Square number

If the number is $\overline{AABB} = 1100A+11B = 11(100A + B),$ then $100A+B = \overline{A0B}$ must be the product of 11 and a perfect square. The only multiples of 11 in the form $\overline{A0B}$ are:

$209 = 11\cdot 19$

$308 = 11\cdot 28$

$407 = 11 \cdot 37$

$506 = 11 \cdot 46$

$605 = 11 \cdot 55$

$704 = 11 \cdot 64,$ which satisfies our criteria!

$803 = 11\cdot 73$

$902 = 11 \cdot 82$

Thus, $A=7$ and $B=7,$ so the perfect square is $7744 = (11 \cdot 8)^2.$

If the first two digits and alst two digits are identical, then the number must be a multiple of 11. Since it is a perfect square, it must be a multiple of 121 (Euler's theorem of prime factor representation of an integer).

Perfect squares end in 0,1,4,5,6,9. If the last digit is an odd number then the tens digit of a perfect square should be even (Any odd perfect square=(10x+1)^2=100x^2+20x+1 or (10x+3)^2=100x^2+60x+9 or (10x+5)^2=100x(x+1)+25 or (10x+7)^2= 100x^2+140x+9 or (10x+9)^2=100x^2+180x+9 in each case 10s digit is even). Since last two digits are identical, the number cannot be odd

Hence last two digits must be 00, 44 or 66. An even perfect square is of the form (2x)^2=4x^2 and is a multiple of 4. Hence the number cannot end in 66

If the number ends in 00 then we have a multiple of 11 which is a perfect square. The smallest such multiple of 11 is 121 which is 11 * 11 and hence no single digit multiple satisfies it. Therefore, last two digits have to be 44. Then x is of the form AA44. x/11 is of the form A04 which is also a multiple of 11 (since x is a multiple of 121). Hence A=7, i.e., x=7744

I solved it by first seeing the range of numbers which could be squared to obtain a 4 digit number which satisfies the conditions stated in the question. The range, hence, starts from above 30 (whose square is 900). According to the question, the square has the first two digits same, and the last two same. This satisfies the rule of being divisible by 11. Thus the number (which is squared to obtain a 4 digit number) has to be divisible by 11 too. Thus starting from 30 we can take all the numbers which are divisible by 11, into consideration, and by trial and error, we can quickly find that 88 satisfies. The square of 88 is 7744. Thus that is the answer.