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?

Explanation:-

Any four digit number in which first two digits are equal and last two digits are also equal will be in the form 11*(11a+b) i.e. it will be the multiple of 11 like 1122, 3366, 2244, . . . . Now, let the required number be aabb. Since aabb is a perfect square, the only pair of a and b that satisfy the above mentioned condition is a = 7 and b = 4.
Hence, 7744 is a perfect square.