$f(x)$ is linear if and only if $f(x) + f(x - a - b) = f(x - a) + f(x - b)$ I'm trying to show that a function $f:\mathbb{R} \rightarrow \mathbb{R}$ is linear if and only if $$f(x) + f(x - a - b) = f(x - a) + f(x - b)$$ for all $a, b, x\in\mathbb{R}$. 
The forward direction is trivial and I'm having some problems with the backwards direction. I've thought of using the continuous Cauchy functional equation $$f(x + y) = f(x) + f(y) \Longrightarrow f(x) = cx$$
but I'm having trouble with getting the equation into a usable form and also I'm not sure how to show that the function is continuous.
Any help would be appreciated.
Edit Linear in the sense of $ax + b$ for some constants $a, b$.
 A: I assume by linear you mean "of the form ax+b", as otherwise the statement is false (see Brett's answer). Suppose that $f$ satisfies $$f(x) + f(x - a - b) = f(x - a) + f(x - b)$$ but is not linear, so we have some $x,y$ such that $f(x)+f(y)\neq f(x+y)+f(0)$. Then $$f(x+y)+f(0)\neq f(x)+f(x-x-(-y))= f(x-x)+f(x-(-y))=f(0)+f(x+y)$$
a contradiction, so $f$ must be linear in this sense.
Edit: Because of your mention of the continuous Cauchy functional equation, and because the statement you are trying to prove is false without the assumption of continuity, I assume that $f$ is continuous. To see why any nonlinear $f$ must have some $x,y$ such that $f(x)+f(y)\neq f(x+y)+f(0)$, suppose that $f$ is such that $f(x)+f(y)= f(x+y)+f(0)$ for all $x,y$. Let $a=f(1)-f(0)$ and $b=f(0)$. Then $f(1)=f(\frac{m}{m})=mf(\frac{1}{m})-(m-1)f(0)$ so $f(\frac{1}{m})=\frac{1}{m}(f(1)-f(0))+f(0)=a\frac{1}{m}+b$, which means $$f(\frac{n}{m})=nf(\frac{1}{m})-(n-1)f(0)=a\frac{n}{m}+nb-(n-1)b=a\frac{n}{m}+b$$ and so by continuity and the fact that the rationals are dense in the reals we see that $f(x)=ax+b$ for all real numbers $x$.
A: Proof of cauchy continuity condition:
We have $f(0+0)= f(0)+f(0)$ this gives $f(0)= 0$.
And we can show that $f(-x)=-f(x)$.
For integer value $n$, we can show that $f(nx)= nf(x)$  Also for $n\neq 0$ we can show that $f(\frac{x}{n})= \frac{1}{n}f(x)$
This gives for $m\neq 0$ integer, we have:
$$f(\frac{n}{m}x)=\frac{n}{m}f(x)$$
Now if $x\in \mathbb R$, then there is sequence of continuous function of rational $q_n$ such that $\lim q_n\to x$.
Now $f(x)=\lim f(q_n)$ as $f$ is continuous. Hence
$$f(x)= \lim  q_n f(1)= x.f(1)= cx$$
Where $f(1)=c$
