How to find period of a real function $f$ given the functional equation $\sqrt{3}f(x) = f(x-1) + f (x+1) $? If a periodic function satisfies the equation $\sqrt{3}f(x) = f(x-1) + f (x+1) $ for all real $x$  then prove that fundamental period of the function is $12$.
Here fundamental period means the smallest positive real for which function repeats its value for all $x$.
I tried replacing $x$ by $x \pm 1$ then try to find $f(x)$ in terms of other but always end up with it in terms of sum of other two arguments in the function eg $f(x-2)$ + $f (x+2)$  etc. 
Please provide a general method and also especially do give the thought process or reasoning for all the steps ie why you are doing these particular steps or what led you to thinking that doing these steps would give you the period of f.

Define the linear operator $T$ by
$$Tf(x):=- √3f(x) + f(x-1) + f(x+1)$$
and $E$ by
$$Ef(x)=f(x+1).$$
One can see that $T=E-\sqrt{3}+E^{-1}$.
Consider solving $$x-\sqrt{3}+x^{-1}=0$$
$$\implies x^2-2x\frac{\sqrt{3}}{2}+1=0$$
$$\implies\left(x-\frac{\sqrt{3}}{2} \right)^2=-\frac{1}{4}$$
$$\implies x = \frac{\sqrt{3}\pm i}{2}=\exp\left(\frac{\pm i\pi}{6}\right)=\omega_{\pm}$$
One can thus see that $T=E^{-1}(E-\omega_+)(E-\omega_-)$.
Note that the $\omega_\pm$ are primitive $12^{th}$ roots of unity, so the smallest $n$ such that $T$ will factor exactly into $E^n-1$ is $12$.
It thus follows that the fundamental period of such a function must be $12$.
