Suppose I have a group homomorphism $f: (\mathbb{R}^{2}, +) \rightarrow (\mathbb{R}, +)$ defined by $$f(x,y)= f(\frac{x+y}{2}, \frac{x+y}{2})+f(\frac{x-y}{2}, \frac{y-x}{2}) \text{,}$$ where $x, y \in\mathbb{R}$. Is it true then that $f(x, y)= (x+y)^n, n \text{ odd}$ satisfies the above equation? I get the feeling that this should work $\forall{n} \in \mathbb{N}$.
Here's what I tried:
If $f(x,y)=(x+y)^n$ on the LHS of the above equation, then on the RHS we have that $(\frac{x+y}{2}+ \frac{x+y}{2})^{n} + (\frac{x-y}{2}+ \frac{y-x}{2})^{n}= (x+y)^n +0= (x+y)^n$. Am I correct?
Your solution seems to be pretty simple and correct, I agree with it as well as with the fact that the equality is true $\forall n \in \mathbb N$