The reverse of a two-digit integer is the number obtained by reversing the order of its digits. For example, 34 is the reverse of 43. How many numbers exist such that when added to their reverse, they give a perfect square?
Let's denote by $a b \mathrm{e}$ ba the number and its reverse. We have that

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

On the other hand, $a \leq 9$ and $b \leq 9$, so, $a+b \leq 18$. Since 11 is a prime number and $a+b \leq 18$, for $11(a+b)$ to be a perfect square, we can only have $a+b=11$.

Thus, we have 8 numbers satisfying the condition of

Reminder: two-digit numbers where $a$ is the tens digit and $b$ is the units digit are of the form

$10 \times a+b$.

Ex: $47=4 \times 10+7$

problem: 29, 38, 47, 56, 65, 74, 83 and 92.