Find three numbers, in an arithmetic progression with a common difference of 2, such that the sum of their squares is a number formed of four identical digits.
Since the common difference of the arithmetic progression is 2, the three numbers can be denoted by $n-2, n$, and $n+2$. The sum of their squares is a four-digit number with all digits the same, say, $k k k k$, where $k$ is some integer between 1 and 9, that is,

$$
k k k k=(n-2)^{2}+n^{2}+(n+2)^{2}=3 n^{2}+8
$$

From this point, we present two possible solutions.

Solution 1: Since $k k k k-8=3 n^{2}$ is a multiple of 3, the remainder of the division of $k k k k$ by 3 is equal to the remainder of the division of 8 by 3, which is 2. But $k k k k=k \times 1111$ and the remainder of the division of 1111 by 3 is 1, so the remainder of the division of $k$ by 3 is 2. Since $1 \leq k \leq 9$, we can only have $k$ equal to 2, 5, or 8. As in Solution 1, the cases $k=2$ and $k=8$ are impossible, and $k=5$ is the only option, in which case $n=43$ and the three numbers sought are 41, 43, and 45, which constitute the only solution to the problem.

Solution 2: Since $k k k k=3 n^{2}+8$, we obtain

$$
3 n^{2}=k k k k-8=(k k k \times 10+k)-(9-1)=(k k k \times 10-9)+(k+1)
$$

But $k k k \times 10-9=k k(k-1) 1$ is a multiple of 3, so $k+1$ is also a multiple of 3. Since $1 \leq k \leq 9$, we can only have $k$ equal to 2, 5, or 8. In the case $k=2$, we get

$$
n^{2}=\frac{2222-8}{3}=738=2 \times 369
$$

which is impossible, since 738 is not a perfect square. Similarly, if $k=8$, we get

$$
n^{2}=\frac{8888-8}{3}=2960=2^{4} \times 5 \times 37
$$

which, again, is impossible, since this last number is not a perfect square. Therefore, the only remaining option is $k=5$. In this case,

$$
n^{2}=\frac{5555-8}{3}=1849=43^{2}
$$

so $n=43$ and the three numbers in arithmetic progression sought are 41, 43, and 45, which constitute the only solution to the problem.