Here is the problem:
Fix $n\in\mathbb{N}$. Find all monotonic solutions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(x+f(y))=f(x)+y^n$.
I've tried to show that $f(0)=0$ and derive some properties from that but have been unable to do so.
A solution would be appreciated.
If $|f(0)|>0$, $f$ is a monotonic periodic function and thus is constant; however, the functional equation admits no constant solutions so we must have
$f(0)=0\;\;\;(1).$
Now, substituting $x=0$ into the functional equation gives us
$f(f(y))=y^n\;\;\;(2).$
Note that $(2)$ allows to write the functional equation as
$f(x+f(y))=f(x)+y^n=f(x)+f(f(y))\;\;\;(3).$
Also note that by the functional equation and $(1),$ the range of $f$ contains all non-negative numbers (fix $x=0$ and vary $y$). Hence any non-negative number may be substituted for $f(y).$
Thus the functional equation can be written as
$f(x+y)=f(x)+f(y),x\in\mathbb{R}, y\geq0\;\;\;(4).$
Since $f$ is monotonic, we must have that for $x\geq0, f(x)=Cx$ for some constant $C$.
Substituting this partial solution into the original functional equation, we see that this is only possible if $n=1;$ that is, if $n\neq1$ the functional equation does not admit solutions.
If $n=1$, $f$ is clearly surjective by the original functional equation and thus $(4)$ holds $\forall x,y\in\mathbb{R}.$
It follows that the only potential solutions are of the form $f(x)=Cx$ for some constant $C$, since monotonic additive functions are linear.
Substituting into the functional equation, we find that the only solutions are $f(x)=\pm x\;\;\;(\forall x\in\mathbb{R}).$