Wow!! Only such number!!!

The number should be of the form $\overline{AABB}$ where $A$ and $B$ are digits and $A\neq 0$ .

It's clear from the divisibility test of $11$ that

$\overline{AABB}$ gives remainder $0$ when divided by $11$

So, it should be a square of any of the $2$ -digit multiples of $11$ greater than $22$ ( $121$ and $484$ are $3$ -digit numbers.)

Now, you can simply check it :

$33^2 = 1089$

$44^2 = 1936$

$55^2 = 3025$

$66^2 = 4356$

$77^2 = 5929$

$88^2 =\boxed{7744}$

$99^2 = 9801$

Amazingly, there's only one such number!

Let each of the first two digits be $x$ and each of the last two digits be $y$ . Then the number is $1100x+11y=11(100x+y)=33^2\times x+11(x+y)$ .

Since $0<x\leq 9$ and $0\leq y\leq 9$ , therefore $0<x+y\leq 18$ .

Since $100x+y$ and hence $x+y$ must be a multiple of $11$ , therefore $x+y=11\implies 100x+y=99x+11=11(9x+1)\implies 11(100x+y)=121(9x+1)$ .

So $9x+1$ must be a perfect square, say, $a^2$ . Then

$x=\dfrac{(a-1)(a+1)}{9}$ .

Since both $a-1$ and $a+1$ can't be multiples of $3$ , $a+1$ must be $9$ (since $x\leq 9$ ). That is, $a=8$ , and $x=7\implies y=11-7=4$ .

Hence the required number is $\boxed {7744}$ .

All perfect squares are either divisible by $4$ or leaves a remainder $1$ when divided by $4$

Let the number be $AABB$

Now $BB$ is either divisible by $4$ or leaves a remainder of $1$ when divided by $4$ . If you were to check, $BB$ is either $33, 44, 77$ or $88$ . However, square numbers do not have unit digits as $3, 7,$ or $8$ . Hence, $BB$ is fixed as $44$ .

Therefore, the numbers can be anything from $1144, 2244, 3344, 4444, 5544, 6644, 7744, 8844$ or $9944$

Further, $AABB$ is divisible by $11$ ; therefore it should also be divisible by $11^2$

It means the number is $121 × x$ , where $x$ is a square number, which ends with $4$ .

There are only two numbers possible, that is, $4$ or $64$ .

$4 × 121 = 484$ and $64 × 121 = 7744$

Therefore, the only number with the above mentioned property is $\boxed {7744}$

Let us see what can be last digit of a perfect square.It can be 0,1,4,5,6 or 9

Let us suppose number is aabb, then either this number should be divisible by 4 or leaves remainder as 1.

What can be the last two digits from first observation.

These are 00,11,44,55,66 and 99. But,only 44 and 00 follows the second rule.

let us take 00 as last two digit,then number will be aa00 which is equal to 1000 a+100 a+0 10+0=1100a which will not be perfect square for any digit a ( 100 11a)

So,our last option is 44 and number will be aa44 which will equal to 1000 a+100 a+40+4=1100a+44=44(25a+1) = 4*11(25a+1),therefore 25a+1 should be factor of 11 . 25a+1=22a+3a+1,since 22a is divisible by 11 so,we have make to 3a+1 divisible by 11 which will happen at a=7 So,our number comes out 7744.

check:-

7744=77*100+44=44(175+1)=44*176=44*44*4...so,its square root will be 88.

$\boxed{88}^2 = 7744$