If $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(x+1)+f(x-1)=\sqrt 3 f(x), \forall x$ then $f$ is periodic.
I tried to replace $x$ by $x+1, x-1$ in the equality,to get something like $f(x + k)=f(x)$ but without success.
Any help is appreciated.
Fix some $x \in \Bbb R$.
Define $a_n = f(x+n)$, for $n \in \Bbb N$.
Note that $a_n$ satisfies the recurrence relation $a_{n+1} = \sqrt{3} a_n - a_{n-1}$. By the general theory of recurrence relations it follows that $a_n=c_1 r_1^n + c_2 r_2^n$ where $r_1,r_2$ are the roots of the quadratic equation $r^2-\sqrt{3}r+1=0$.
But note that $r_1,r_2$ are complex numbers with norm 1 (specifically, $\pm e^{i\pi/6}$). Hence it follows that $a_n = A\cos (\pi n/6) + B \sin(\pi n /6)$. Thus you can conclude that $a_{n+12}=a_n$ for all $n$, i.e, $f(x+12)=f(x)$.
From the given equation, $$f(x+4)=\sqrt{3}f(x+3)-f(x+2),\\\sqrt{3}f(x+3)=3f(x+2)-\sqrt{3}f(x+1),\\f(x+2)=\sqrt{3}f(x+1)-f(x).$$ Adding those, we obtain that $$f(x+4)=f(x+2)-f(x),$$ and also $$f(x+6)=f(x+4)-f(x+2),$$ therefore $f(x+6)=-f(x)$. Hence, $$f(x+12)=-f(x+6)=f(x),$$ therefore $f$ is periodic with period $12$.
$$f(x+1) +f(x-1)=\sqrt3 f(x)$$ Let $x=x+1$ and $x=x-1$ $$f(x+2) + f(x)=\sqrt3 f(x+1)$$ $$f(x) + f(x - 2)=\sqrt3 f(x-1)$$ Adding the above we get $$f(x+2) + f(x - 2)+ 2f(x)=\sqrt3(f(x+1) +f(x-1))$$ $$=3f(x)$$ $$f(x+2) + f(x - 2)=f(x)$$ Letting $x=x+2$ in above $$f(x+4) + f(x)=f(x+2)=f(x)-f(x-2)$$
