Problem 3. A two-digit number added to the number written with the same digits but in reverse order gives a number which is a square of some natural number. Find all such two-digit numbers.

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Solution. Let $\overline{a b}$ be the desired number. From the condition, we get

$$
10 a+b+10 b+a=n^{2}, \quad n \in \mathbb{N}
$$

from which $11(a+b)=n^{2}$. Since $a+b<19$, it follows that $a+b=11$, hence $\overline{a b} \in\{29,38,47,56,65,74,83,92\}$.