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.

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 $a$. 
