Solutions of $f(x+y^{n})=f(x)+[f(y)]^{n}$. Consider the functional equation $f(x+y^{n})=f(x)+[f(y)]^{n}$ where $f:\mathbb R \to \mathbb R$ and $n$ is given integer $>1$. This equation was discussed yesterday and it was shown that $f$ is necessarily additive. Assuming continuity it was concluded that $f(x)\equiv cx$ for some $c$. [ Necessarily $c$ is an n-th root of unity]. If $n$ is even then  the given functional equation gives $f(x+y^{n}) \geq f(x)$ which easily leads to the conclusion that $f$ is an increasing function. It follows that $f$ is Borel measurable; since any Borel measurable additive function if of the type $f(x)\equiv cx$ the assumption that $f$ is continuous is not necessary. My question is what can be said for $n$ odd? Can one use some trick to prove that $f$ is necessarily Borel measurable? Or is there a counter-example? Discontinuous additive functions are constructed using Hamel basis but I am unable to use this method to construct a counter-example. I would appreciate receiving any ideas about this question.
 A: This is to show that if $n=5$ (so that $f(x+y^5)=f(x)+f(y)^5$ for all $x,y\in\mathbb{R}$), then $f(x)=0$, $f(x)=x$ or $f(x)=-x$. I think you can generalize the argument, so I leave the general cases to you. 
As you mentioned, $f(x+y)=f(x)+f(y)$ for all $x, y\in \mathbb{R}$. Additionally, we have $f(x^5)=f(x)^5$. Since $f$ is additive, $f(qx)=qf(x)$ for all $q\in\mathbb{Q}$, $x\in\mathbb{R}$. 
Since $f( (x+y)^5)=  f(x+y)^5 = ( f(x)+f(y))^5$, we have 
$$
\begin{align}
f & (x^5+5x^4 y + 10x^3y^2 + 10x^2y^3+5xy^4+y^5)\\
&=f(x)^5+5f(x)^4 f(y) + 10 f(x)^3f(y)^2+10f(x)^2f(y)^3+5f(x)f(y)^4+f(y)^5.
\end{align}
$$
Then, the first and the last term cancel due to $f(x^5)=f(x)^5$, $f(y^5)=f(y)^5$. 
Since $5x^4 y + 10x^3y^2 + 10x^2y^3+5xy^4=(x+y)(5x^3y+5x^2y^2+5xy^3)$, we have
$$
\begin{align}
f((x+y)(5x^3y+5x^2y^2+5xy^3))=(f(x)+f(y))(5f(x)^3f(y)+5f(x)^2f(y)^2+5f(x)f(y)^3)
\end{align}
$$
Now, impose $x+y=q\in\mathbb{Q}\backslash\{0\}$ and substitute $y=q-x$, then 
$$
f(5x^3y+5x^2y^2+5xy^3)=f(1)\left(5f(x)^3f(y)+5f(x)^2f(y)^2+5f(x)f(y)^3\right),
$$ 
hence
$$
\begin{align}
f & (x^3(q-x)+x^2(q-x)^2+x(q-x)^3)\\
&=f(1)\left(f(x)^3(qf(1)-f(x))+f(x)^2 (qf(1)-f(x))^2+f(x)(qf(1)-f(x)^3)\right)
\end{align}
$$
This is a polynomial identity in $q$, and holds for infinitely many values of $q$. Thus, it should also hold for $q=0$. Thus, we obtain
$$
f(-x^4)=-f(1)f(x)^4
$$
So, $f(x^4)=f(1)f(x)^4$. 
Since $f(1)=f(1)^5$, $f(1)=0, 1, \mathrm{or} \ -1$. If $f(1)=0$, then $f(x)=0$. If $f(1)=1$, then $f$ is increasing, so $f(x)=x$. If $f(1)=-1$, then $f$ is decreasing, so $f(x)=-x$. 
A: Here's a generalization of i707107's argument that is actually a bit simpler, as long as I didn't make any mistakes:
You have
$$f(x+y)=f(x)+f(y)$$
and
\begin{align}
\sum_{i=0}^n \binom{n}{i}f(x^iy^{n-i})
&=f((x+y)^n)\\
&=f(x+y)^n\\
&=(f(x)+f(y))^n\\
&=\sum_{i=0}^n \binom{n}{i}f(x)^if(y)^{n-i}.
\end{align}
Taking $y$ rational, we have $f(x^iy^{n-i})=y^{n-i}f(x^i)$ and $f(y)=yf(1)$, so
$$\sum_{i=0}^n\binom{n}{i}y^{n-i}\left[f(x^i)-f(1)^{n-i}f(x)^i\right]=0$$
As this is a polynomial of degree $n$ that is $0$ for all rationals, it is identically $0$, so
$$f(x^i)=f(1)^{n-i}f(x)^i$$
for all $0\leq i\leq n$. Originally, we had $f(1)=f(1)^n$, so $f(1)\in\{-1,0,1\}$. If $f(1)=0$, we have $f(x^i)=0$, so $\boxed{f(x)\equiv 0}$. Otherwise, we have
$$f(x^2)=f(1)^{n-2}f(x)^2=f(1)f(x)^2$$
$$f(x+y^2)=f(x)+f(y^2)=f(x)+f(1)f(y)^2.$$
If $f(1)=1$, this means $f$ is increasing, and if $f(1)=-1$ this means $f$ is decreasing. Either way, $f$ is not everywhere dense, so $f(x)=cx$ for some $c$ and all $x$. The observation that $f(1)=\pm 1$ means $\boxed{f(x)=x}$ and $\boxed{f(x)=-x}$ are our only other solutions. 
