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.

(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.
