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 that, when added to their reverse, give a perfect square?
Let's recall that two-digit numbers $ab$, where $a$ is the tens digit and $b$ is the units digit, are given by $ab = a \times 10 + b$. For example, $47 = 4 \times 10 + 7$.

If $ab$ is a two-digit number, then its reverse is $ba$. We have that

$$
ab + ba = a \times 10 + b + b \times 10 + a = (a + b) \times 11
$$

On the other hand, $a, b \leq 9$, so $a + b \leq 18$. Since 11 is a prime number and $a + b \leq 18$, 11 does not divide $a + b$ and, therefore, the product $(a + b) \times 11$ is only a perfect square if $a + b = 11$. Thus, we have 8 numbers satisfying the condition of the problem: $29, 38, 47, 56, 65, 74, 83$ and 92.