What $z\in \mathbb{C}$ satisfy \begin{eqnarray*} \wp(z) = \wp(nz) \end{eqnarray*} if $n\in \mathbb{N}$?
The Weierstrass p-function is defined as \begin{eqnarray*} \wp(z)=\frac{1}{z^2}+\sum_{\omega\in \Lambda^*}\left( \frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\right) \end{eqnarray*} where $\Lambda^*$ is the lattice for complex numbers $\omega_1$ and $\omega_2$ excluding zero.
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$\begingroup$It is "well-known" that every complex value of the Weierstrass function appears exactly two times (including multiplicities) inside the fundamental parallelogram. Further, this function is even. Therefore the set of solutions consists of two parts: \begin{align} nz=z\;\mathrm{mod}\;\Lambda,\qquad nz=-z\;\mathrm{mod}\;\Lambda. \end{align} Or, in a more explicit form, $$ z=\frac{j\omega_1+k\omega_2}{n\pm1},\qquad j,k\in\mathbb{Z}.$$
$\endgroup$ 0 $\begingroup$Does this work?
Write \begin{eqnarray*} \wp (z) &=& \dfrac{1}{z^2} + \sum_{k=1}^m(2k+1)S_{2k+2}z^{2k}\\ \wp(nz) &=& \dfrac{1}{(nz)^2} + \sum_{k=1}^m(2k+1)S_{2k+2}(nz)^{2k} \end{eqnarray*}
Then look at it term-by-term, \begin{eqnarray*} \dfrac{1}{z^2}=\dfrac{1}{(nz)^2} \Rightarrow (nz)^2-z^2=0\Rightarrow (n+1)z(n-1)z=0 . \end{eqnarray*}
Similarly, \begin{eqnarray*} z^{2k}=(nz)^{2k} \Rightarrow (n^{2k}-1)z^2 = 0 \Rightarrow (n^k+1)z(n^k-1)z=0 \end{eqnarray*} Then write \begin{eqnarray*} \left( \sum_{i=0}^{k-1} n^{i} \right)\left( \sum_{i=0}^{k-1} (-n)^{i} \right) (n+1)z(n-1)z=0. \end{eqnarray*}
Therefore, $z\in \mathbb{C}$ that satisfies $(n\pm 1)z=0$ would satisfy the original equation.
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