Inspired by Aakash Khandelwal

We can claim that $a_n = \sin^2 \left(\dfrac{2^n \pi}{45} \right)$ and prove it by induction.

1. For $n=0$ , $a_n = \sin^2 \left(\dfrac{2^0 \pi}{45} \right) = \sin^2 \left(\dfrac{\pi}{45} \right)$ as given. The claim is true for $n=0$ .
2. Assuming the claim is true for $n$ , then

$\begin{aligned} \quad \quad a_{n+1} & = 4a_n (1-a_n) \\ & = 4\sin^2 \left(\frac{2^n \pi}{45} \right) \left(1-\sin^2 \left(\dfrac{2^n \pi}{45} \right) \right) \\ & = 4\sin^2 \left(\frac{2^n \pi}{45} \right) \cos^2 \left(\dfrac{2^n \pi}{45} \right) \\ & = \sin^2 \left(\dfrac{2^{n+1} \pi}{45} \right)\end{aligned}$

$\quad \space \space$ The claim is also true for $n+1$ . Therefore, the claim is true for all $n$ .

For $a_n = a_0$ ,

$\begin{aligned} \Rightarrow \sin^2 \left(\frac{2^n \pi}{45} \right) & = \sin^2 \left(\frac{\pi}{45} \right) \\ & = \sin^2 \left(k \pi \pm \frac{\pi}{45} \right) \quad \quad \small \color{#3D99F6}{\text{where } k \text{ is a positive integer.}} \\ \Rightarrow 2^n & \equiv 45 k \pm 1 \\ \Rightarrow 2^n & = \pm 1 \pmod{45} \end{aligned}$

The smallest $n$ is found to be $\boxed{12}$ , as $2^{12} \equiv 1 \pmod{45}$ .