Consider four-digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

The number we can write $1000 \; \text{a} + 100 \; \text{a} + 10 \; \text{b} + \text{b}. = 1100 \; \text{a} + 11 \; \text{b} = 11(100 \; \text{a} + \text{b}) \longrightarrow(1)$

This number will be perfect square if a number in the form of $\text{a0b}$ is divisible by $11$ and when $\text{a0b}$ is divided by $11$ perfect square will be get.

$\text{a0b}$ will be divisible by $11$ when $a+b=11$