= Solution
Work first where all three coordinates are finite and the joining line is not vertical. By the <chord-and-tangent group law>, the third intersection of $y=\lambda x+\mu$ with the <elliptic curve> is $-(P_1+P_2)$, whose first coordinate is also $x_3$. Substitution gives the monic <polynomial>
$$
x^3-\lambda^2x^2+(a-2\lambda\mu)x+(b-\mu^2).
$$
Its roots are $x_1,x_2,x_3$, counted with <intersection multiplicity>. Comparing its coefficients with the <elementary symmetric polynomials> proves
$$
s_1=\lambda^2,\qquad s_2=a-2\lambda\mu,\qquad s_3=\mu^2-b,
\qquad \boxed{(s_2-a)^2=4s_1(s_3+b).}
$$
This <symmetric-coordinate identity for elliptic-curve addition> is an identity of <rational functions>. It includes tangent cases by specialization. Literally assigning finite $s_i$ when a point is $O$ would be meaningless; such cases must be interpreted on the projective curve. A nonsingular equation in the printed short form has characteristic different from two.
Here is how the identity yields the <quadratic form> for the <degree of an isogeny>. With $u=x(P)$, $v=x(Q)$, write $s=u+v$ and $q=uv$. As an equation for $t=x(P+Q)$, it becomes
$$
(u-v)^2t^2-\{2s(q+a)+4b\}t+(q-a)^2-4bs=0.
$$
Both $x(P+Q)$ and $x(P-Q)$ satisfy it, because replacing $Q$ by $-Q$ does not change $v$. Thus its two branches are the sum and difference maps. Their common denominator is the square of $u-v$. The following <divisor on an algebraic curve> calculation keeps track of cancellations and exceptional points rigorously, instead of assuming degrees of displayed numerators always add.
On $E_2\times E_2$, equality of the two first coordinates means $Q=P$ or $Q=-P$. Each coordinate has a double pole at $O$. Consequently the <principal divisor> identity is
$$
\operatorname{div}(x(P)-x(Q))
=D_++D_- -2(\{O\}\times E_2)-2(E_2\times\{O\}),
$$
where $D_+$ is the diagonal and $D_-$ is the graph of negation. Equivalently, the two zero divisors are the inverse images of $O$ under difference and sum, exactly the two branches identified above. Their generic multiplicities are one; the divisor identity also accounts for their intersections at <2-torsion> points.
For nonzero <isogenies of elliptic curves> $\phi,\psi:E_1\to E_2$ with $\phi\ne\pm\psi$, pull this identity back by $(\phi,\psi)$. Taking degrees of the resulting <principal divisor> on $E_1$ gives the <divisor proof of the degree parallelogram law>:
$$
\boxed{\deg(\phi+\psi)+\deg(\phi-\psi)=2\deg\phi+2\deg\psi.}
$$
If one map is zero, this is immediate. If $\phi=\pm\psi$, it follows from $\deg([2]\circ\phi)=4\deg\phi$. This also proves the <parallelogram law> for a general <Weierstrass equation of an elliptic curve> in characteristic two: the same divisor identity uses the quotient by negation, even though the particular symmetric-coordinate formula above is unavailable.
Put $q(\phi)=\deg\phi$ and $q(0)=0$. Applying the <parallelogram law> to $n\phi,\phi$ gives the recurrence $q((n+1)\phi)+q((n-1)\phi)=2q(n\phi)+2q(\phi)$, and hence $q(n\phi)=n^2q(\phi)$ for every integer $n$. Its polarization $B(\phi,\psi)=q(\phi+\psi)-q(\phi)-q(\psi)$ is symmetric and integer-valued. For fixed $\psi$, the <parallelogram law> makes $B(\cdot,\psi)$ odd and gives $B(x+y,\psi)+B(x-y,\psi)=2B(x,\psi)$. Interchanging $x,y$ and adding proves additivity, since integers have no two-torsion. Thus $B$ has <bilinearity>, and $q$ is a positive <quadratic form> on $\operatorname{Hom}(E_1,E_2)$, with $q(\phi)>0$ whenever $\phi\ne0$.
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