2. The square of the sum of the digits of the number $A$ is equal to the sum of the digits of the number $A^{2}$. Find all such two-digit numbers $A$.
2. Note that $A^{2} \leqslant 99^{2}=9801<9999$. Therefore, the sum of the digits of the number $A^{2}$ is less than $9 \cdot 4=36$. Since it is equal to the square of the sum of the digits of the number $A$, the sum of the digits of the number $A$ is less than $\sqrt{36}=6$.

But if the sum of the digits is less than 6, then it is less than or equal to 5. It remains to check all two-digit numbers whose sum of digits is no more than 5 - there are 15 such numbers: 10, $11,12,13,14,20,21,22,23,30,31,32,40,41,50$. It is not difficult to verify that only nine numbers satisfy the condition of the problem, as indicated in the answer.

Comment. A natural question arises: how can we find all (not just two-digit) numbers that satisfy the condition of the problem? The author is not aware of a simple answer to this question. It can be proven (try to do this) that any number $A$, the square of the sum of whose digits is equal to the sum of the digits of $A^{2}$, has the following properties:

- in its decimal representation, only the digits 0, 1, 2, and 3 can appear;
- threes cannot stand next to each other or next to twos. Unfortunately, these conditions are not sufficient: for example,

$$
222^{2}=49284
$$

but the sum of the digits of the number 49284 is 27 and does not match $6^{2}$.