I have the functional equation $$F(x)+F(y)=F(\sqrt {x^2+y^2})$$
I know that one solution to this equation is $F(x)=a\cdot x^2$. However, I am told that this is the unique solution, but I don't know how to show this?
How do we know that this is the unique solution?
Replacing $x$ by $-x$ we see that $F(x)=F(-x)$: $F$ is an even function. Define $G(u)=F(\sqrt{u})$ for $u\ge0$. Then $$G(u)+G(v)=F(\sqrt u)+F(\sqrt v)=F(\sqrt{u+v})=G(u+v).$$ This functional equation has the only continuous solutions $G(u)=au$. But it also has non-continuous solutions. As $\Bbb R$ is an infinite-dimensional $\newcommand{\Q}{\Bbb Q}\Q$-vector space, there is a nonzero $\Q$-linear map $\phi:\Bbb R\to\Q$. Define $F(x)=\phi(x^2)$. Then $F$ is non-continuous and satisfies the functional equation.
The trouble is, that functions like $\phi$ can only be "constructed" by some form of the Axiom of Choice.