$f\left( x-1 \right) +f\left( x+1 \right) =\sqrt { 3 } f\left( x \right)$ Let f be defined from real to real
$f\left( x-1 \right) +f\left( x+1 \right) =\sqrt { 3 } f\left( x \right)$ 
Now how to find the period of this function f(x)?
Can someone provide me a purely algebraic method to solve this problem please?
Update:My method
An elementary algebraic approach to the problem :
$f(x-1)+f(x+1)=\sqrt { 3 } f(x)$
Replace $x$ with $x+1$ and $x-1$ respectively.
We get $f(x)+f(x+2)=\sqrt { 3 } f(x+1)$ and 
$f(x-2)+f(x)=\sqrt { 3 } f(x-1)$
From these three equations we get $f(x-2)+f(x+2)=0$
Putting $x=x+2$ and adding with last equation we get $f(x-2)+f(x+4)=0$....(1)
Similarly $f(x-4)+f(x+2)=0$.....(2)
Put $x=x-6$ in (1)
We get $f(x-8)+f(x-2)=0$.....(3)
From (1) and (3) we get $f(x-8)=f(x+4)$
So the period of $f(x)$ is 12

For each fixed $y \in [0,1)$, your equation is a second order linear recurrence relation with constant coefficients. This suggests that the first step should be to solve the characteristic equation $\lambda^2-\sqrt{3} \lambda + 1 = 0$. You find $\lambda_1,\lambda_2 = \frac{\sqrt{3} \pm i}{2}=e^{\pm i \pi/6}$. This means that, for $y \in [0,1)$ and $n \in \mathbb{Z}$:
$$f(y+n)=c_1 e^{i n \pi/6} + c_2 e^{-i n \pi/6}$$
where $c_1,c_2$ are specified by the choice of $f(y)$ and $f(y-1)$. This tells you that the period is no larger than $\frac{2 \pi}{\pi/6} = 12$. It can be smaller: for instance your equation has a trivial constant solution $f \equiv 0$. 
