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For a prime $p\equiv 3(\mathrm{mod}8)$, apply two-isogeny descent to the congruent number elliptic curve $y^2=x^3-p^{2}x$ and its companion $y^2=x^3+4p^{2}x$. The first square-class image is exactly $\{1,-1,p,-p\}$, already supplied by 2-torsion. The second has only positive candidates $1,2,p,2p$. The quartic covering in a two-isogeny descent equations for $p$ and $2p$ are impossible modulo $p$, because $-1$ is not a quadratic residue. For $2$ the equation is $W^2=2U^4+2p^2{V}^4$. Coprimality forces $U,V$ odd and $W=2Z$; modulo sixteen this gives $2Z^2=1+9=10$, which is impossible. Thus the second image is trivial and the square-class index formula for two-isogeny descent gives rank zero. The same prime-reduction argument bounds its rational torsion by four, and it already has four rational 2-torsion points, so it has no rational point corresponding to a right triangle of area $p$.

For this integral Weierstrass equation of an elliptic curve, computeIf an odd prime divides $m$ or $m^2-1$, then $c_4$ is a unit there and $\Delta$ is not. The j-invariant of an elliptic curve consequently has negative valuation. Since good reduction implies an integral j-invariant, such a prime cannot be a prime of good reduction, regardless of whether the displayed equation is minimal. Conversely, if it divides neither factor, this equation already has unit discriminant.

At five the good possibilities have $m\equiv \pm 2$, so $m^2\equiv 4$. At seven they have $m^2\equiv 2$ or $4$. Direct elliptic-curve point count over a finite field gives the following nonzero ordinates, together with the three zero-ordinate points and $O$:At either prime, torsion-freeness of the formal group over Qp for odd p makes reduction injective on all of $E(\mathbb{Q}{)}_{\mathrm{tors}}$, so its order is at most eight. Using only prime-to-$p$ injectivity here would leave an unjustified possible $p$-primary component.

All three nonzero 2-torsion points $(0,0),(-1,0),(-m^2,0)$ are rational. The rational point $P=(m,m(m+1))$ lies on the curve, since $m(m+1)(m+m^2)=m^2(m+1{)}^2$. For an equation $y^2=x(x^2+ax+b)$ the elliptic-curve addition formula givesHere $a=1+m^2$, $b=m^2$, so $2P=(0,0)$ and $P$ has order four. The point $(-1,0)$ is not in $⟨P⟩$, whose only nonzero point of order two is $(0,0)$. Therefore $P$ and $(-1,0)$ generate a subgroup of order eight isomorphic to $\mathbb{Z}/4\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$. The upper bound proves the rational torsion in the family x times x plus one times x plus m squared:

Galois cohomology and the finite division quotient. Let $K$ be a number field, $G_K=\mathrm{Gal}(\overline{K}/K)$ with its profinite topology, and $M$ a discrete continuous $G_K$-module. Its Galois cohomology is continuous group cohomology. In degrees zero and one,Thus a crossed homomorphism measures the failure of a choice to be Galois-invariant, and changing the choice changes the cocycle by a coboundary. A short exact sequence of such modules gives a long exact sequence in group cohomology. The connecting map sends an invariant element to the cocycle obtained by lifting it and comparing its Galois translates.

For an elliptic curve $E/K$ and $n\ge 2$, multiplication by $n$ is surjective on $E(\overline{K})$ and has kernel $E[n]$. The Kummer exact sequence of an elliptic curve therefore givesExplicitly, choose $nQ=P$ and set $\kappa (P)(\sigma )=\sigma Q-Q$. Replacing $Q$ changes this by a coboundary; replacing $P$ by $P+nR$, $R\in E(K)$, does not change it. If its class is zero, some torsion translate of $Q$ is Galois-invariant, so $P\in nE(K)$. This proves injectivity of the Kummer map of an elliptic curve.

The whole group $H^1(K,E[n])$ need not be finite. The crucial restriction is ramification. Let $S$ contain the finite primes of bad reduction of an elliptic curve and those dividing $n$. At $v\notin S$, good reduction and prime-to-residue-characteristic multiplication imply that every division point of a local point is in an unramified extension, by the formal-group lifting argument. Thus every global Kummer class restricts trivially to inertia there.

Choose a finite Galois extension $L/K$ containing $E[n]$ and the $n$th roots of unity, and enlarge $S$ to include its ramified primes. Over $L$, the torsion module is constant and can be identified with ${\mu }_n^2$. By Kummer theory,For classes unramified outside the primes $S_L$ above $S$, valuations outside $S_L$ must be divisible by $n$. Their restriction images therefore lie in $L(S_L,n{)}^2$, the square of an S-unramified power class group. Its finiteness can be seen without assuming the Mordell-Weil theorem: the finiteness of S-unramified Kummer classes follows fromThe S-unit group is finitely generated by the Dirichlet unit theorem and its valuations at $S_L$. The localized ideal class group is a quotient of the ordinary ideal class group of $L$, which is finite by the finiteness of the ideal class group. The Minkowski bound for ideal classes supplies an integral ideal of bounded norm in each class, and only finitely many integral ideals have bounded norm.

The kernel of restriction from $H^1(K,E[n])$ to $H^1(L,E[n])$ is also finite: a cocycle trivial after restriction can, after subtracting a coboundary, be made to factor through the finite group $\mathrm{Gal}(L/K)$. There are only finitely many maps from that group to the finite module $E[n]$. This is the finite-extension kernel of a Kummer map mechanism. Consequently the possible Kummer classes form a finite set, provingThis is the Weak Mordell-Weil theorem. Imposing membership in the local Kummer image at every place refines the finite unramified collection to the n-Selmer group. Its relation with the obstruction to a globally rational point iswhere the last group is the $n$-torsion of the Tate–Shafarevich group. Thus local solubility gives a finite computable upper bound but can leave a genuine global obstruction.

Heights and finite generation. For the other essay, normalize the absolute values on a field $K$ so that the product formula holds. The Absolute logarithmic Weil height of $[x_0:x_1]\in {\mathbb{P}}^1(K)$ isAt the infinite places the local degrees are one or two and the usual real or complex modulus is used. The product formula makes this independent of scaling; it is also unchanged by extending $K$. Over $\mathbb{Q}$, using coprime integral coordinates gives $h([u:v])=\log \max (|u|,|v|)$, the naive height on the projective line. The Northcott theorem says that points of bounded height and bounded field degree form a finite set; in particular this holds over the fixed field $K$. One way to see the finiteness is the height-Mahler measure formula: for an algebraic number of degree $d$, the coefficients of its primitive minimal integer polynomial are bounded by binomial factors times $\exp (dh)$. Bounds on $d$ and $h$ therefore leave only finitely many integer polynomials and hence only finitely many algebraic numbers.

Set $h_E(P)=h(x(P))$ for $P\ne O$, and $h_E(O)=0$. Since $x:E\to {\mathbb{P}}^1$ has degree two, bounded $h_E$ gives only finitely many points of $E(K)$. The height growth under a morphism of the projective line states that a degree-$d$ morphism gives $h(f(t))=dh(t)+O(1)$. The upper bound follows from its homogeneous coordinate polynomials; the lower bound follows from their having no common zero, using a resultant identity at each place. Applied to the degree-four duplication map, it gives a uniform constant $C$ withConsequently $4^{-j}{h}_E(2^{j}P)$ is Cauchy: the absolute difference of consecutive terms is at most $C{4}^{-(j+1)}$. Define the canonical height of an elliptic curve byIt is nonnegative and satisfies $\hat{h}(2P)=4\hat{h}(P)$. The factor one-half is the conventional normalization for the divisor $(O)$, since the first-coordinate height belongs to $2(O)$.

The addition-divisor identity also gives the bounded-error height identityuniformly in $P,Q$. This is the height form of the identity for the sum and difference pullbacks of the line bundle associated to $2(O)$; it includes points where the affine addition formulas have a zero denominator. Apply it to $2^{j}P,2^{j}Q$, divide by $2\cdot 4^j$, and take the limit. Thus $\hat{h}$ satisfies the parallelogram law, and its polarization is a bilinear form, the height pairing. Nonnegativity implies the Cauchy-Schwarz inequality for that pairing, by evaluating its nonnegative quadratic polynomial on $mP+nQ$ for all integers $m,n$ and approximating real ratios by rational numbers. In particularThese properties do not presuppose finite generation. Bounded canonical height gives a finite set by the bounded difference and the Northcott theorem. Torsion points have canonical height zero. Conversely, if $\hat{h}(P)=0$, all $2^{j}P$ have bounded naive height, so two coincide; their difference makes $P$ torsion. Thus the height detects precisely the free part, once finite generation is proved.

Use the Weak Mordell-Weil theorem with $n=2$ and choose finitely many coset representatives $R_i$. Write $P=2Q+R_i$, and put $H={\max }_i\hat{h}(R_i)$. The height inequality yieldsIteration enters the finite set $\{P:\hat{h}(P)\le H+1\}$, since after $j$ steps the height is at most $H+2^{-j}\max (\hat{h}(P)-H,0)$. Unwinding $P=2Q+R_i$ expresses every point using that finite set and the $R_i$. This proves the canonical-height proof of Mordell-Weil finite generation and henceThe Fundamental theorem of finitely generated abelian groups supplies this decomposition. The Mordell-Weil theorem combines a finite quotient from arithmetic Galois cohomology with a contracting height descent; either ingredient alone would not establish finite generation.

Translate the rational 2-torsion point to $(0,0)$ and clear denominators to obtain the integral Weierstrass equation of an elliptic curve. Nonsingularity requires $b\ne 0$ and $b^{\prime }=a^2-4b\ne 0$. The two-isogeny descent usesand its dual isogenyThese formulas extend across their missing affine points as degree-two isogenies of elliptic curves, with kernels $\{O,(0,0)\}$ on their respective curves; substitution gives $\psi \varphi =[2]$ and $\varphi \psi =[2]$.

Define homomorphisms to the square-class group of a field ${\mathbb{Q}}^{\times }/({\mathbb{Q}}^{\times }{)}^2$ byFor ordinary intersections with a line, the product of the three first coordinates is the square of its intercept, proving the homomorphism identity; tangent cases use the same product with multiplicities. For $T=(0,0)$, direct addition gives $x(P+T)=b/x(P)$, so $\alpha (P+T)=[b]\alpha (P)$. Also $\alpha (P)\alpha (-P)=1$ and $\alpha (T{)}^2=1$, which handle inverse points and $T+T=O$. These verify the exceptional values as well. Their kernels are $\psi E^{\prime }(\mathbb{Q})$ and $\varphi E(\mathbb{Q})$. One can check the first assertion directly: if $x=s^2\ne 0$, solving for the first coordinate of a preimage under $\psi$ giveswith rational corresponding ordinate. Conversely, on $E^{\prime }$ one has ${\psi }_x=Y^2/(4X^2)$, a square. The exceptional point $(0,0)$ has a rational preimage exactly when $b$ is a square. The other kernel assertion follows identically, since the double companion curve is isomorphic to $E$ by scaling its coordinates by four and eight.

We now justify the rank factor in the square-class index formula for two-isogeny descent, rather than forgetting a torsion correction. Put $A=E(\mathbb{Q})$, $B=E^{\prime }(\mathbb{Q})$, andThe preimage of $\psi \varphi A$ under $\psi$ is $\varphi A+\ker \psi$, soThe Mordell-Weil theorem gives $[A:2A]=2^r\#A[2]$, where $r$ is the rank of an abelian group of $A$. If $b^{\prime }$ is a square, $\#A[2]=4$ and $(0,0)\in \varphi A$, so $\delta =1$. If it is not a square, $\#A[2]=2$ and $\delta =2$. Hence $\delta \#A[2]=4$ in both cases, and

To bound these two images, let $\ell \nmid b$ and take a point with $x\ne 0$. If $v_{\ell }(x)>0$, then $x^2+ax+b$ is a unit, so $2v_{\ell }(y)=v_{\ell }(x)$ is even. If $v_{\ell }(x)<0$, the leading term $x^3$ uniquely has least valuation in the equation, so $2v_{\ell }(y)=3v_{\ell }(x)$ and again $v_{\ell }(x)$ is even. Thus every image class has a signed square-free integer representative supported on the prime divisors of $b$. The exceptional class $[b]$ also has that property. There are at most $2^{\nu (b)+1}$ such classes. Applying the same reasoning to $E^{\prime }$ gives $\#{\alpha }^{\prime }(B)\le 2^{\nu (b^{\prime })+1}$. The prime-support bound in two-isogeny descent is thereforeHere $\nu$ counts the distinct prime divisors of the absolute value of a nonzero integer; $\nu (1)=\nu (-1)=0$. The nonsingularity hypotheses exclude the otherwise undefined case $\nu (0)$.

For the unheaded practical procedure, enumerate signed square-free divisors $d$ of $b$, and of $b^{\prime }$ on the companion curve. A class $d$ is represented by a point of $E$ exactly when its quartic covering in a two-isogeny descenthas a rational solution with $(U,V)\ne (0,0)$. For $UV\ne 0$, reconstruct $x=d{U}^2/V^2$, $y=dUW/V^3$; substitution verifies the equivalence. Solutions with $V=0$ or $U=0$ account for the identity class or the class of $(0,0)$ respectively. Clearing denominators allows integral $U,V,W$ with $\gcd (U,V)=1$.

Test these finitely many coverings over the real numbers and local fields, especially at two and the primes dividing $b{b}^{\prime }$, to exclude impossible classes. Search the survivors for rational points, close the witnessed classes under multiplication, and use the index formula once both images are determined. Equivalently, local solubility gives a descent upper bound and independent rational points, certified for example by their canonical height of an elliptic curve pairing, give a lower bound. When the bounds coincide the rank is determined. This procedure often succeeds, but local solubility alone does not prove global solubility: a nonzero Tate–Shafarevich group can leave surviving coverings without rational points. In that case the computation gives a rigorous bound rather than a falsely certified exact rank.

Take the odd congruence class $p\equiv 3(\mathrm{mod}8)$. The congruent number elliptic curve in the standard scaled coordinates isThese are the two curves in two-isogeny descent. On $E_p$, the prime-support bound in two-isogeny descent restricts the first image to $\{1,-1,p,-p\}$. All four classes occur: the 2-torsion points give $\alpha ((0,0))=[-p^2]=[-1]$, $\alpha ((p,0))=[p]$, and $\alpha ((-p,0))=[-p]$. Thus $\#\alpha (E_p)=4$.

On $E_p^{\prime }$, the equation forces $x>0$ whenever $x\ne 0$. Its exceptional value is $[4p^2]=1$. The second image is therefore contained in $\{1,2,p,2p\}$. Use the quartic covering in a two-isogeny descent to exclude its three nontrivial candidates.

For $d=p$, a rational solution can be written with coprime integral $U,V$, and must satisfyThen $p\mid W$, so $U^4+(2V^2{)}^2\equiv 0(\mathrm{mod}p)$. Since $p\equiv 3(\mathrm{mod}4)$, $-1$ is not a quadratic residue; a sum of two squares is zero only if both are zero. This forces $p\mid U,V$, contradicting coprimality. For $d=2p$, the same reasoning applies to $W^2=2p(U^4+V^4)$ and again forces $p\mid U,V$.

For $d=2$, the equation isIt implies $W=2Z$, whence $2Z^2=U^4+p^2{V}^4$. Exactly one of $U,V$ odd is impossible by parity, while both even contradict coprimality. Thus both are odd. Since $p\equiv 3(\mathrm{mod}8)$ implies $p^2\equiv 9(\mathrm{mod}16)$, reduction modulo sixteen gives $2Z^2\equiv 1+9=10$. But the possible residues of $2Z^2$ modulo sixteen are $0,2,8$. This contradiction eliminates $d=2$. Consequently ${\alpha }^{\prime }(E_p^{\prime })=\{1\}$ and the square-class index formula for two-isogeny descent givesThe reduction argument for $y^2=x^3+kx$ used earlier bounds rational torsion by four for every nonzero integer $k$: its point-count cancellation at $q\equiv 3(\mathrm{mod}4)$ does not depend on the sign of $k$, and the same prime choices apply. Here $k=-p^2$ and all four 2-torsion points are rational. Hence $E_p(\mathbb{Q})$ consists exactly of $O,(0,0),(p,0),(-p,0)$.

To relate this calculation to congruent numbers, a rational point $(x,y)$ with $y\ne 0$ produces the right triangle with side lengthsThe identities $A^2+B^2=C^2$ and $AB/2=p$ follow from $y^2=x(x^2-p^2)$. Conversely, a positive rational right triangle of area $p$ with legs $A,B$ and hypotenuse $C$ givesDirect substitution uses $AB=2p$ and $C^2=A^2+B^2$. No such point exists here. This proves the noncongruent primes congruent to three modulo eight:

For an algebraic variety $X$ defined over a field $K$, a K-rational point is a morphism $\mathrm{Spec}K\to X$ over $K$. In affine coordinates it is a solution of the defining equations with every coordinate in $K$; in projective coordinates it is a nonzero coordinate vector over $K$, considered up to common nonzero scaling. The set is denoted $X(K)$. Over a number field, having points over every completion need not imply having a rational point; for genus-one coverings this failure is recorded by the Tate–Shafarevich group.