Let's determine the distinct digits $A, B, C, D$ in the square number $A B A B C D C D$.

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Translate the above text back to Hungarian, keeping the original formatting and line breaks, and output the result directly.
However, since the instruction is to translate the initial Hungarian text to English, the above statement is a note for the translator and should not be included in the final translation. Here is the correct translation:

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Határozzuk meg az egymástól különböző $A, B, C, D$ számjegyeket az $A B A B C D C D$ négyzetszámban.

Let's determine the distinct digits $A, B, C, D$ in the square number $A B A B C D C D$.
Given the repeating pairs of digits $A, B$ and $C, D$, the sought $N^2$ number can be written as:

$$
101 \cdot \overline{A B} \cdot 10^{4} + 101 \cdot \overline{C D} = 101 \left( \overline{A B} \cdot 10^{4} + \overline{C D} \right) = 101 \cdot K
$$

Since 101 is a prime number, and in the factorization of a square number into powers of different bases, all exponents must be even, the number $K$ must be 101 times a square number. Furthermore, $10^{4} = (10^{4} - 1) + 1 = 101 \cdot 99 + 1$, so

$$
K = 101 \cdot 99 \cdot \overline{A B} + (\overline{A B} + \overline{C D}) = 101 \cdot M^2
$$

(where $M = N / 101$), thus $\overline{A B} + \overline{C D}$ must also be a (positive integer) multiple of 101. However, since $1 \leq \overline{A B} < 100$ and $1 \leq \overline{C D} < 100$, it follows that $2 \leq \overline{A B} + \overline{C D} < 200$, and 101 times 2 is already greater than 200, so

$$
\overline{A B} + \overline{C D} = 101, \quad \text{thus} \quad M^2 = 99 \cdot \overline{A B} + 1 \\
(M-1)(M+1) = 3^2 \cdot 11 \cdot \overline{A B}
$$

Since $1 \leq \overline{A B} < 100$, $M^2$ must be at least a three-digit number and at most a four-digit number, i.e., the square of a real two-digit number.

One of the factors on the left side of (2) must be divisible by 11. Similarly, one of the factors must be divisible by 9, because both factors cannot be divisible by 3, as their difference is 2.

Let's try a solution where the same factor on the left side of (2) is divisible by both 9 and 11, i.e., by 99. This, given $M < 100$, can only be $M+1$ if $M = 98$, in which case $M-1 = 97 = \overline{A B}$, and from (1) $\overline{C D} = 4 = 04$. Thus, $A = 9, B = 7, C = 0, D = 4$, which are distinct, giving us a solution. (Indeed, in this case $N = 101 \cdot M = 9898$, and $N^2 = 97970404$, which follows necessarily from the above.)

Looking for a solution where one factor on the left side of (2) is divisible by 11 and the other by 9, it suffices to check the divisibility of the numbers $11k \pm 2$ by 9, since the difference between the factors is 2. Only the cases $k = 1$ and $k = 8$ apply: 9 and 11, or 88 and 90, but from the first, $\overline{A B} = 1 = 01$, $\overline{C D} = 100$ does not fit; from the second, $\overline{A B} = (88 \cdot 90) / 99 = 80$, $\overline{C D} = 21$, and this is also a solution: $A = 8, B = 0, C = 2, D = 1 (M = 89, N^2 = 8989^2 = 80802121)$. There are no more solutions.

Sándor Horváth (Budapest, I. István g. III. o. t.)

Remarks. 1. We can also arrive at the solutions by eliminating $\overline{A B}$ from (1): $M^2 = 99(101 - \overline{C D}) + 1 = 100^2 - 99 \cdot \overline{C D}, (100 - M)(100 + M) = 3^2 \cdot 11 \cdot \overline{C D}$.

2. The same equation can also be written as: $M^2 = 100^2 - (100 - 1) \cdot \overline{C D} = (100 - \overline{C D}) \cdot 100 + \overline{C D}$, which means the sum of the two two-digit parts of $M^2$ is 100. Checking all two-digit squares, only $89^2 = 7921$ and $98^2 = 9604$ have this property, $\overline{C D} = 21$ or 04.

Fewer trials are sufficient if we consider that $\overline{C D}$ is a two-digit square, so it can take only 22 different values. By forming $100 - \overline{C D} - \text{e.g.,} \overline{C D} = 56$ in this case 44, we only need to check in the table whether 4456 is a square number.

Zsigmond Nagy (Budapest, Kaffka M. g. III. o. t.)
3. The splitting of $M^2$ into two digits in the hundred-based number system can also be stated as follows: the two digits of $M^2$ in the hundred-based number system must sum to 100.

The divisibility rule for "hundred-one" in the hundred-based number system is the same as for eleven in the ten-based system. (According to the latter, a number is divisible by 11 if and only if the sum of its digits, with minus signs before every second digit from right to left, is divisible by 11. Applying this to the above $K = \overline{A B 00 C D}$, we get (1).