## Task 1 - 230921

Determine all two-digit natural numbers $x$ for which the sum of $x$ and the number $y$ obtained by swapping the digits of $x$ is a perfect square!
If $a$ and $b$ are the digits of a two-digit number $x$, then

$$
1 \leq a \leq 9 ; \quad 0 \leq b \leq 9 ; \quad \text { and } \quad x=10 a+b
$$

By swapping the digits, we get $y=10 b+a$. The sum

$$
x+y=11 a+11 b=11(a+b)
$$

is a perfect square if and only if the prime factor 11 and any other prime factors are each present in even numbers in $a+b$. Due to (1), i.e., $1 \leq a+b \leq 18$, $a+b$ can contain no other prime factors besides 11. Therefore, $x+y$ is a perfect square if and only if $a+b=11$. This condition is met under the constraints (1) exactly for the value table $(a ; b)$ in the following table. Hence, exactly the numbers $x$ listed here have the required property.

| $a$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $b$ | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 |
| $x$ | 29 | 38 | 45 | 56 | 65 | 74 | 83 | 92 |