A Singular Square

$\begin{aligned} \overline{AABB} & = 1100A + 11B \\ & = {\color{#3D99F6}11}(100A+B) & \small \color{#3D99F6} \text{Since 11 is a factor and }\overline{AABB} \text{ is a} \\ & = 11(99A + A+ B) & \small \color{#3D99F6} \text{perfect square, }11^2 \text{ must be a factor.} \\ & = 11^2 \left( 9A + \frac {A+ B}{11} \right) & \small \color{#3D99F6} \text{For integral RHS, } A+B = 11 \\ & = 11^2 \color{#3D99F6}(9A + 1) & \small \color{#3D99F6} (9A+1) \text{ must be a perfect square.} \\ & = 11^2 \color{#3D99F6}(9\times 7 + 1) & \small \color{#3D99F6} \implies A=7, \ B = 11-7=4 \\ &=11^28^2 \\ & = 88^2 \\ & = \boxed{7744} \end{aligned}$

This can be solved by checking whether each of the possible numbers of the form $\overline{AABB}$ are square. However, I think we would prefer to be a little more clever about it. Firstly, a good start with problems of this type is to write our number algebraically:

$n=1000A+100A+10B+1B$

$n=1100A+11B$

$n=11(100A+B)$

So we can see that our number, $n$ , must be divisible by $11$ . Also, as $n$ is a perfect square, we can see that it must be divisible by $11^2=121$ . Let:

$n=m^2$

As we know that $n$ is divisible by $121$ we can see that $m$ must be divisible by $11$ . At this point we can try all the multiples of $11$ from $33$ to $99$ , and we will discover our answer. This has reduced our search size by more than a factor of $10$ . However, we can do better.

$n=11(100A+B)$

As $n$ is divisible by $121$ , we can see that $100A+B$ is divisible by $11$ . However,

$100A+B=99A+(A+B)$

So, $A+B$ must be divisible by $11$ . As $A$ and $B$ are both less than $10$ , $A+B=11$ . From here we have,

$n=11(99A+11)$

$\frac{n}{121}=9A+1$

And as n is a perfect square, $9A+1$ must also be a perfect square. The only possibility is $A=7 \implies 9A+1=64$ . This gives us,

$n=121\times 64=7744=88^2$

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When I first solved this, I stopped at the search of all multiples of $11$ between $33$ and $99$ , which I think is a legitimate strategy.