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Here is a very basic comment no one has made yet: If $f(u)$ is a rational function of $u$, then there will be some nonzero polynomial $G$ such that $G(f(u), f(v), f(u+v))=0$. That's because $\mathbb{C}(u, v, u+v)$ has transcendence degree $2$ over $\mathbb{C}$.
The same argument applies if $f$ is a rational function of $e^u$, or if $f$ is a rational function of $\wp(u)$ and $\wp'(u)$, where $\mathcal{P}$ \wp$is the Weierstrauss$\wp$-function. Can we show that every example is of one of these forms? 2 \wp-ify Here is a very basic comment no one has made yet: If$f(u)$is a rational function of$u$, then there will be some nonzero polynomial$G$such that$G(f(u), f(v), f(u+v))=0$. That's because$\mathbb{C}(u, v, u+v)$has transcendence degree$2$over$\mathbb{C}$. The same argument applies if$f$is a rational function of$e^u$, or if$f$is a rational function of$\mathcal{P}(u)$\wp(u)$ and $\mathcal{P}'(u)$, \wp'(u)$, where$\mathcal{P}$is the Weierstrauss P-function.$\wp$-function. Can we show that every example is of one of these forms? 1 Here is a very basic comment no one has made yet: If$f(u)$is a rational function of$u$, then there will be some nonzero polynomial$G$such that$G(f(u), f(v), f(u+v))=0$. That's because$\mathbb{C}(u, v, u+v)$has transcendence degree$2$over$\mathbb{C}$. The same argument applies if$f$is a rational function of$e^u$, or if$f$is a rational function of$\mathcal{P}(u)$and$\mathcal{P}'(u)$, where$\mathcal{P}\$ is the Weierstrauss P-function.