= Solution
An <entire function> that is doubly <periodic> is bounded: its absolute value has a finite maximum on the compact square $[0,1]+i[0,1]$, and periodic translations carry every point of $\mathbb C$ into that square. The <Liouville theorem> therefore proves \b[every such analytic function is constant].
For the zero-pole count, suppose the <meromorphic function> is not identically zero; otherwise isolated zero multiplicities are not defined. It has only finitely many <zeros> and <poles> modulo the lattice $\Lambda=\mathbb Z+i\mathbb Z$, because a compact square can be covered by finitely many local meromorphic neighbourhoods. Choose a translated unit square with no <zeros> or <poles> on its boundary. The <logarithmic derivative> $f'/f$ is doubly <periodic>, so integrals on opposite sides cancel, with their opposite orientations. The <argument principle> gives
$$
\boxed{N-P=\frac1{2\pi i}\oint\frac{f'(z)}{f(z)}\,dz=0.}
$$
Both counts include multiplicities. Each half-open <fundamental parallelogram> contains exactly one representative of each lattice orbit, and translation by a lattice vector preserves the multiplicity of a <zero> or <pole>. Consequently the counts agree in the prescribed half-open square too, even if that particular square has boundary <zeros> or <poles>. Nonzero constants have both counts zero.
For the series, let $\Lambda=\mathbb Z+i\mathbb Z$ and write it as the <Weierstrass elliptic function> $\wp(z)$. On a compact set avoiding $\Lambda$, choose $R$ with $|z|\leq R$. For $|w|>2R$,
$$
\frac1{(z-w)^2}-\frac1{w^2}=\frac{2wz-z^2}{w^2(z-w)^2},\qquad \left|\frac1{(z-w)^2}-\frac1{w^2}\right|\leq\frac{C_R}{|w|^3}.
$$
There are $O(j)$ lattice points with $j\leq|w|<j+1$, hence $\sum_{w\ne0}|w|^{-3}$ converges by comparison with $\sum_jj^{-2}$. The <Weierstrass M-test> proves absolute <locally uniform convergence> of the tail, and the finite remaining terms define <holomorphic functions> off their <poles>. It follows that $\wp$ is a <holomorphic function> off $\Lambda$. Near each $w_0\in\Lambda$, separate the term $(z-w_0)^{-2}$; the remaining terms converge locally uniformly to a <holomorphic function>. Thus $\wp$ is <meromorphic> on $\mathbb C$, with a double <pole> at each lattice point and no other <poles>.
It remains to prove the required <periods>, rather than merely assert them from the lattice. Termwise <differentiation>, justified by <locally uniform convergence> of these holomorphic series, gives
$$
\wp'(z)=-2\sum_{w\in\Lambda}\frac1{(z-w)^3}.
$$
This series is absolutely locally uniformly convergent off $\Lambda$. For either generator $\omega=1,i$, reindexing $w\mapsto w-\omega$ proves $\wp'(z+\omega)=\wp'(z)$. Therefore $\wp(z+\omega)-\wp(z)$ has zero <derivative> off $\Lambda$. Its principal parts at lattice points cancel, so it extends to an entire constant $c_\omega$.
The original convergent series is even: reindexing $w\mapsto-w$ gives $\wp(-z)=\wp(z)$. At $z=-\omega/2$, which is not a lattice point for either generator, evenness gives
$$
c_\omega=\wp(\omega/2)-\wp(-\omega/2)=0.
$$
Hence \b[the series defines a meromorphic function with periods $1$ and $i$]. This <termwise derivative proof of Weierstrass periodicity> uses an absolutely convergent derivative sum; splitting the original series into two separate inverse-square sums would not be justified.
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