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For nonzero integral $D$, the elliptic curve $E_D:y^2=x^3-D^{2}x$ has rational torsion precisely $O,(0,0),(D,0),(-D,0)$. At every good prime $p\equiv 3(\mathrm{mod}4)$, cancellation of the Legendre symbols at $x$ and $-x$ gives $\#E_D({\mathbb{F}}_p)=p+1$. The torsion-freeness of the formal group over Qp for odd p makes the rational torsion subgroup inject into this group. The Dirichlet theorem on primes in arithmetic progressions excludes odd factors of its order and bounds its two-primary part by four. The four displayed rational 2-torsion points then give equality.

For positive integral $k$, the rational torsion subgroup of $y^2=x^3+kx$ has order dividing four. Indeed, torsion-freeness of the formal group over Qp for odd p, reduction of torsion points on an elliptic curve, and the point-count criterion for y squared equals x cubed plus k x bound its order by $p+1$ at all good primes $p\equiv 3(\mathrm{mod}4)$. The Dirichlet theorem on primes in arithmetic progressions excludes every odd torsion prime and bounds the two-primary part by four, by choosing $p\equiv 3(\mathrm{mod}8)$. There is exactly one nonzero rational 2-torsion point, namely $(0,0)$. A rational half of it must have $x^2=k$ by the elliptic-curve addition formula. Writing $k=s^2$ with $s>0$, its equation is $y^2=2s^3$, so $s=2t^2$ and $k=4t^4$. Conversely $(2t^2,4t^3)$ has order four. Thus the cyclic group of order four occurs exactly for $k=4t^4$, $t\ge 1$. The order bound itself holds for every nonzero integral $k$ of either sign: positivity was used only to classify the rational points of orders two and four.

When an elliptic curve has full rational 2-torsion, its Kummer map embeds $E(\mathbb{Q})/2E(\mathbb{Q})$ into a product of square-class groups. Valuation and local-solubility conditions reduce the image to finitely many classes and thereby bound the rank of the Mordell-Weil group.

Let $E:y^2=(x-\alpha )(x-\beta )(x-\gamma )$ with distinct algebraic integers in a number field $K$. For a rational point $P$, its halving field is unramified outside the finite set $S$ of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At $v\notin S$, the elliptic-curve discriminant is a unit, so there is good reduction. A half of the reduction of $P$ is defined over a finite extension of the residue field. Pass to the corresponding unramified extension of the completion and lift this half using the Hensel lemma. The doubling error lies in the formal kernel of a minimal Weierstrass equation. Since two is a unit, prime-to-residue-characteristic multiplication on a formal group corrects this error uniquely. Every half differs by rational 2-torsion, so all halves are unramified at $v$. The halving cocycle with rational two-torsion has image of order at most four, making the halving field a Galois extension of degree at most four. The bounded-degree extensions with restricted ramification are finite in number; there are only finitely many homomorphisms from their finite Galois groups to $E[2]$. Thus $E(K)/2E(K)$ is finite, without assuming any form of the Mordell-Weil theorem.

At an odd good prime, use a model $y^2=x^3-d^{2}x$ with $d$ a unit. If $p\equiv 3(\mathrm{mod}4)$, pairing $x$ with $-x$ cancels their Legendre symbols, so the trace is zero. If $p\equiv 1(\mathrm{mod}4)$, the four distinct rational 2-torsion points force the group order to be divisible by four, whereas $p+1$ is two modulo four. The trace cannot vanish. This proves the converse without an explicit complex-multiplication character-sum evaluation.