= Solution
<Arithmetic heights> measure arithmetic complexity. For a reduced <rational number> $x=a/b$, its <multiplicative height> is $H(x)=\max(|a|,|b|)$ and its <logarithmic height> is $h(x)=\log H(x)$. Over a <number field> $K$, use the absolute <projective height>
$$
h([x_0:\cdots:x_d])=\frac1{[K:\mathbb Q]}\sum_v[K_v:\mathbb Q_v]\log\max_i|x_i|_v.
$$
Normalize the <absolute values on a field> by the <product formula>. That formula makes <arithmetic height> independent of the chosen homogeneous representative, and the normalization makes it independent of enlarging $K$. The <Northcott theorem> says that algebraic points of bounded degree and bounded <arithmetic height> form a <finite set>. For <rational numbers> this follows immediately by bounding the numerator and denominator. In a fixed <number field>, the same finiteness follows by bounding the coefficients of the <minimal polynomial> using its conjugates and the <arithmetic height>.
For an <elliptic curve> put $h_x(P)=h([x(P):1])$ and $h_x(O)=0$. The two points above a given $x$ show that bounded $h_x$ gives finitely many $K$-rational points. More importantly, multiplication expands this <arithmetic height>. To see the mechanism, choose a short <Weierstrass equation of an elliptic curve> $y^2=x^3+Ax+B$ over $K$. Doubling gives
$$
x(2P)=\frac{x(P)^4-2Ax(P)^2-8Bx(P)+A^2}{4(x(P)^3+Ax(P)+B)}.
$$
After homogenization the numerator and denominator are quartics with no common zero. At infinity the numerator is nonzero; at a root $t$ of the cubic the numerator becomes $(3t^2+A)^2$, nonzero by nonsingularity. For a degree-$d$ morphism of the <projective line>, fixed homogeneous coefficients give the upper <arithmetic height> bound $h(f(x))\leq d h(x)+C$. A nonzero resultant supplies the reverse bound: at each place it bounds the common cancellation of the two homogeneous values, with a nontrivial constant needed at only finitely many places. Applying these bounds to duplication gives a constant $C$ independent of $P$ with
$$
|h_x(2P)-4h_x(P)|\leq C.
$$
This includes points mapping to $O$, by the projective formulation.
The <canonical height> removes the bounded errors. With the conventional factor one half, define
$$
\widehat h(P)=\frac12\lim_{n\to\infty}4^{-n}h_x(2^nP).
$$
The successive terms before multiplying by one half differ by at most $C4^{-n-1}$. The limit exists, and summing this geometric bound gives
$$
|\widehat h(P)-\tfrac12h_x(P)|\leq C/6,\qquad \widehat h(2P)=4\widehat h(P),\qquad\widehat h(P)\geq0.
$$
Hence bounded <canonical height> still gives finitely many points over the fixed <number field>.
To obtain the pairing, the quadratic nature comes from a uniform <arithmetic height> parallelogram estimate, not just from doubling. If $x=x(P)$ and $z=x(Q)$, the unordered coordinates $x(P+Q),x(P-Q)$ are the roots of the binary quadratic with coefficients
$$
C_2=(x-z)^2,\quad C_1=2(xz+A)(x+z)+4B,\quad C_0=(xz-A)^2-4B(x+z).
$$
These formulas follow by adding and multiplying the two chord expressions with slopes $(y(P)-y(Q))/(x-z)$ and $(y(P)+y(Q))/(x-z)$. Their homogenizations have bidegree $(2,2)$ and no simultaneous zero. Off the diagonal $C_2\ne0$; on the finite diagonal $x=z=t$, $C_1=4f(t)$ and $C_0=f'(t)^2-8tf(t)$ for $f(t)=t^3+At+B$, so simultaneous vanishing would contradict nonsingularity. At the double point at infinity the homogeneous $C_0$ is nonzero. Local coefficient norms, or the corresponding resultant bounds, therefore give
$$
h([C_0:C_1:C_2])=2h_x(P)+2h_x(Q)+O(1).
$$
The <arithmetic height> of a binary quadratic's coefficient vector differs by $O(1)$ from the sum of the heights of its two roots. At finite places this follows from multiplicativity of the maximum coefficient norm of products; at infinite places the fixed-degree norms are comparable. Thus
$$
h_x(P+Q)+h_x(P-Q)=2h_x(P)+2h_x(Q)+O(1).
$$
Apply this to $2^nP,2^nQ$, divide by $2\cdot4^n$ and take limits. The errors vanish and give the exact <height parallelogram identity>. Polarization consequently makes
$$
\langle P,Q\rangle=\frac12(\widehat h(P+Q)-\widehat h(P)-\widehat h(Q))
$$
a symmetric <bilinear> pairing. Nonnegativity of $\widehat h(aP+bQ)$ for <integers> $a,b$, and density of rational ratios, give the <Cauchy-Schwarz inequality>. In particular $\widehat h(P-Q)\leq2\widehat h(P)+2\widehat h(Q)$. Also $\widehat h(mP)=m^2\widehat h(P)$ for <integers> $m$. <Torsion points of an elliptic curve> have <canonical height> zero because their doubling orbit is finite; conversely, if a point has <canonical height> zero, all its <integer> multiples have bounded <naive height>, and finiteness forces two multiples to coincide. Thus <canonical height> vanishes exactly on torsion.
There is an arithmetic finiteness input before <height descent>: $E(K)/2E(K)$ is finite, the <Weak Mordell-Weil theorem>. Here is its mechanism. The multiplication <exact sequence> gives an injective Kummer map $E(K)/2E(K)\to H^1(K,E[2])$: choose a half-point $Q$ and use the cocycle $\sigma\mapsto\sigma Q-Q$; changing $Q$ changes a coboundary, and a trivial class means a half-point can be chosen over $K$. Let $L=K(E[2])$, a <Finite Galois extension>. Over $L$ the <torsion module> is $(\mathbb Z/2\mathbb Z)^2$, so <Kummer theory> identifies its first <Galois cohomology> with two copies of $L^\times/L^{\times2}$.
Only classes unramified outside a fixed <finite set> $S$ occur. Include the <primes> over $2$, bad-reduction <primes> and the ramified <primes> of $L/K$. At a good <prime> away from $2$, multiplication by two on the proper smooth elliptic model is <finite étale morphism>, so the torsor of halves of a point is unramified. In square-class language, <valuations> outside $S$ are even. Such classes form a <finite group>: write the <principal ideal> of a representative as a square times an ideal supported on $S$. The ideal class gives an element of the finite two-torsion of the localized <ideal class group>; after fixing that class, ambiguity is an $S$-unit modulo squares. The $S$-unit theorem makes the latter finite. The <group kernel> of restriction from $K$ to $L$ is contained in $H^1(\operatorname{Gal}(L/K),E[2])$, which is finite because both the group and module are finite. This proves the weak theorem, without assuming finite generation of $E(K)$.
Finally choose representatives $R_1,\ldots,R_s$ of $E(K)/2E(K)$ and put $M=\max_i\widehat h(R_i)$. Every $P$ has the form $2Q+R_i$, and
$$
\widehat h(Q)=\frac14\widehat h(P-R_i)\leq\frac12\widehat h(P)+\frac12M.
$$
If $\widehat h(P)>M+1$, this reduces <canonical height> by more than one half. Iterate until a point of <canonical height> at most $M+1$ is reached. There are only finitely many such points, by Northcott and the bounded difference of heights. Reconstructing $P=2Q+R_i$ at every stage proves that these finitely many small points and the finitely many representatives generate $E(K)$. Therefore
$$
\boxed{E(K)\cong E(K)_{\mathrm{tors}}\oplus\mathbb Z^r\quad\text{for some finite }r.}
$$
The canonical pairing is positive definite on the free part. It turns rank computations into an arithmetic lattice problem, supplies independence tests for proposed generators, and bounds searches once a <arithmetic height> bound is available. Weak finite-quotient finiteness alone would not prove this result: the <arithmetic height> contraction is the step that converts it into finite generation.
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