Everything is function of time period

$\begin{bmatrix} f(x+1) \\ f(x) \end{bmatrix}=\begin{bmatrix} \sqrt{3} & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} f(x) \\ f(x-1) \end{bmatrix}$

The eigenvalues of the matrix $A$ are $e^{\pm\pi i/6}$ , twelfth roots of unity, so that $A^{12}=I_2$ , while $A^m$ fails to have nonzero fixed points for $0<m<12$ . Thus $f(x+12)=f(x)$ for all $x$ , and the fundamental period is $\boxed{12}$ .

$\sqrt{3}f(x)=f(x+1)+f(x-1)$ $\sqrt{3}f(x+2)=f(x+3)+f(x+1)$

$\sqrt{3}(f(x+2)+f(x))=f(x+3)+2f(x+1)+f(x-1)$ $\sqrt{3}(\sqrt{3}f(x+1))=f(x+3)+2f(x+1)+f(x-1)$ $3f(x+1)=f(x+3)+2f(x+1)+f(x-1)$ $f(x+1)=f(x+3)+f(x-1)$ $f(x+3)=f(x+5)+f(x+1)$

$f(x+1)+f(x+3)=f(x+3)+f(x-1)+f(x+5)+f(x+1)$ $f(x+5)+f(x-1)=0$ $f(x+11)+f(x+5)=0$

$f(x+11)+f(x+5)=f(x+5)+f(x-1)$ $f(x+11)=f(x-1)$

$f(x+12)=f(x)$

Hence, period of $f(x) =12$

Since the equation given is a linear recurrence relation , we can find $f(x)$ by solving the characteristic polynomial which is as follows:

$\begin{aligned} x^2 - \sqrt{3}x + 1 & = 0 \\ \Rightarrow x & = \frac{\sqrt{3} \pm i}{2} = e^{\pm i \frac{\pi}{6}} \quad \text{where } i^2 = -1 \end{aligned}$

Therefore, the function is of the form

$\begin{aligned} f(x) & = ae^{i\frac{\pi x}{6}} + be^{-i\frac{\pi x}{6}} \\ & = (a+b) \cos \left(\frac{\pi x}{6}\right) + i (a-b) \sin \left(\frac{\pi x}{6}\right) \\ \Rightarrow f(x+12) & = (a+b) \cos \left(\frac{\pi (x+12)}{6}\right) + i (a-b) \sin \left(\frac{\pi (x+12)}{6}\right) \\ & = (a+b) \cos \left(\frac{\pi x}{6} + 2 \pi \right) + i (a-b) \sin \left(\frac{\pi x}{6} + 2 \pi \right) \\ & = (a+b) \cos \left(\frac{\pi x}{6}\right) + i (a-b) \sin \left(\frac{\pi x}{6}\right) \\ & = f(x) \end{aligned}$

Therefore, the fundamental period of the function is $\boxed{12}$ .