I got some problems when solving the functional equation $f(x+y^n)=f(x)+[f(y)]^n, (x,y\in\mathbf{R}) $ for all positive integar $n$. I tried to solve it as following:
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Let $y=0$, then $f(x)=f(x)+[f(0)]^n$ implies $f(0)=0$;
Let $x=0, y=1$, then $f(1)=[f(1)]^n$ implies $f(1)=0$ or $1$.
Let $x=0$, then $f(y^n)=[f(y)]^n$.
Then the equation can be re-written as
$f(x^n+y^n)=f(x^n)+f(y^n), x, y\in\mathbf{R}$. ------- [1]
According to the continuous solution of Cauchy equation, the solution of the funciton (1) is $f(x)=f(1)x=0$ or $x$.
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However, I want to know about:
(1) Whether my solution above is correct?
(2) Maybe I shall discuss the parity of $n$, or not. And how to solve the equation $f(x^n+y^n)=f(x^n)+f(y^n)$ for a general positive integar $n$.
(3) If all solutions of equation [1] have been found, then does the equation $f(x+y^n)=f(x)+[f(y)]^n$ have other solution which is not the solution of [1]?
As Yves Daoust mentions, any such $f$ will be additive. Over $\mathbb{R}$, without some condition $f(x)=cx$ is one solution but not the only one - there are infinitely many pathological solutions. For more information, see this Wikipedia page which gives some insight as to why nonlinear additive functions exist.