All solutions to functional equation $f(x+1)-f(x)=1$

Question

I was thinking of the possibility of finding all solutions other than $f(x)=x$ for the functional equation:

$f(x+1)-f(x)=1$

If there are other solutions, what will be some restrictions for the equation to have just one solution?

Answer 1

First, think about all functions $g$ of period 1, we have f$(x+1)$=f$(x)$, we can make one period constant, then let value of function in last period substract 1, and let value of function in following period add 1. Repeat it, and you can get a new function.

Answer 2

For all function g of period 1 the function f(x) = [x] + g(x) works; for example f(x) = [x] + sin 2$\pi$x