 Solution

The local estimate gives $\wp (z)=z^{-2}+O(z)$ at zero and, by periodicity, at every lattice point. If two candidates existed, their difference would extend to an entire doubly periodic function: the principal parts cancel at every possible pole. It is bounded on a fundamental parallelogram, hence on the plane, so Liouville theorem makes it constant. The local $O(z)$ difference tends to zero, making the constant zero. Thus the function is unique.

The function $z\mapsto \wp (-z)$ has the same periodicity, poles and local estimate. Uniqueness therefore proves $\wp (-z)=\wp (z)$. Its Laurent expansion is consequentlywith no constant term by the given estimate. Periodicity and evenness imply $\wp (1/2+t)=\wp (1/2-t)$, so ${\wp }^{\prime }(1/2)=0$. The elliptic zero-pole counting theorem says a nonconstant elliptic function has, in one period cell, equally many zeros and poles counted with multiplicity. Applied to $\wp (z)-\wp (1/2)$, whose total pole order is two, it gives precisely two zeros counted with multiplicity. The zero at $1/2$ already has multiplicity at least two. It therefore has exactly two, provingFinally ${\wp }^{\prime \prime }-6{\wp }^2$ is periodic, and the Laurent expansion shows cancellation of all possible pole terms: ${\wp }^{\prime \prime }=6z^{-4}+2c_2+O(z^2)$ and $6{\wp }^2=6z^{-4}+12c_2+O(z^2)$. The difference extends to an entire bounded function and is constant. HenceEvery property of the Weierstrass elliptic function used here has been obtained from the stated local and periodic properties, together with the explicitly stated standard elliptic-function counting theorem.