49(971). A two-digit number, when added to the number written with the same digits but in reverse order, gives the square of a natural number. Find all such two-digit numbers.
Solution. Let's have a two-digit number $\overline{a b}=10 a+b$. Then, according to the condition, $\overline{a b}+\overline{b a}=10 a+b+10 b+a=11(a+b)$. Since 11 is a prime number, the number $11(a+b)$ can only be a square of a natural number if $a+b=11$. Considering that $a$ and $b$ are digits, from the condition $a+b=11$ we can easily find all two-digit numbers that satisfy the condition of the problem.

Answer: $29,38,47,56,65,74,83,92$.