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 $f(x)=\pm x\;\;\;(\forall x\in\mathbb{R}).$
It can be shown that instead of monotonicity, we can assume other properties for $f$ and get the same result. For this, we may first find some properties of $f$ without assuming monotonicity and then apply the extra assumption.
Suppose that for some positive integer $ n $ we have: $$f\big(x+f(y)\big)=f(x)+y^n\tag0\label0$$ Letting $x=y=0$ in \eqref{0} we get: $$f\big(f(0)\big)=f(0)+0^n=f(0)$$ $$\therefore f\Big(f\big(f(0)\big)\Big)=f\big(f(0)\big)=f(0)$$ Again, letting $x=0$ and $y=f(0)$ we get: $$f\Big(f\big(f(0)\big)\Big)=f(0)+f(0)^n$$ $$\therefore f(0)=f(0)+f(0)^n$$ $$\therefore f(0)=0$$ Now, letting $x=0$ in \eqref{0} we get: $$f\big(f(y)\big)=y^n\tag1\label1$$ So by applying $f$ on the both sides of \eqref{1} we have: $$f\Big(f\big(f(y)\big)\Big)=f\big(y^n\big)$$ On the other hand, substituting $f(y)$ for $y$ in \eqref{1} we get: $$f\Big(f\big(f(y)\big)\Big)=f(y)^n$$ Combining the last two results, we find that: $$f\big(y^n\big)=f(y)^n\tag2\label2$$ Now, by \eqref{0}, \eqref{1} and \eqref{2} we have: $$f\Big(f\big(x+f(y)\big)\Big)=f\big(y^n+f(x)\big)$$ $$\therefore\big(x+f(y)\big)^n=f\big(y^n\big)+x^n=f(y)^n+x^n$$ $$\therefore\Big(x+f\big(f(y)\big)\Big)^n=f\big(f(y)\big)^n+x^n$$ $$\therefore\big(x+y^n\big)^n=x^n+y^{n^2}$$ Letting $x=y=1$ in the last equation we get $2^n=2$ and so $n=1$. Thus, substituting $f(y)$ for $y$ in \eqref{0}, by \eqref{1} we have: $$f(x+y)=f(x)+f(y)\tag3\label3$$ $$f\big(f(x)\big)=x\tag4\label4$$ So, $f$ satisfies \eqref{0} iff $n=1$ and $f$ satisfies \eqref{3} and \eqref{4}.
It's well-known that given a Hamel basis, one can construct wild functions satisfying \eqref{3} and \eqref{4}. On the other hand, it's also well-known that if $f$ satisfies \eqref{3} and it's also continuous, monotone, Lebesgue measurable or bounded when restricted to a bounded interval, then there is a constant $c$ such that for every real number $x$ we have $f(x)=cx$. For example, see Overview of basic facts about Cauchy functional equation. By \eqref{4} you can see that $c=\pm1$.
