Let's determine the digits $A, B, C, D$ if

$$
\overline{A B^{2}}=\overline{A C D B}, \quad \text { and } \quad \overline{C D^{3}}=\overline{A C B D}
$$
I. solution. The number $\overline{A B}$ is at least 32, because the square of a smaller natural number has at most three digits. With a little trial and error, we realize that $A$ cannot be less than 9. Indeed, the square of a two-digit number, whose tens digit is $3,4,5,6,7$, is respectively less than $40^{2}=1600,50^{2}=2500$, $60^{2}=3600,70^{2}=4900$, and $80^{2}=6400$, while the square of those starting with 8 is at most $89^{2}=7921$. Thus, the square of these numbers starts with a smaller digit than the base.

Consequently, the number $\overline{C D^{3}}$ can only fall between 9000 and 10000. The only cube number between these two limits is $21^{3}=9261$, because $20^{3}$ is still only 8000, while $22^{3}$ is already 10648. Therefore, the only solution is $A=9, C=2, B=6, D=1$. From this, $\overline{A B}=96$, and its square is 9216, which satisfies the first condition.

Uray Szabolcs (Szentendre, Móricz Zs. g. I. o. t.)

Remark. We can arrive at the single possible value of $A$ with fewer trials by the following path. In $\overline{A B^{2}}=(10 A+B)^{2}=$ $100 A^{2}+20 A B+B^{2}$, since $A, B \leqq 9$, we have $20 A B+B^{2} \leqq 1701$, meaning that at most 17 hundreds can be added to the $A^{2}$ hundreds. In $\overline{A B^{2}}$ in the form $\overline{A C D B}$, the number of hundreds is $\overline{A C}=10 A+C$. Therefore, $10 A+C \leqq A^{2}+17$, and thus even more so $10 A \leqq A^{2}+17$, or $A^{2}-10 A+17 \geqq 0$. The left side is negative for values of $A=5 \pm \sqrt{8}$, so either $A \leqq 2$ (which we have already ruled out), or $A \geqq 8$. Since $89^{2}<8000$, only $A=9$ can lead to a solution.

Szentai Judit (Budapest, Kanizsay D. lg. I. o. t.)

II. solution. $\overline{C D^{3}}-\overline{A B^{2}}=\overline{B D}-\overline{D B}=(B-D) 9$, is divisible by 9. We will show that $\overline{A B}$, and thus $\overline{C D}$, is divisible by 3. If $e$ is an integer, $e^{3}-e=(e-1) e(e+1)$ is divisible by 3, because the product of three consecutive integers is divisible by 3. Applying this to $e=C D$:

$$
\overline{C D^{3}}-\overline{C D}=\overline{A C B D}-\overline{C D}=1000 A+10 B+90 C=90(10 A+C)+10 \cdot \overline{A B}
$$

The left side and the first term on the right side are divisible by 3, so the second term on the right side must also be divisible by 3, which is only possible if $\overline{A B}$ is divisible by 3. In this case, however, $\overline{C D^{3}}-\overline{A B^{2}}$ can only be divisible by 9 if $\overline{C D}$ is also divisible by 3. On the other hand,

$$
10^{3}=1000 \leqq \overline{C D^{3}}<10000<22^{3}(=10648)
$$

so $10 \leqq \overline{C D} \leqq 21$. The numbers divisible by 3 in this range are $12,15,18,21$.

We also know that the last digit of $\overline{C D^{3}}$ is $D$, which matches the last digit of the base. This is only true for numbers ending in $1,4,5,6,9$, so $\overline{C D}$ can only be 15 or 21. Among these, $15^{3}$ ends in 75, and in the case of $\overline{C D}=15$, $\overline{A B^{2}}$ would have to end in 57, which is impossible because a square number cannot end in 7.

The only remaining possibility is $\overline{C D}=21$. For this,

$$
\overline{C D^{3}}=21^{3}=9261(=\overline{A C B D}) \text { and }(\overline{A C D B}=) 9216=96^{2}=\overline{A B^{2}}
$$

satisfies the conditions.