f(f(x))=x+a^k x^k +y + 2a^k=f(f(x^k+y+a^k)=f(f(x^k+f(f(y)))=f(f(y)+f(x)^k)=f(f(x))^k + y = [x+a^k]^k + y so we get that a=0. hence f(f(x))=x. (*) also f(x^k)=f(x)^k (**). f(x) is one to one and onto. one can easily see that f(x+y)=f(x)+f(y) (***) so f(x)=x for all x in Q.

The additivity, along with $f(x^k)=f^k(x)$ for some $k > 1$, implies at once that $f(x)=x$ for all $x \in \mathbb{R}$. Indeed, for $n \in \mathbb{Z}$, we have $f(x+n)=f(x)+n$, so \[ \sum_j \binom{k}{j}f^j(x)n^{k-j} = (f(x)+n)^k = f^k(x+n) = f\big( (x+n)^k \big) = \sum_j \binom{k}{j}f(x^j)n^{k-j}. \] Fixing $x$, we may view this as a polynomial identity with respect to $n$. Comparing the coefficients of $n^{k-1}$, we see that $f(x^2) = f(x)^2 \geq 0$, so $f(x) \geq 0$ for $x \geq 0$ which, along with the additivity, implies that $f(x)$ increases monotonously. Hence, $f(x) \equiv x$.

A slightly more difficult problem is this: find all functions $f: \mathbb{R}^+ \to \mathbb{R}$ satisfying this condition.

Satisfying which condition? (Edit: never mind. The functional equation is the same, but for functions of positive reals rather than all real numbers.)

I think he means the same condition above

Yes, the problem I suggested is this: for a fixed integer $k>1$, find all functions $f: \mathbb{R}^+ \to \mathbb{R}$ satisfying $f(x^k+f(y)) = y + f(x)^k$ for all $x,y \in \mathbb{R}^+$ for which $x^k + f(y) > 0$. (So that we may not plug in $y=0$ or $x=0$ to make the problem trivial ).

Is this generalisable to say, n variables.

.For k odd you miss f(x)=-x

vess wrote: The additivity, along with $f(x^k)=f^k(x)$ for some $k > 1$, implies at once that $f(x)=x$ for all $x \in \mathbb{R}$. Could you prove that?