Four digit square with just Two digits!

$x^2= \overline{aabb} \\ x^2=11 (\overline{a0b}) \\ x^2= 11 (100a+b) \\ x^2 = 11( 99a + a+b) \implies a+b = 11 \text{ as } a+b<20 \\ x^2 = 11^2(9a+1) \implies 9a+1 = y^2\\ \text{ check, we have } 9a+1=9(7)+1 = 64 = 8^2 \\ \therefore, a=7, b=4, x=88 \\ \overline{aabb} = \boxed{7744}$

$aabb$ must be divisible by $11$ if you use the divisibility rule. So it also has to be divisible by $121$ .

Then you can just try $33^2, 44^2, 55^2 \ldots 99^2$ and see which one gives a number of form $aabb$ .

Simple calculations, can be done on paper, answer is $88^2 = \boxed{7744}$

Let $N = \overline{aabb} = 1100a + 11b$ . This means that $N$ is a multiple of $11$ . Since $N$ is a perfect square, it is also a multiple of $11^2$ , Let $N = 11^2n^2$ , where $n$ is an integer. Then

$\begin{aligned} 1100a + 11b & = 11^2n^2 & \small \blue{\text{Divide both sides by }11} \\ 100 a + b & = 11n^2 & \small \blue{\implies 100 a + b \equiv a + b \equiv 0 \text{ (mod 11)}} \\ 100a + \blue{11 - a} & = 11 n^2 & \small \blue{\implies a + b = 11 \implies b = 11-a} \\ 99a + 11 & = 11 n^2 & \small \blue{\text{Divide both sides by }11} \\ 9a + 1 & = n^2 \\ \implies a & = 7 & \small \blue{\text{The only single-digit integer solution,}} \\ \implies b & = 11-a = 4 \\ \implies n & = \sqrt{9a+1} = 8 \\ \boxed{7744} & = 11^2n^2 = 88^2 & \small \blue{\text{which is true.}} \end{aligned}$

Similar approach to @Percy Jackson 's: say $x^2=\overline{aabb}$ , that is $x^2=1100a+11b=11(100a+b)$

Now, if a prime $p$ divides a square number, then so does $p^2$ . So we have $11|100a+b$ ; using the usual rule for divisibility by $11$ , this means $a+b=11$ .

A square can only end in one of $\{0,1,4,5,6,9\}$ ; combining these two facts the list of possible squares is $\{2299,5566,6655,7744\}$

A square has to be a multiple of $4$ or one more than a multiple of $4$ ; it's easy to check that this only leaves $7744=88^2$ as a possibility.

aabb = 11 × a0b

Since the above is a square, a0b also is a multiple of 11, so a - 0 + b = a + b = 11.

b can only be 0, 1, 4, 6 or 9. 5 is impossible for the fixed 25 as *last two digits of squares, and in the same way we can eliminate 1, 6 & 9 (since all these numbers have their tens in an opposite parity making them inapplicable in the original square number aabb). This leave 0 and 4, but we already know that a + b = 11, so b = 0 will make the sum impossible.

Answer : b = 4 & a = 11 - b = 7

• (10k + 5)²
  = (10k)² + 2(10k)(5) + 5²
  = 100k² + 100k + 25
  = 100k(k + 1) + 25
  
  
  

100k(k + 1) will leave the units and tens places to 25 alone being a multiple of the hundred there.

I did a little of work with square numbers. I once found out that 88 x 88 = 7742, but I soon realized I made a mistake in my calculations. I changed the equation to make it true and found out that 88 x 88 = 7744. Thus, the answer is 7744 , which is 88 x 88, or 88 squared.