if $x+y+z=0$ then $(\zeta(x) +\zeta(y) + \zeta(z))^2= \wp(x) + \wp(y) +\wp(z)$?

Question

I found that if $x+y+z=0$, one can prove that $(\zeta(x) +\zeta(y) + \zeta(z))^2= \wp(x) + \wp(y) +\wp(z)$ using Liouville's theorem (here $\wp$ is the Weierstrass elliptic function)

How one can do this?

Answer 1

$L=w_1\Bbb{Z}+w_2\Bbb{Z}$.

$\wp_L$ is the unique $L$-periodic meromorphic function whose poles are double at $L$ and $\wp_L(z)=z^{-2}+O(z)$.

$\zeta_L'=-\wp_L$ and $\zeta_L(z)=z^{-1}+O(z)$. It is the unique meromorphic function whose poles are simple at $L$, $\zeta_L(z)=z^{-1}+O(z)$ and $\zeta_L(z+w_j)=\zeta_L(z)+c_j$.

$$f_{L,y}(z)=(\zeta_L(-z-y) +\zeta_L(y) + \zeta_L(z))^2$$ is meromorphic doubly periodic with double poles at $L$ and $L-y$ where $f_{L,y}(z)=z^{-2}+O(1),f_{L,y}(z-y)=z^{-2}+O(1)$, thus $$g_{L,y}(z)=f_{L,y}(z)-\wp_L(z)-\wp_L(y-z)$$ is entire and $L$-periodic, it is constant.

Since $\zeta_L'=-\wp_L'$ and $\wp_L$ is even then $\zeta_L(z)=z^{-1}+O(z^2)$ and $$g_{L,y}(z)= (\wp_L(y) z+z^{-1}+O(z^2))^2-z^{-2}-\wp_L(-y)+O(z)=\wp_L(y)+O(z)$$ and hence $$g_{L,y}(z)=\wp_L(y)$$