Classique Number

Let the license number be $N$ , and its first two and last two digits be $a$ and $b$ respectively. Then $N = 1000a + 100a + 10b + b = 11(100a+b)$ . Since $N$ is a perfect square, $100a+b$ must be a multiple of 11 and must be of the form $100a+b = 11n^2$ , where $n$ is a positive integer. And we have:

$\begin{aligned} \implies 100a + b & \equiv a + b \equiv 0 \text{ (mod 11)} \\ \implies a + b & = 11 \end{aligned}$

Therefore,

$\begin{aligned} 100a + b & = 11n^2 \\ 100a + 11 - a & = 11n^2 \\ 99a + 11 & = 11n^2 \\ 9a + 1 & = n^2 \\ \implies a & = \frac {n^2 -1}9 = 7 & \small \color{#3D99F6} \text{when }n = 8 \\ b & = 11-7 = 4 \\ \implies N & = \boxed {7744} = 88^2 \end{aligned}$