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For the period lattice $\Lambda =\mathbb{Z}{\omega }_1+\mathbb{Z}{\omega }_2$, define the Weierstrass elliptic function byOn a compact set away from $\Lambda$, the summand is $O(|\omega {|}^{-3})$ uniformly for sufficiently large $|\omega |$. The permitted lattice convergence estimate therefore gives Normal convergence of the Weierstrass elliptic-function series. Consequently $\wp$ is a meromorphic function, with a double pole of principal part $(z-\omega {)}^{-2}$ at each lattice point and no other poles. Termwise differentiation gives the normally and absolutely convergent seriesPairing $\omega$ with $-\omega$ shows that $\wp$ is even and ${\wp }^{\prime }$ is odd. Reindexing the absolutely convergent derivative series gives ${\wp }^{\prime }(z+\lambda )={\wp }^{\prime }(z)$ for every $\lambda \in \Lambda$. Thus $\wp (z+\lambda )-\wp (z)$ is constant, with removable singularities at the lattice points. For either basis period $\lambda ={\omega }_1,{\omega }_2$, evaluate at $z=-\lambda /2$: evenness makes this constant zero. These two periods imply all periods in $\Lambda$. Hence $\wp$ is a nonconstant elliptic function with period lattice $\Lambda$.

To obtain the Laurent coefficients of the Weierstrass elliptic function, take $0<|z|<{\min }_{\omega \ne 0}|\omega |$ and expandThe expansion and lattice summation can be interchanged uniformly on smaller disks, using the same convergent majorant. The odd powers cancel in pairs. In terms of the lattice Eisenstein sums, the answer isOnly the absolutely convergent lattice Eisenstein sums of weights at least four occur here; no conditionally defined $G_2$ is needed.

Put $g_2=60G_4$ and $g_3=140G_6$. The Laurent series just found givesTherefore $F={\wp }^{\prime 2}-4{\wp }^3+g_2\wp +g_3$ has no pole at zero, and its value there is zero. By periodicity it has no poles anywhere. An entire elliptic function is bounded on a closed fundamental parallelogram, hence on the plane, and is constant by Liouville theorem. Since $F(0)=0$, this proves the Weierstrass elliptic differential equationWe still need to identify the three roots without assuming their distinctness. For a nonconstant elliptic function $u$, the argument principle applied to $u^{\prime }/u$ around a fundamental parallelogram makes the opposite-edge integrals cancel. Thus its zeros and poles, counted with multiplicity, have equal total number. In particular $\wp -a$ has exactly two zeros in a period cell for every finite $a$.

Let $h_i={\omega }_i/2$, with ${\omega }_3=-{\omega }_1-{\omega }_2$. Their classes are the three nonzero two-torsion points of a complex torus. Oddness and periodicity give ${\wp }^{\prime }(h_i)=-{\wp }^{\prime }(h_i)$, so ${\wp }^{\prime }(h_i)=0$. Thus $\wp -e_i$, where $e_i=\wp (h_i)$, has a zero of multiplicity at least two at $h_i$, and the preceding count says its multiplicity is exactly two. If two of the half-period values of the Weierstrass elliptic function were equal, the same function would have at least four zeros, a contradiction. Evaluating the Weierstrass elliptic differential equation at the three half-periods now gives three distinct roots of its cubic. Consequently

For the remaining identity, fix $i$ and put $A_i=(e_i-e_j)(e_i-e_l)$, where $\{i,j,l\}=\{1,2,3\}$. The elliptic functionhas possible poles only at $z=0,-h_i$ modulo $\Lambda$. At each, the other factor has an exact double zero, so these poles are removable. Thus $H_i$ is constant, and its limit at zero is ${\wp }^{\prime \prime }(h_i)/2$. Differentiating the cubic differential equation first where ${\wp }^{\prime }\ne 0$, then extending the resulting identity holomorphically, gives ${\wp }^{\prime \prime }=6{\wp }^2-g_2/2$. Since $e_i$ is a root of $P(X)=X^3-g_{2}X/4-g_3/4$, we obtainThis proves the Weierstrass half-period translation formula by pole cancellation. Set $u_i=h_i/2={\omega }_i/4$. Since $u_i+h_i\equiv -u_i(\mathrm{mod}\Lambda )$, evenness gives $\wp (u_i+h_i)=\wp (u_i)$, and thereforeThe denominator here is nonzero: the only zero of $\wp -e_i$ modulo $\Lambda$ is $h_i$, whereas $u_i$ is not congruent to $h_i$. Finally differentiate the Weierstrass half-period translation formula and use $z-h_i\equiv z+h_i(\mathrm{mod}\Lambda )$ to obtain the Weierstrass quarter-period derivative identityThe equality holds wherever these quotients are initially defined, and hence as an identity of meromorphic functions at all remaining points.

At $z=h_i/2$, evenness and periodicity give $\wp (z+h_i)=\wp (z)$. The Weierstrass half-period translation formula therefore identifies $(\wp (h_i/2)-e_i{)}^2$ with $(e_i-e_j)(e_i-e_l)$. Differentiating the Weierstrass half-period translation formula proves the identity, interpreted as equality of meromorphic functions, including the apparent singularities of its quotients.