Finding period of $f$ using the functional equation $f(x) + f(x+a+b) = f(x+a) + f(x+b)$ 
Assume that $f:\mathbb R\to\mathbb R$, and $f(x) + f(x+a+b) = f(x+a) + f(x+b)$ where $|f(x)|\le 1 \ \forall x \in \mathbb R$ ($a > b$).
Find the period of the function $f(x)$.

My approach:
$$f(x+a+b) - f(x+a) = f(x+b) - f(x)$$
After this, I tried replacing $x $ with $x+a $and several other substitutions but couldn't reach a conclusive result. Also, I'm not sure how to use $ |f(x)|\le 1 $ to get the function's period.
Of course, the aim is to establish a result of the form $f(x+T) = f(x)$.
It'd be great if someone could give a detailed solution. Thanks a lot!
$f(x)$ is not a constant function. If it is, prove it to be so.
Edit:
The problem is a modified IMO 1996 shortlisted problem. In that problem, $a$ and $b$ had specific values ($\frac16$ and $\frac17$), but I wish to generalize the problem for any two constants $a$ and $b$.
 A: Let $$f(x+a+b) - f(x+a) = f(x+b) - f(x)\tag1$$ for all $x\in R$. Then,
$$f(x+a+nb) - f(x+a) = f(x+nb) - f(x)\tag2$$ for all $x\in R$ and all natural $n$, as we can show by induction: it's trivially true for $n=0$. Let's assume (2) is true for $n=k$:
$$f(x+a+kb) - f(x+a) = f(x+kb) - f(x)\tag3$$ for all $x\in R$. Replacing $x$ by $x+kb$ in (1), we obtain $$f(x+a+(k+1)b) - f(x+a+kb) = f(x+(k+1)b) - f(x+kb)\tag4$$ for all $x\in R$, and adding (3) and (4) gives
$$f(x+a+(k+1)b) - f(x+a) = f(x+(k+1)b) - f(x)$$ for all $x\in R$, i.e. (2) for $n=k+1$, proving the induction step.
With exactly the same reasoning, exchanging the roles of $a$ and $b$, we can extend that to $$f(x+ma+nb) - f(x+ma) = f(x+nb) - f(x)\tag5$$ for all $x\in R$ and all natural $m,n$.
Now let's assume $b/a$ is rational, i.e. $b/a=m/n$ for some natural numbers $m,n$. That means $$ma=nb=c\tag{period},$$ and for those $m,n$, (5) becomes
$$f(x+2c)-f(x+c)=f(x+c)-f(x)\tag6.$$ Adding $f(x+c)-f(x)$ on both sides of (6), we obtain $f(x+2c)-f(x)=2[f(x+c)-f(x)]$, and using induction, again, we can derive
$$f(x+nc)-f(x)=n[f(x+c)-f(x)].$$ So we have
$$|f(x+c)-f(x)|=\frac1n|f(x+nc)-f(x)|\le\frac2n$$ from the assumption $|f(x)|\le1$, and since that is true for all natural $n$, we must have 
$$f(x+c)=f(x)$$  for all $x\in R$, i.e. $f$ has period $c$.
The assumption that $b/a$ is rational is essential: the function $\displaystyle f(x)=\frac12\left(\cos\frac{2\pi}ax+\cos\frac{2\pi}bx\right)$ obviously satisfies (1) and $|f(x)|\le1$ for all $x\in R$, but it's periodic if and only if $b/a$ is rational, as can be seen by assuming $f(T)=f(0)=1$, implying that both $\cos\frac{2\pi}aT$ and $\cos\frac{2\pi}bT$ have to be $1$, i.e. $T/a$ and $T/b$ have to be integers.
