$f(x+1/2)+f(x-1/2)= f(x)$ 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.
 A: We may conclude that $f(x)=f(x+3)$ holds for all $x$.
Following @Mike Earnest's suggestion,
\begin{align}
f(x)&=f(x+1/2)+f(x-1/2),\\
f(x-1/2)&=f(x)+f(x-1).
\end{align}
Add up these two equations, and you will have
$$
f(x+1/2)+f(x-1)=0,
$$
or, due to the arbitrariness of $x$,
$$
f(x+3/2)+f(x)=0.
$$
Now let $x\to x+3/2$, and
$$
f(x+3)+f(x+3/2)=0.
$$
The difference of the last two equations gives
$$
f(x)=f(x+3).
$$
Of course, as @Przemysław Scherwentke has mentioned, this might not mean that $3$ is the period of $f$, as we do not know if this $3$ is the smallest non-negative value $T$ that ensures $f(x)=f(x+T)$. All in all, this is what we can get from the given relation.
A: 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.
A: $f(x+1) + f(x) = f(x + \frac 12)= f(x) - f(x - \frac 12)$
$f(x+1) = -f(x - \frac 12)$
So $f(x - \frac 12) = -f(x - \frac 12 - 1\frac 12) = -f(x-2)$ 
So $f(x+1) = -f(x-\frac 12) = f(x-2)$.
So period is at most $3$.
Not sure off the top of my head how to show it doesn't necessarily need to be smaller.
A: The period may not exist. For example a constant function $f\equiv0$ is an example with no (minimal) period.
A: hint:
$$
\eqalign{
  & f(x - 1/2) + f(x + 1/2) = f(x)  \cr 
  & x = y/2\quad  \Rightarrow   \cr 
  & f\left( {\left( {y - 1} \right)/2} \right) + f\left( {\left( {y + 1} \right)/2} \right) = f\left( {y/2} \right)  \cr 
  & f\left( {y/2} \right) = g\left( y \right)\quad  \Rightarrow   \cr 
  & g\left( {y + 1} \right) = g\left( y \right) - g\left( {y - 1} \right) \cr} 
$$
which is a Finite Difference equation of 2nd degree, whose solution depends on two starting conditions ...
A: $f(x+1/2)=f(x+1)+f(x)$; $f(x-1/2)=f(x)+f(x-1)$; substituting in the original equation: $2f(x)+f(x+1)+f(x-1)=f(x)$, $f(x)=-f(x-1)-f(x+1)$. Using the last identity, $f(x+1)=-f(x)-f(x+2)$ and substituting in the last equation $f(x+1)$, we obtain $f(x)=-f(x-1)+f(x)+f(x+2)$ or $f(x-1)=f(x+2)$ and the period is $3$
