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.


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. 

Start with:
$f(x+1) = \sqrt{3}\,f(x) - f(x-1) \tag{1}$
Using $(1)\,$:
$$
f(x+3) = \sqrt{3}\,\big(\sqrt{3}\,f(x+1) - f(x)\big) - f(x+1) = 2 f(x+1) - \sqrt{3}\,f(x)\tag{2}
$$
Using $(1)$ and $(2)\,$:
$$\require{cancel}
\begin{align}
f(x+6) &= 2 f(x+4) - \sqrt{3}f(x+3) \\
 &= 
2 \big(2 f(x+2) - \sqrt{3} f(x+1)\big) - \sqrt{3} \big(2 f(x+1) - \sqrt{3}\,f(x)\big) \\
 &= 4 f(x+2) - 4 \sqrt{3} f(x+1) + 3 f(x) \\
 &= 4 \big(\cancel{\sqrt{3} f(x+1)} - f(x)\big) - \cancel{4 \sqrt{3} f(x+1)} + 3 f(x) \\
 &= -f(x) \tag{3}
\end{align}
$$
Therefore $f(x+6)=-f(x)\,$, so $\,f(x+12) = -f(x+6) = f(x)\,$.

[ EDIT ]   To address this part of the question:


Please ... do give the thought process ... what led you to thinking ...

You wrote I have no idea what recursive relation is in a comment, which precludes getting the hint from the roots of the characteristic polynomial, like other answers used.
One possible intuition would be to recognize something reminiscent of trig in the relation and, with a leap of imagination, think at the identity:
$$
\sin\left(x+ \frac{\pi}{6}\right)+\sin\left(x- \frac{\pi}{6}\right) = \sqrt{3} \,\sin(x)
$$
Other than that, by brute force, you can express $f(x+k)$ in terms of $f(x+1)$ and $f(x)\,$ by recursively applying the given relation, so you could try and calculate $f(x+2)\,$, $f(x+3)\,$, $f(x+4)\,$ etc in hope that the term $f(x+1)$ cancels out at some point, and you get a relation between $f(x+k)$ and $f(x)$ alone for some $k\,$, which is what happened in my answer at $k=6\,$.

[ EDIT #2 ]  To elaborate on the last paragraph as asked in a comment, $\,f(x+k)\,$ can be expressed in terms of $\,f(x+1)\,$ and $\,f(x)\,$ by applying the given relation repeatedly:


*

*$\,k=2\,$: $\;\;f(x+2) = \sqrt{3}\,f(x+1) - f(x)$

*$\,k=3\,$: $\;\;f(x+3) = 2 f(x+1) - \sqrt{3}\,f(x)$

*$\,k=4\,$: $\;\;f(x+4) = \sqrt{3}f(x+1)-2f(x)$

*$\,\cdots$
In other words $\,f(x+k) = a_k f(x+1) + b_k f(x)\,$ where $a_k,b_k$ are some constants. If a $k=n$ is found such that $\,a_n=0\,$ then that leaves $\,f(x+n)=b_n f(x)\,$. If furthermore $b_n^{\,m}=1$ for some $\,m\,$, then $\,f(x+ m \cdot n)=b_n f(x+(m-1) \cdot n)=\cdots=b_n^{\,m}f(x)=f(x)\,$ so $m \cdot n$ is a period for $f(x)$. In the given problem, this happens for $n=6$ with $b_6=-1$ and $m=2\,$, so $2 \cdot 6 =12$ is a period.
