Functional equation $f(x + y) = f(x)^m + f(y)^{m + 1}$

Question

Let $m \in \mathbb{N}$.

How i can find all functions $f$ such that $f(x + y) = f(x)^m + f(y)^{m + 1} \forall x,y \in R$?

Thank you in advance.

Answer 1

$$f^{m+1}(x)+f^m(y)=f(x+y)=f^{m}(x)+f^{m+1}(y).$$

So you'd need $f^{m+1}(x)-f^{m}(x)=f^{m+1}(y)-f^{m}(y)$, for all $x,y$, which means that you need $f^{m+1}(x)-f^{m}(x)=c$ for some constant $c$. This means that $f(x)$ can only take one of at most $m+1$ values.

If it has to be continuous, then it has to be constant. If constant, it must be a solution of $c=c^{m+1}+c^{m}$, so $c=0$ or a root of $c^{m}+c^{m-1}-1$.

It seems likely we can prove it is constant without continuity as a condition. Need a little more thinking to figure that out.

To not be constant, you'd need a $c$ so that there is subset $S$ of the roots of $z^{m+1}+z^{m}-c$ with more than one element such that if $u,v\in S$, then $u^m+v^{m+1}=u^m-v^{m}+c\in S$. But also, $v^{m}-u^{m}+c\in S$. So there is some weird symmetry around $c$ going on with some of these roots.