A function $f:\mathbb{R} \to \mathbb{R}$ is such that $f(2)=2$ and $$f(x+1)+f(x-1)=\sqrt{3}f(x) \tag{1}.$$ Find the values of $f(0)$, $f(4)$, $f(6)$ and $f(18)$.
My approach: replace $x$ with $x+1$ in $(1)$ we get
$$f(x+2)+f(x)=\sqrt{3}f(x+1) \tag{2}.$$
Replace $x$ with $x+2$ in $(1)$ and using $f(x+2)$ from $(2)$ we get
$$f(x+3)+f(x+1)=\sqrt{3}f(x+2)=\sqrt{3}\left(\sqrt{3}f(x+1)-f(x)\right) $$ $\implies$
$$f(x+3)=2f(x+1)-\sqrt{3}f(x)=2f(x+1)-f(x+1)-f(x-1)=f(x+1)-f(x-1).$$ So $$f(x+3)=f(x+1)-f(x-1) \tag{3}.$$
Replace $x$ with $x+2$ in $(3)$ we get
$$f(x+5)=f(x+3)-f(x+1)=-f(x-1).$$
So
$$f(x+5)=-f(x-1)$$ $\implies$
$f(x+6)=-f(x)$ and $f(x+12)=f(x).$
So $f(x)$ is periodic with Fundamental Period $12$.
Using $x=1$ $(3)$ we get $f(0)+f(4)=2$ but I could not find the value of $f(0)$.
Without any additional conditions on $f$, it is possible to pick $x_0, f(x_0), f(x_0+1)$ arbitrary and then use $(1)$ (i.e. $f(x) = \sqrt 3f(x-1)-f(x-2)$ to go forward $f(x) = \sqrt 3f(x+1)-f(x+2)$ to go backwards) to extend this uniquely to $x_0+\mathbb Z$. Likewise, we can prescribe arbitrary values for $f|_{(0,2]}$ and extend this uniquely to all of $\mathbb R$. Even if we want $f$ to be smooth, we can prescribe $f$ on $(\epsilon,2]$ arbitrarily. This gives us $f|_{(\epsilon-1,0]}$. If we "glue" these parts smoothly together, the resulting $f\colon\mathbb R\to\mathbb R$ is smooth.
With $a:=f(1)$ we find by the recursion that $f(0)=a\sqrt 3-2$, $f(18)=f(6)=-f(0)=2-a\sqrt 3$ and $f(4)=4-a\sqrt 3$. However, as just seen $a$ can be picked arbitrarily, even if we assume $f$ is smooth.