23. The sequence $\left(x_{n}\right)_{n-1}^{\infty}$ is defined recursively by
$$
x_{n+1}=\frac{x_{n}+(2-\sqrt{3})}{1-x_{n}(2-\sqrt{3})}
$$
with $x_{1}=1$. Determine the value of $x_{1001}-x_{401}$.
23. Answer: 0
Solution. Let $x_{n}=\tan \alpha_{n}$. Since $2-\sqrt{3}=\tan \left(\frac{\pi}{12}\right)$, it follows that
$$
x_{n+1}=\tan \alpha_{n+1}=\frac{\tan \alpha_{n}+\tan \left(\frac{\pi}{12}\right)}{1-\tan \alpha_{n} \tan \left(\frac{\pi}{12}\right)}=\tan \left(\alpha_{n}+\frac{\pi}{12}\right)
$$

So, $x_{n+12}=\tan \left(\alpha_{n}+\pi\right)=\tan \alpha_{n}=x_{n}$, implying that this sequence has a period of 12 . Observe that $1001 \equiv 5(\bmod 12)$ and $401 \equiv 5(\bmod 12)$. Consequently,
$$
x_{1001}-x_{401}=x_{5}-x_{5}=0
$$