How to solve the functional equation $f(x+a)=f(x)+a$ Are there any other solutions of the functional equation $f(x+a)=f(x)+a$ ($a=\mathrm{const}$, $a\in\mathbb{R}\setminus\left\{0\right\}$) apart from $f(x)=x+C$ ($C=\mathrm{const}$)?
Edit: $a$ is a fixed number here.
 A: There are continuous functions apart from $x+C$. 
Given $a\not=0$ let $f(x)=x+\sin(\dfrac{2\pi x}{a})$.  
Then $f(x+a)=x+a+\sin(\dfrac{2\pi (x+a)}{a})=x+a+\sin(\dfrac{2\pi x}{a}+2\pi)= x+\sin(\dfrac{2\pi x}{a}) +a=f(x)+a$. 
As indicated in another answer you may get many other (discontinuous) functions by partitioning the reals. But you may also get many continuous functions, following the above model using any continuous periodic function with period $a$. 
A: No if the equation holds for all $a \in \mathbb{R} \setminus \{0\}$. The functional equation $f(x + a) = f(x) + a$ implies that
$$\frac{f(x + a) - f(x)}{a} = 1.$$
Taking the limit of both sides as $a \to 0$ shows that $f'$ exists and $f'(x) = 1$, so the only solutions are $f(x) = x + C$. as you stated.
A: (This is with the understanding that $f(x + a) = f(x) + a$ is true for some fixed $a$, but all $x$.)
There are many solutions, with very poor behavior.
If you call $x$ and $y$ equivalent if $x = na + y$ for some integer $n$, then you can partition $\mathbb{R}$ into the equivalence classes. Pick a representative $x$ from each equivalence class $[x]$ and define $f(x)$ to be whatever you like, then extend to the rest of the equivalence classes using the defining equation $f(y) = f(na + x) = f(x) + na$.
Riffing off Mirko slightly, you can think about it this way. You can define $f$ however you like on the interval, say, $[0,a)$ (assuming $a>0$). Then define $f$ on $[a,2a)$ by $f(y) = f(x+ a) = f(x) + a$ for the unique $x\in[0,1)$ with $x+a = y$, and continue inductively.
A: The given equation is equivalent to $g(x+a)=g(x)$ where $g(x)=f(x)-x$. Take any periodic function g with period a and take $f(x)$ to be $g(x)+x$. This gives a complete solution to your question.
