Let's determine the digits $A, B, C, D$ if the following equalities hold:

$$
(A+B \sqrt{3})^{2}=\overline{C D}+\overline{B C} \sqrt{3}, \quad(A+C \sqrt{3})^{2}=\overline{D C}+\overline{C D} \sqrt{3}
$$
I. solution. Let's factor the left side of the first equation and rearrange the terms containing $\sqrt{3}$ to one side:

$$
A^{2}+3 B^{2}-(10 C+D)=(10 B+C-2 A \cdot B) \sqrt{3}
$$

This can only hold if both sides are 0, because otherwise an integer would stand on the left side, and a non-integer on the right side. Thus, we get two equations for the digits, and from the second conditional equation, we get two more:

$$
\begin{aligned}
& A^{2}+3 B^{2}=10 C+D \\
& A^{2}+3 C^{2}=10 D+C \\
& 2 A \cdot B=10 B+C \\
& 2 A \cdot C=10 C+D
\end{aligned}
$$

On one hand, none of the digits can be 0, because they appear independently or as leading digits in the original equations. On the other hand, we consider different letters to represent different digits.

We will show that all four sought digits are even. In (3), the left side is even, so $C$ must be even. Thus, the left side of (4) is divisible by 4, and since $10 C$ is also divisible by 4 on the right, $D$ is divisible by 4. In (2), the right side and the $3 C^{2}$ term are even, so $A^{2}$, and thus $A$, is even, and the left side is divisible by 4, so $C$ is divisible by 4. Therefore, in (1), the right side and the $A^{2}$ term are divisible by 4, so $B^{2}$ is also divisible by 4, and $B$ is even. Thus, the values of $C$ and $D$ can only be 4 and 8 in some order, and the values of $A$ and $B$ can be 2 and 6 in some order. From (3),

$$
A=\frac{10 B+C}{2 B}=5+\frac{C}{2 B}>5
$$

thus only $A=6$ is possible, so $B=2, C / 2 B=1, C=2 B=4$, and finally $D=8$. This set of values indeed satisfies both conditions:

$$
(6+2 \sqrt{3})^{2}=48+24 \sqrt{3}, \quad(6+4 \sqrt{3})^{2}=84+48 \sqrt{3}
$$

Miklós Bély (Györ, Révai M. Gymnasium, 2nd grade)

Note. It is easily seen that if any digit is 0, then all are, so allowing 0 digits only gives this trivial solution to the above. The following solution will also show that allowing different letters to possibly represent the same digit does not result in any new solutions other than this.

II. solution. We start again from equations (1)-(4) of the first solution. From (3) and (4),

$$
C=2(A-5) B, \quad D=2(A-5) C=4(A-5)^{2} B
$$

If $B=0$ here, then $C=D=0$, and from (2), for example, $A=0$ follows, which we do not consider a solution. If $B \neq 0$, then $A=5$ or 6, because only then can $D$ be a digit. The possibility $A=5$ leads to a contradiction, because then $C=D=0$ as well, but then the second initial equality cannot hold.

Therefore, only $A=6$ is possible. In this case, $C=2 B, D=4 B$, and substituting these into equations (1) and (2),

$$
36+3 B^{2}=24 B, \quad 36+12 B^{2}=42 B
$$

Subtracting the second from four times the first,

$$
108=54 B, B=2, \text { and thus } C=4, D=8
$$