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

Algebraic (non-coanstant)?  NO.
But, for example
$$
f(x) = \sin\frac{\pi x}{6}
$$
is a solution.
