I would like to know the solutions of the functional equation:
$$f(x+f(y))+f(y+f(x))=2f(f(x))f(f(y)), \forall x,y\in\mathbb{R}$$
where $f:\mathbb{R}\rightarrow\mathbb{R}$. I have already determined that $f\equiv 0$ is a solution, and I would not be surprised if it was the only solution, but I have been unsuccessful in both proving this and finding a counterexample. Letting $a=f(0)$, I know that any other solution must satisfy
- $f(a)=1$
- $f(f(x))=f(x+a)$ (and more generally $f^{n+1}(x)=f(x+na)$)
- $f(x+f(x))=f(f(x))^2=f(x+a)^2$ (and hence $f(x+f(x))>0$)
- $f(2a)=f(1)$ (and more generally $f((n+2)a)=f(na+1)=f^{n+1}(1)$)
- $f(3a)=f(a+1)=f(1)^2$
- $f(2f(x))=f(f(x+a)^2)$
However, I cannot see how to derive a reasonable description of another solution/a proof that such a solution is impossible from this information. I'd appreciate some help with this. Thanks!