Special Perfect Square

It is clear that 31< $\sqrt{aabb}$ <100 and $aabb$ is divisible by 11[because $a-a+b-b=0$ ]
11 is a prime number so $\sqrt{aabb}$ is also divisible by 11.
So, now we have 7 numbers {33,44,55,66,77,88,99} squares of which may be a number of the form $aabb$ .
Now $8\times8$ =64 and 6+4 = 10 So $88^{2}$ will be the answer.
$88^{2}=\boxed{7744}$

Let $x = \sqrt{\overline{aabb}} = \sqrt{11(100a+b)}$ for $1 \leq a \leq 9$ and $0 \leq b \leq 9$ . The only possible value of $(a,b)$ satisfying $x$ as the integer is $(a,b) = (7,4)$

Here the number $n$ = $\sqrt{aabb}$ $=$ $1000a+100a+10b+b$ $=$ $11(100a+b)$

Now it is clear, that $n$ is divisible by $11$

As $n$ is a square number , it is divisible by $n^{2}$

Again, $100a+b$ $=$ $11*9a+(a+b)$

So, $(a+b)$ is divisible by $11$ .

But the last digit of a square number can not be $2,3,7,8$ . So, possible value of $(a,b)$ is $(2,9),(5,6),(7,4)$ .

Again n= $11^{2}*(\frac{100a+b}{11}$ ). S0 $\frac{100a+b}{11}$ is also a positive integer .

Now substituting the possible value of $(a,b)$ , for $(7,4)$ $n$ be a square number.

So $aabb$ = $7744$

aabb=1100a+11b =11 (100a+b)=11 [99a+a+b].here a+b must be divisible by 11 .and a,b various from 1to9.i.e. a+b=7+4=11.so required number is 7744.

Little JS program and it is done:

for (var i = 0; i < 100; i++) { var square = i * i; console.log(square); }

Then you just scan for the number you want. Or:

for (var i = 0; i < 100; i++) { var square = i * i; var squareByEleven = Math.floor(square / 100); if (squareByEleven % 11 === 0) { console.log(square); } }

This last one narrows it down because it will only print the numbers that, once divided by 100 and having its decimal places removed, are divisible by 11, which means its first two digits are the same. Then you just look for the final answer: 7744

Square of 88 is 7744