Prove that if $f$ is periodic, continuous, there are no $r\ne 0$ and $c\ne 0$ constants satisfying $f(x+r)=f(x)+c$ (for all $x$)
Let's assume that exist $r$ and $c$ satisfying $f(x+r)=f(x)+c$. Thus, by induction, $f(x+ nr)=f(x)+nc$ , where $n \in Z$.
If $r$ is a period of $f$, then $c=0$, it's okay.
If exist $n,m \in Z$: $np=mr$, where p is a period of $f$, then it is done, because $f(x+mr)=f(x)+mc$ and $f(x+m r)=f(x+np)=f(x)$, thus $c=0$.
I tried in this way, but I have stuck: let's assume that exist $r$ and $c$ satisfying $f(x+r)=f(x)+c$. Thus, by induction, $f(x+ nr)=f(x)+nc$ , where $n \in Z$. So, if $f$ would be bounded, I would be done, because $f(x)+nc$ is not bounded. But $\tan(x)$ is periodic, continuous and not bounded, it's a counter-example.
I need some hint, thanks for advance.
Since $f$ is continuous it is bounded.
We have $f(x+nr) = f(x) + nc$, hence letting $n \to \infty$, we must have $c=0$.
Since $f$ is periodic, we can choose $r$ to be the period which gives $f(x+nr) = f(x)$.
So the most we can conclude is that $c = 0$.