$\sqrt{3} f(x + 1) = f(x) + f(x+2)$

$\sqrt{3} f(x - 1) = f(x - 2) + f(x)$

--- From this:

$\sqrt{3} ( f(x - 1) + f(x + 1)) = 2f(x) + f(x - 2) + f(x+2)$

$\sqrt{3} (\sqrt{3} f(x)) = 2f(x) + f(x - 2) + f(x+2)$

$f(x) = f(x-2) + f(x + 2)$

--- Do it again

$f(x + 1) = f(x-1) + f(x + 3)$

$f(x - 1) = f(x-3) + f(x + 1)$

We get:

$-f(x - 3) = f(x + 3)$

$f(x) = -f(x+6)$

therefore $f(x) = f(x + 12)$

I found it easiest to just solve the recurrence relation over the whole numbers; we can then keep shifting the interval(*) over which we solved the equation to make it hold for all numbers:

$\sqrt(3)f(x)=f(x-1)+f(x+1)$

Characteristic equation:

$\displaystyle r^{2}-\sqrt{3}r+1=0$

$\displaystyle r=\frac{\sqrt{3}\pm i}{2} = e^{\pm i\theta}$ .

Thus, since $f(x)=a\cos{x\theta}+b\sin{x\theta}$ , we have the period $\displaystyle \frac{2\pi}{\theta}$ = $\displaystyle \frac{2\pi}{\pi/6}$ = $12$ .

Edit: as it turns out, other (smaller) fundamental periods could be generated for differing $f(x)$ -s. For example, if one uses $\theta=\frac{11\pi}{6}$ instead of $\frac{\pi}{6}$ , the general solution yields a function with period $\frac{12}{11}$ . Similarly, any period $\frac{12}{12k\pm 1}$ can be achieved. Furthermore, one can combine two different $\theta$ -s satysfying the equation (and yielding by themselves a solution with a period of the forementioned form $\frac{12}{12k\pm 1}$ by using them for different parts of the interval(*) from the beginning to yield a function with any period $\frac{12}{k}$ , as long as $(12;k)=1$ . Credit to @Will Heierman.

Fourier analysis: $f(x)$ restricted to domain $\mathbb Z$ is the sum of harmonics based on a period with integer value $T$ ; it is the real part of: $f(x) = \sum_{n = 0}^{T - 1} A_n \exp\left( \frac{2n\pi i x}T\right).$ Then $f(x \pm 1) = \sum_{n = 0}^{T - 1} A_n \exp\left( \frac{2n\pi i x}T\right)\exp\left(\pm \frac{2n\pi i}T \right),$ so that (with $\exp(ai) + \exp(-ai) = 2\cos a$ ) $f(x+1) + f(x-1) = 2\sum_{n = 0}^{T - 1} A_n \exp\left( \frac{2n\pi i x}T\right)\cos\left(\frac{2n\pi}T \right),$ which for all integers $x$ should be equal to $\sqrt 3 f(x)$ : $2\sum_{n = 0}^{T - 1} A_n \exp\left( \frac{2n\pi i x}T\right)\cos\left(\frac{2n\pi }T \right) = \sqrt 3 \sum_{n = 1}^{T-1} A_n \exp\left( \frac{2n\pi i x}T\right).$ This requires that for every $n$ , either $A_n = 0$ or $2\cos(2n\pi/T) = \sqrt 3.$ This means that $\cos\frac{2n\pi}T = \tfrac12\sqrt 3\ \ \ \ \therefore\ \ \ \ \frac{2n\pi}T = \pm \frac\pi 6\ \ \ \therefore\ \ \ n = \frac T{12}.$ Thus only the term with $n/T = 1/12$ will be non-zero. Therefore the function has period $\boxed{12}$ and must be of the form $f(x) = \mathrm{Re}\ A \exp\left( \frac{2\pi i x}{12}\right) = A_1\cos\left( \frac{2\pi x}{12}\right) + A_2\sin\left( \frac{2\pi x}{12}\right),$ where $A = A_1 + A_2i$ .

Note that we have only considered the restriction of $f$ to the whole numbers $\mathbb Z$ . We can give the same reasoning for the restriction to any subdomain of the form $x_0 + \mathbb Z$ , with $0 \leq x_0 < 1$ . In general, then, $f(x) = A_{1,\{x\}}\cos\left( \frac{2\pi x}{12}\right) + A_{2,\{x\}}\sin\left( \frac{2\pi x}{12}\right),$ where $\{x\} := x - \lfloor x\rfloor$ and $(A_{1,x},A_{2,x})_{x \in [0,1\rangle}$ is an uncountable collection of parameters, not all zero.

Let us make the bold assumption that $f(x)=C a^{x}$ where C and a are constants.

Substituting this into the equation and dividing by C gives

$\sqrt{3} a^{x}=a^{x-1}+a^{x+1}$

Dividing throughout by $a^{x-1}$ gives the quadratic equation

$\sqrt{3}a=1+a^{2}$

with the easy solution

$a=\frac{\sqrt{3} \pm i}{2}=e^{\pm\frac{i \pi}{6}}$

And so the general solution to the functional equation is

$f(x)=C_{1}e^{\frac{i \pi x}{6}}+C_{2}e^{-\frac{i \pi x}{6}}$

For a full solution we would need boundary conditions to determine the arbitrary constants. However all we want is the period, T, of the function. Knowing how the imaginary exponentials relate to the trig functions which have period $2 \pi$ allows us to see from the general solution that

$\frac{\pi}{6}T=2 \pi$

which implies that $T=\boxed{12}$

as usual I took a simple minded approach. Sine is periodic, so I let f(x) = sinks, with k to be determined. It turns out to be Pi/6 to satisfy the relation for f(x). Next we want sin(Pi x/6)=sin (Pi/6 (x+T)), which gives T= 12. Probably the rt(3) made me think of a trig function. I have to admire the other solutions which made no assumptions, well done all!