I am trying problems from Tom M Apostol Modular Functions and Dirichlet Series in number theory and I am struck on this problem which is from Chapter - Elliptic Functions.
Problem is - Prove that $\wp(u) = \wp (v) $ iff $u-v$ or $u+v$ is a period of $\wp$ .
My attempt - assuming $u-v, u+v$ to be period of $\wp(z) $ , it is very easy to see $\wp (u)=\wp(v) $ .
But if I assume $\wp (u) =\wp (v) $ , then I get $\frac {u+v} {u^2 v^2}= \sum_\omega\neq 0 \frac { u+v - 2 \omega} { ( u- \omega) ^2 × ( v-\omega ) ^2 } $ .
I don't know how to proceed after this. Can somebody please give some hint.
The easy part was already known to the OP, but since I wrote it let me keep it here: if $u-v = \omega_i$ with $\omega_i$ one of the periods, then it follows immediately that $$ \wp(u) = \wp(\omega_i + v) = \wp(v), $$ using that $\wp$ is periodic with period $\omega_i$. If we start from $u+v = \omega_i$ we use the fact that $\wp(-x) = \wp(x)$ to get rid of the extra minus sign.
The converse is slightly trickier: suppose $\wp(u) = \wp(v)$. Note that $\wp$ has a second order pole only, which implies it attains every complex number twice on the fundamental parallelogram. The evenness of $\wp$, i.e. $\wp(u) = \wp(-u)$, now implies that if $u,v$ are both in the fundamental parallelogram $$ v = -u \text{ mod } \mathbb{L}, $$ which is equivalent to $$ v+u = 0 \text{ mod } \mathbb{L}. $$ The only other possibility occurs when $u$ and $v$ are not in the same quadrant. The preceding implies that in that case $v=\pm u \text{ mod } \mathbb{L}$, so we find $$ v\pm u = 0 \text{ mod } \mathbb{L}. $$ This proves the asked.