$f(x+1/2)+f(x-1/2)= f(x)$ Then the period of $f(x)$ is? 
$f(x+1/2)+f(x-1/2)= f(x)$. Then the period of $f(x)$ is:
a)$1$
b)$2$
c)$3$
d)$4$?

Attempt: 
I substituted $x= x \pm1/2$ but the equations I got didn't help at all. 
How do I go about solving such a question? I am just looking for a hint and not the entire solution.

Let $L$ be the operator on functions on $\mathbb{R}$ such that $(Lg)(x) = g(x+1/2)$. 
In terms of $L$, the equation at hand $f(x+1/2) + f(x-1/2) = f(x)$ is equivalent to $(L^2-L+1)f = 0$. 
Notice the factorization of polynomial 
$$(t^6-1) = (t^3-1)(t^3+1) = (t^3-1)(t+1)(t^2-t+1)$$
Any solution of $(L^2-L+1)f = 0$ automatically satisfy $(L^6-1)f = 0$ or equivalently $f(x+3) = f(x)$. This means $3$ is a period (through not necessary the smallest period) of $f$.
For non-trivial examples, take $f(x) = \sin\left(\frac{2\pi}{3}(6k+1 )x\right)$
where $k \in \mathbb{Z}$, the smallest period of $f(x)$ is $\frac{3}{|6k+1|}$ and you can verify they are solutions to the equation at hand.
