2. 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. Determine all such two-digit numbers?
Solution. Given a number is $10a + b$, and the number written with the same digits but in reverse order is $10b + a$. Their sum is equal to

$$
10a + b + 10b + a = 11(a + b)
$$

If $11(a + b)$ is the square of some number, then $a + b = 11k^2$ and since $a$ and $b$ are digits, we get $a + b = 11$. The possible cases are:

$$
\begin{aligned}
& a=2, b=9 ; a=3, b=8 ; a=4, b=7 ; a=5, b=6 \\
& a=6, b=5 ; a=7, b=4, a=8, b=3, a=9, b=2
\end{aligned}
$$

Accordingly, the required numbers are: $29, 38, 47, 56, 65, 74, 83$ and 92.