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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?
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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 $\mathcal{P}(u)$ \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?
show/hide this revision's text 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.

Can we show that every example is of one of these forms?