8.3. 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 $A$ and explain why there are no others.
Answer: $10,20,11,30,21,12,31,22,13$.

Solution: Let's consider what the sums of the digits of the original number and the number squared can be. The square of a two-digit number can result in no more than a four-digit number. The sum of the digits of a four-digit number is no more than $9 \cdot 4=36$, and 36 is only achieved in the case where the number is 9999. However, 9999 is not a square (it is divisible by 11 but not by 121), so the sum of the digits of the squared number is strictly less than 36. On the other hand, it is equal to the square of the sum of the digits of the original number. There are five natural number squares less than 36. For each of them, we will consider the sum of the digits of the original number and all possible original numbers to find those that fit us.

| sum of digits $A$ | sum of digits $A^{2}$ | $A$ | $A^{2}$ | sum of digits $A^{2}$ | check |
| :--- | :--- | :--- | :--- | :--- | :--- |
| 1 | 1 | 10 | 100 | 1 | fits |
| 2 | 4 | 20 | 400 | 4 | fits |
|  | 4 | 11 | 121 | 4 | fits |
| 3 | 9 | 30 | 900 | 9 | fits |
|  | 9 | 21 | 441 | 9 | fits |
|  | 9 | 12 | 144 | 9 | fits |
| 4 | 9 | 40 | 1600 | 7 |  |
|  | 16 | 31 | 961 | 16 | fits |
|  | 16 | 22 | 484 | 16 | fits |
|  | 16 | 13 | 169 | 16 | fits |
| 5 | 25 | 50 | 2500 | 7 |  |
|  | 25 | 41 | 1681 | 16 |  |
|  | 25 | 32 | 1024 | 7 |  |
|  | 25 | 23 | 529 | 16 |  |
|  | 25 | 14 | 196 | 16 |  |

Criteria: answer, answer with verification - 3 points. Proof that the sum of the digits of the original number does not exceed 6 - 2 points. These points are cumulative! Deduct 1 point for each missed case, and no less than 3 points for a larger number of missed cases.