ must be a four-digit number composed of the same digits, i.e., one of the numbers 1111, 2222, 3333, 4444, 5555, 6666, 7777, 8888, 9999. The difference between these numbers and the number 11 must be a number divisible by 12. From the numbers $1100,2211,3322,4433,5544,6655,7766,8877,9988$, we will select those divisible by 12. The only one is 5544. For it, we have $5544=12 \cdot 21 \cdot 22$ and

$$
5555=(2 \cdot 21-1)^{2}+(2 \cdot 21+1)^{2}+(2 \cdot 21+3)^{2}
$$

According to this, the required consecutive odd numbers are $41,43,45$.