Let's determine those numbers $\overline{A B C D}$ in the decimal system for which

$$
\overline{A B C D}=\overline{A B} \cdot \overline{C D}+\overline{A B}^{2}
$$
The four-digit number $\overline{A B C D}$ can be rewritten in the form $100 \overline{A B}+\overline{C D}$, which we write in (1), and transform it as follows:

$$
\begin{aligned}
100 \overline{A B}+\overline{C D} & =\overline{A B}(\overline{C D}+\overline{A B}), \text { further rearranging } \\
99 \overline{A B}+\overline{A B}+\overline{C D} & =\overline{A B}(\overline{C D}+\overline{A B}), \\
99 \overline{A B} & =(\overline{A B}-1)(\overline{A B}+\overline{C D})
\end{aligned}
$$

With the notation $(\overline{A B}-1)=x$, we get

$$
99=x(\overline{C D}+x-98)
$$

which means $x$ is a divisor of 99. Furthermore, since $x=\overline{A B}-1$ (where $\overline{A B}$ is a two-digit number, and as such, at least 10 and at most 99), we have $9 \leq x \leq 98$. There are three such numbers among the divisors of 99: 9, 11, and 33.

If $x=9$, then $\overline{C D}=100$ would result, which is impossible.

If $x=11$, then $\overline{A B}=12$ and $\overline{C D}=96$, making the four-digit number 1296.

If $x=33$, then $\overline{A B}=34$ and $\overline{C D}=68$, making the four-digit number 3468.

It is easy to verify that the last two numbers satisfy the original equation.