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Suppose an elliptic curve $E/K$ has all four geometric 2-torsion points rational. For $P\in E(K)$ choose $Q\in E(\overline{K})$ with $2Q=P$. The function $c_P(\sigma )=\sigma (Q)-Q$ takes values in $E[2]$ and is a group homomorphism from the absolute Galois group to $E[2]$: its usual cocycle law becomes additivity because the action on $E[2]$ is trivial. Replacing $Q$ by another half, or $P$ by $P+2R$ with $R\in E(K)$, leaves it unchanged. Adding chosen halves proves additivity in $P$. Its kernel in $E(K)/2E(K)$ is zero, since $c_P=0$ precisely when the chosen half is rational. This constructs the injective map directly, without assuming any theorem of two-descent on an elliptic curve or finite generation of the Mordell-Weil group.

For $E:y^2=x(x+1)(x+4)$,The points $(-2,2)$ and $(-1,0)$ generate the two factors, and a two-descent on an elliptic curve proves that the Mordell-Weil group has rank zero.

A congruent number is a positive integer which occurs as the area of a right triangle with positive rational side lengths. Scaling a triangle by a nonzero rational factor multiplies its area by a rational square, so the essential parameter is its positive square class; integer parameters can be reduced to square-free integers. For example $6$ is congruent by the triangle $(3,4,5)$, and $5$ is congruent by $(3/2,20/3,41/6)$.

The connection with elliptic curves comes from the congruent number elliptic curve in the modelFrom a rational point with $y\ne 0$, defineDirect squaring gives $a^2+b^2=c^2$, and its area isConversely, from a positive rational triangle $(a,b,c)$ of area $N$, putSubstitution using $c^2=a^2+b^2$ and $ab=2N$ gives $y^2=x^3-N^{2}x$ and $y\ne 0$. ThusThe standard rational torsion of a congruent number curve consists of $O,(0,0),(N,0),(-N,0)$. One way to see the torsion restriction is to use good primes $p\equiv 3(\mathrm{mod}4)$. The character sum cancels at $x$ and $-x$, giving $\#E_N({\mathbb{F}}_p)=p+1$. For any odd prime $\ell$, the Dirichlet theorem on primes in arithmetic progressions supplies good primes with $p\equiv 3(\mathrm{mod}4)$ and $p\equiv 1(\mathrm{mod}\ell )$, so the reduction of torsion points on an elliptic curve excludes $\ell$-torsion. Good primes $p\equiv 3(\mathrm{mod}8)$ similarly bound the two-primary torsion by four, and the four displayed points give equality. Consequently the Mordell-Weil theorem turns the criterion into positive rank of $E_N(\mathbb{Q})$. Arithmetic two-descent on an elliptic curve and local obstructions are therefore tools for the congruent number problem, and a nontorsion point gives infinitely many rational right triangles from a nontorsion point by taking multiples.

We now give elementary proofs for $1$ and $2$, including the descent steps. Recall the primitive Pythagorean parametrizationwhere $m>n>0$ are coprime and of opposite parity. To derive it, in a primitive Pythagorean triple put the even leg at $A$. The relatively prime integers $(C+B)/2,(C-B)/2$ have product $(A/2{)}^2$, so each is a square, say $m^2,n^2$. This gives the formulas and the parity condition. The area is $mn(m-n)(m+n)$, and its four factors are pairwise coprime: their common divisors reduce to divisors of $m,n$ or of two, while $m\pm n$ are odd.

Area one is impossible. Clearing denominators of a hypothetical rational triangle of area one gives an integer triangle with square area. Dividing out its common factor leaves a primitive integer triangle whose area is still a rational square; an integer which is a rational square is an integer square. Choose such a triangle with smallest hypotenuse. Pairwise coprimality in the area product forcesThe positive integers $u,v$ are odd and coprime. If $m$ were even, then $n$ would be odd and $u^2-v^2=2s^2$ would be $2$ modulo eight, whereas two odd squares have difference zero modulo eight. Hence $m$ is odd, $n$ is even and $s$ is even.

Now set $a=(u+v)/2$ and $b=(u-v)/2$. These are positive coprime integers andThus $(a,b,r)$ is another integer right triangle with square area and hypotenuse $r$, strictly smaller than $m^2+n^2$. This contradicts the minimal choice and proves the Fermat right triangle theorem. In particular $1$ is not a congruent number.

For area two we first establish the auxiliary descent fourth-power sum cannot be a square. Suppose $x^4+y^4=z^2$ with positive integers, and choose a primitive solution with smallest $z$. Two odd fourth powers sum to $2$ modulo sixteen, so after exchange $x$ is odd and $y$ is even. Parametrize the primitive triangle $(x^2,y^2,z)$:The case $m$ even, $n$ odd makes $x^2\equiv 3(\mathrm{mod}4)$, so $m$ is odd and $n$ even. Coprimality then gives $m=u^2$ and $n=2v^2$. ConsequentlyThis second triangle is primitive, since $x$ is odd and $\gcd (x,v)=1$. Parametrize it as $x=r^2-s^2$, $2v^2=2rs$, $u^2=r^2+s^2$, with $r,s$ coprime. The equality $v^2=rs$ makes $r=a^2$, $s=b^2$. Thusa new positive solution with smaller hypotenuse $u<z=u^4+4v^4$. This contradicts infinite descent, proving the auxiliary assertion.

Area two is also impossible. Clearing denominators and removing a common factor now gives a primitive integer triangle with area twice an integer square. Indeed, if an integer is $2(a/b{)}^2$ with $\gcd (a,b)=1$, then $b^2$ divides two, forcing $b=1$. Its area product would satisfyThe odd factors must be squares and the unique even factor twice a square. As in the preceding modulo-eight argument, $m$ even would force $m+n=c^2$, $m-n=d^2$ and $n=b^2$ odd, with $c^2-d^2=2b^2$, impossible modulo eight. ThusThe integers $u=(c+d)/2$ and $v=(c-d)/2$ are positive and coprime, and $uv=(c^2-d^2)/4=b^2$. Hence $u=r^2$, $v=s^2$. Butcontradicting the auxiliary descent. This proves two is not a congruent number, and concludesThe two obstructions are global arithmetic descents. The elliptic-curve correspondence organizes the wider problem, but by itself is not a substitute for proving these particular nonexistence assertions.

All geometric 2-torsion is rational: its points are $O,(\alpha ,0),(\beta ,0),(\gamma ,0)$. We will construct the required finite quotient directly, without assuming any theorem about finite generation or two-descent on an elliptic curve.

Choose $Q\in E(\overline{K})$ with $2Q=P$ and define the halving cocycle with rational two-torsionSuch a half exists because the nonconstant multiplication morphism $[2]$ on a smooth projective elliptic curve is surjective over an algebraic closure. Since $E[2]$ is rational, $c_P(\sigma \tau )=c_P(\sigma )+c_P(\tau )$. A different half is $Q+T$ with rational $T\in E[2]$ and gives the same cocycle. Replacing $P$ by $P+2R$, with $R\in E(K)$, allows the half $Q+R$ and again changes nothing. Adding chosen halves shows additivity in $P$. If the cocycle is zero, the half is fixed by the absolute Galois group and lies in $E(K)$, so $P\in 2E(K)$. This proves, rather than assumes, an injectionThe fixed field of $\ker c_P$ is exactly $K(Q)$. Its Galois group is the image of $c_P$, so its degree is at most four. This also follows from the permitted biquadratic description.

Let $S$ be the finite set of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At a place $v\notin S$ the displayed integral cubic has unit elliptic-curve discriminant, hence good reduction of an elliptic curve, and the residue characteristic is odd. Choose a half $\overline{Q}$ of the reduction of $P$ over the algebraic closure of the residue field. Its coordinates belong to a finite residue extension; let $L_v/K_v$ be the corresponding finite unramified extension. Smoothness and the Hensel lemma lift $\overline{Q}$ to some $Q_0\in E(L_v)$. The difference $P-2Q_0$ reduces to zero and belongs to the formal kernel of a minimal Weierstrass equation. In its integral formal group law the doubling series is $2t+O(t^2)$, with unit linear coefficient. The prime-to-residue-characteristic multiplication on a formal group gives a unique $R$ in this kernel such that $2R=P-2Q_0$. Thus $Q_0+R$ is a half of $P$ defined over an unramified local extension. Every other half differs by rational 2-torsion, so it too is unramified. We have proved the unramified halving fields for a split cubic assertion at every $v\notin S$.

There are only finitely many bounded-degree extensions with restricted ramification of $K$ of degree at most four. This is the permitted number-field finiteness consequence of the Hermite–Minkowski theorem: bounded local degrees bound the discriminant exponents at the finitely many allowed ramified primes, so absolute discriminants are bounded. Each possible halving field $L$ is Galois, and there are only finitely many homomorphisms $\mathrm{Gal}(L/K)\to E[2]$. The injection constructed above therefore has finite image. Consequently $E(K)/2E(K)$ is finite. No elliptic-curve finiteness or descent theorem was used in this argument.

For $K=\mathbb{Q}$, write $s=\#S$. Each coordinate of $c_P$ in a basis of $E[2]\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$ is a quadratic Galois character unramified outside $S$. Its quadratic extension has a signed square-free integer representative $d=\pm {\prod }_{p\in S}{p}^{e_p}$, $e_p\in \{0,1\}$: an odd valuation outside $S$ would ramify there. There are at most $2^{s+1}$ choices, including the trivial character. Hence the pair of characters gives at most $2^{2s+2}$ cocycles. Each prime in $S$ is at most $2m$, because the three root differences have absolute values at most $2m$ and $2\le 2m$. Among the integers up to $2m$ there are at most $m$ primes: two, together with at most $m-1$ odd candidates. The rational halving quotient bound from root size is therefore

Use the two-descent on an elliptic curve map associated with the three rational roots $0,-1,-4$,with the standard limiting values at the 2-torsion. Only the square classes of $-1$, $2$ and $3$ can occur, because all other numerator and denominator valuations in the three factors $x,x+1,x+4$ are even. Checking solubility over $\mathbb{R}$, ${\mathbb{Q}}_2$ and ${\mathbb{Q}}_3$ leaves exactlyThese four classes are represented respectively by $O$, $P_1$, $P_2$ and $P_1+P_2$. Thus $|E(\mathbb{Q})/2E(\mathbb{Q})|=4$. Since part (b) givesthe quotient by doubling has order $2^{r+2}$. Therefore