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For odd $p$ and p-adic units $u,v$, the quadratic Hilbert symbol isThe Legendre symbols record the two unit square classes. For $p=2$, put $ϵ(u)=(u-1)/2\mathrm{mod}2$ and $\omega (u)=(u^2-1)/8\mathrm{mod}2$. ThenThe basis $-1,2,5$ of the square-class group of the 2-adic numbers gives this formula: its only negative basis pairings are $(-1,-1{)}_2$ and $(2,5{)}_2=(5,2{)}_2$. The first is obstructed by sums of squares modulo four, the second by $x^2-2y^2$ modulo eight; the remaining pairings have explicit field norm representatives.

For $y^2=x^3+a{x}^2+bx$ with $a\equiv 8$ and $b\equiv 9(\mathrm{mod}16)$, the square class $-1$ in a two-isogeny descent would require coprime integers $u,v$ and an integer $N$ with $N^2=-u^4+a{u}^2{v}^2-b{v}^4$. If $u$ is odd and $v$ even, the right side is $15$ modulo $16$; if $u$ is even and $v$ odd, it is $7$; if both are odd, it is $14$. None is a square modulo $16$. Therefore $-1$ does not occur in the two-torsion square-class homomorphism image.

This quartic covering in a two-isogeny descent has no solution with coprime integers $U,V$ not both zero. If precisely one is odd, the right side is $3$ modulo four. If both are odd, their fourth powers are one modulo sixteen and $14U^2{V}^2=14$ modulo sixteen, so the right side is twelve modulo sixteen. Neither residue is a square. Coprimality excludes the remaining parity case. Consequently the negative unit square class is absent from the two-torsion square-class homomorphism for $Y^2=X^3+14X^2+X$.

Use two-isogeny descent with $a=-7$, $b=12$, so the partner elliptic curve is $E^{\prime }:Y^2=X^3+14X^2+X$. On $E$, a negative $x$ would make $x(x-3)(x-4)$ negative. Hence the prime-support bound in two-isogeny descent restricts the square class image to $\{1,2,3,6\}$. All four occur: the identity gives $1$, and the rational points $(2,2)$, $(0,0)$, $(6,6)$ give respectively $2$, $12\equiv 3$, and $6$ in ${\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$. Thus $|\alpha E(\mathbb{Q})|=4$.

For the partner, $b^{\prime }=1$, so its square class image is contained in $\{1,-1\}$. A realization of $-1$ would yield coprime integers $U,V$, not both zero, satisfyingIf exactly one of $U,V$ is odd, the right side is $3$ modulo four, which is not a square. If both are odd, $U^4\equiv V^4\equiv 1(\mathrm{mod}16)$ and $14U^2{V}^2\equiv 14(\mathrm{mod}16)$, so the right side is $12$ modulo sixteen, again impossible. Both even is excluded by coprimality. This proves the negative unit obstruction for the fourteen-coefficient isogeny cover; the identity realizes $1$, so $|{\alpha }^{\prime }{E}^{\prime }(\mathbb{Q})|=1$.

The two-isogeny rank formula now gives $2^r=4\cdot 1/4=1$. Consequently this is the rank-zero elliptic curve with roots zero three and four:The displayed nonzero-ordinate points are therefore torsion points of an elliptic curve, rather than evidence of positive rank. In fact doubling either $(2,2)$ or $(6,6)$ gives $(4,0)$, so both have order four.

For $E:y^2=x^3-17x$, the two-isogeny formula gives $E^{\prime }:Y^2=X^3+68X$. By the prime-support bound in two-isogeny descent, the image of the two-torsion square-class homomorphism on $E$ lies in $\{1,-1,17,-17\}$. The identity gives $1$, $T=(0,0)$ gives $-17$, and $P=(-1,4)$ gives $-1$, since $(-1{)}^3-17(-1)=16$. The image is a subgroup of the square-class group, so their product $17$ occurs as well; explicitly $P+T=(17,68)$. HenceOn $E^{\prime }$, a nonzero affine point has $X>0$, because $Y^2=X(X^2+68)$. The only candidate square classes are therefore $1,2,17,34$. The identity gives $1$, $T^{\prime }=(0,0)$ gives $[68]=17$, and $(2,12)$ gives $2$, since $2^3+68\cdot 2=144$. Their product $34$ occurs; adding $T^{\prime }$ gives $(34,-204)$. ThusBoth images have order four. The two-isogeny rank formula provesThis is the rank-two elliptic curve with cubic x cubed minus seventeen x; all candidate classes are represented, so no unresolved local-to-global gap in isogeny descent remains in this computation.

The Kronecker–Weber theorem says that every finite abelian extension of $\mathbb{Q}$ is contained in a cyclotomic field. We prove its quadratic case by identifying the ramification that a quadratic number field must have.

For an odd prime $\ell$, the cyclic Galois group of $\mathbb{Q}({\zeta }_{\ell })/\mathbb{Q}$ has exactly one subgroup of index two. Its fixed field $F_{\ell }$ is quadratic and is unramified at every finite prime other than $\ell$. For a squarefree integer $d$, the discriminant of $\mathbb{Q}(\sqrt{d})$ is $d$ if $d\equiv 1(\mathrm{mod}4)$, and $4d$ otherwise. A nontrivial quadratic discriminant supported on the odd prime $\ell$ must therefore beIndeed $d=\pm 1$ cannot give a nontrivial extension unramified at two, whereas exactly one of $\ell ,-\ell$ is congruent to one modulo four. ThusThis identifies the quadratic subfield of a prime cyclotomic field using ramification, without a Gaussian-sum computation.

Now let $F=\mathbb{Q}(\sqrt{d})$ with $d$ squarefree. Multiply $d$ by ${\prod }_{\ell \mid d,\ell \text{ odd}}{\ell }^{\ast }$. Every odd prime then occurs with even exponent, so its square class is represented by one of $1,-1,2,-2$. Thus adjoining the square roots of the ${\ell }^{\ast }$ removes all possible odd-prime ramification; the remaining quadratic factor is contained inThe original $\sqrt{d}$ is a rational multiple of a product of these square roots. It belongs to the compositum of $\mathbb{Q}({\zeta }_8)$ and the prime cyclotomic fields just used. This compositum is itself contained in a cyclotomic field, proving the theorem for $F$.

The sharper conductor statement follows from the same ramification calculation. For the fundamental discriminant $D$ of $F$, putThe possibilities for $D_2$ follow immediately from the quadratic discriminant formula. The corresponding fields are trivial, $\mathbb{Q}(i)$, $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{-2})$, lying in the cyclotomic fields of conductors $1,4,8,8$. Their conductors are coprime to the odd primes in the product. Hence quadratic Kronecker–Weber theorem by ramification gives the explicit conclusion

For $a,b\in {\mathbb{Q}}_p^{\times }$, the quadratic Hilbert symbol $(a,b{)}_p$ is one if $b$ is a field norm from ${\mathbb{Q}}_p(\sqrt{a})$, and minus one otherwise; a square $a$ gives the trivial extension and symbol one. It depends only on the two square classes. Equivalently, the conic $z^2=a{x}^2+b{y}^2$ has a nonzero rational point precisely when the symbol is one. Interchanging $a$ and $b$ in the conic gives $(a,b{)}_p=(b,a{)}_p$. The Hilbert norm residue symbol is a group homomorphism in each argument; for nonsquare $a$, the cyclic local norm index makes its group kernel a subgroup of index two. The resulting pairing on square classes is nondegenerate. Alsosince $N(\sqrt{a})=-a$. In the language of local class field theory, $(a,b{)}_p={\mathrm{Art}}_p(b)(\sqrt{a})/\sqrt{a}$; in the Brauer group, it records whether the quaternion cyclic algebra $(a,b)$ splits. These are compatible norm-obstruction descriptions.

An unramified extension of local fields of degree two has norm group consisting precisely of elements with even valuation. Indeed,and the norms on units are surjective. For the latter assertion, the norm on the finite residue fields is surjective. On successive higher principal-unit groups, the congruenceallows one to correct a proposed unit norm one digit at a time: the residue field trace is surjective, and completeness makes the corrections converge. An unramified quadratic field is unique up to isomorphism. Therefore, if $d$ is its defining nonsquare class,

For odd $p$, choose the nonsquare unit $u_0$ from part (i). The polynomial $T^2-u_0$ has irreducible separable reduction, so it defines the unramified quadratic extension. ThusThe trivial class is included in ${\Delta }_2$, since a degree-one extension is unramified. The two units pair trivially, while $(u_0,p{)}_p=-1$. Any class outside ${\Delta }_1$ has odd valuation and consequently pairs nontrivially with $u_0$. This proves directly

For $p=2$, the unit square classes from part (i) give ${\Delta }_1=\{1,-1,5,-5\}$. The element $\omega =(1+\sqrt{5})/2$ satisfies $T^2-T-1$; its reduction $T^2+T+1$ is irreducible and separable over ${\mathbb{F}}_2$. This shows that ${\mathbb{Q}}_2(\sqrt{5})$ is unramified of degree two, givingFor a fully explicit pairing check, use the generators $(-1,5,2)$. We have $(-1,5{)}_2=1$ and $(-1,2{)}_2=1$, since $5=N(1+2i)$ and $2=N(1+i)$ from ${\mathbb{Q}}_2(i)$. But $(-1,-1{)}_2=-1$: a norm equal to $-1$ would be a sum of two squares. If either summand had negative valuation, the sum would have negative valuation (equal negative valuations give valuation $1-2r<0$); if both were integral, their squares could not sum to $-1\equiv 3(\mathrm{mod}4)$. Finally, the unramified norm criterion gives $(5,2{)}_2=-1$. The identity $(a,b{)}_2=(b,a{)}_2$ and $(a,a{)}_2=(a,-1{)}_2$ determine the remaining entries. Writing $(a,b{)}_2=(-1{)}^{B(a,b)}$, the exponent matrix isFor $b=(-1{)}^s{5}^t{2}^r$, annihilating both generators $-1,5$ of ${\Delta }_1$ requires $s=r=0$, so $b\in \{1,5\}$. Annihilating $5$ alone requires $r=0$, so $b$ is a unit class. HenceThese are the Hilbert-symbol annihilators of units and unramified square classes. Annihilators here are taken with respect to this pairing, or equivalently in the character group of a finite abelian group after identifying square classes with their Hilbert-symbol characters.

A useful strong form of the Hensel lemma is the following. Let $F$ be a complete discretely valued field, with valuation ring $\mathcal{O}$, normalized valuation $v$, and $P\in \mathcal{O}[T]$. If $a_0\in \mathcal{O}$ satisfiesthen there is a unique root $a\in \mathcal{O}$ in the ball $v(a-a_0)>v(P^{\prime }(a_0))$. When $P(a_0)=0$ the assertion is immediate; otherwise the condition includes $P^{\prime }(a_0)\ne 0$.

Here is a proof by Newton iteration over a valued field. Write $c=v(P^{\prime }(a_0))$ and $t_0=v(P(a_0))>2c$, and setSuppose $v(P^{\prime }(a_j))=c$ and $t_j=v(P(a_j))>2c$. The correction has valuation $t_j-c>c\ge 0$, so $a_{j+1}\in \mathcal{O}$. Taylor expansion over $\mathcal{O}$ cancels the linear term and givesThus the derivative still has valuation $c$, and $t_{j+1}-2c\ge 2(t_j-2c)$. Unless an exact root is reached earlier, the corrections tend to zero with exponentially increasing valuations. Completeness gives $a_j\to a$, and continuity gives $P(a)=0$ and $v(a-a_0)=t_0-c>c$. For uniqueness, if $a,b$ are in this ball, polynomial expansion around $a_0$ givesThe second factor has valuation exactly $c$ and is nonzero, so two roots in the ball must be equal. In particular, a simple residue root lifts uniquely: $v(P(a_0))\ge 1$ and $v(P^{\prime }(a_0))=0$ satisfy the hypothesis.

For the square-class group of a p-adic field, write every element as $p^{r}u$ with $u\in {\mathbb{Z}}_p^{\times }$. Multiplication by a square changes $r$ by an even integer, so valuation parity gives one independent factor $C_2$.

For odd $p$, a unit $u$ is a square if and only if its residue is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction; sufficiency applies the simple-root Hensel lemma to $T^2-u$, whose derivative is a unit at a nonzero residue root. The cyclic group ${\mathbb{F}}_p^{\times }$ has square subgroup of index two. If $u_0$ is a unit with nonsquare residue, thenFor $p=2$, every odd square is one modulo eight. Conversely, if $u\equiv 1(\mathrm{mod}8)$, take $P(T)=T^2-u$ and $a_0=1$. Then $v_2(P(1))\ge 3>2v_2(P^{\prime }(1))=2$, so the strong Hensel lemma gives a square root. Thus unit square classes are exactly their residues $1,3,5,7$ modulo eight. The classes $-1$ and $5$ generate this group of order four independently, and valuation parity supplies the class $2$. Hence

Put $t=\sqrt{-30}$ and $K=\mathbb{Q}(t)$. The ring of integers is $\mathbb{Z}[t]$, and the fundamental discriminant is $D_K=-120$. For a reduced positive definite binary quadratic form $(a,b,c)$ of discriminant $-120$, reduction givesThe coefficient $b$ is even; checking $1\le a\le 6$ and $|b|\le a$, with the usual nonnegative sign choice at boundary equalities, leaves preciselyThe correspondence between proper quadratic forms and ideal classes therefore gives $h_K=4$. All four classes equal their inverses, since inversion replaces $b$ by $-b$. HenceThe three nontrivial classes may be represented by the ramified prime ideals ${\mathfrak{p}}_2=(2,t)$, ${\mathfrak{p}}_3=(3,t)$ and ${\mathfrak{p}}_5=(5,t)$. They square to principal ideals and satisfy ${\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_5=(t)$. None is principal: an integral generator of norm $2$, $3$ or $5$ would require $a^2+30b^2$ to equal that number, which is impossible. The relation then makes the three classes distinct, consistently with the four-form calculation.

ConsiderThe three rational square classes are independent, so $[H:\mathbb{Q}]=8$. Their radical product is $\sqrt{-30}$, so $K\subset H$ and $[H:K]=4$; the extension $H/K$ is abelian. To verify that it is unramified, rather than assuming that radical adjunctions are harmless at $2,3,5$, use the conductor-discriminant formula. The seven quadratic subfields have fundamental discriminantsThe formula givesThe discriminant tower identity now givesThus the relative discriminant is the unit ideal and no finite prime ramifies. The base has no real places, so there is no infinite ramification to check. This unramified abelian degree-four extension has degree $h_K$, and part (i) provesThis is the Hilbert class field of Q of square root minus thirty.

Next translate prime representation into an ideal class. For $p\nmid 30$, we claimIf a representation exists, $\alpha =2x+yt$ lies in ${\mathfrak{p}}_2$ and has $N_{K/\mathbb{Q}}\alpha =2p$. The integral ideal $(\alpha ){\mathfrak{p}}_2^{-1}$ has norm $p$, so it is a prime ideal $\mathfrak{P}$. Its class is $[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]$. Conversely, if $[\mathfrak{P}]=[{\mathfrak{p}}_2]$, the product ${\mathfrak{p}}_2\mathfrak{P}$ is principal, say $(\alpha )$. Since this ideal lies in ${\mathfrak{p}}_2$, its generator has the form $\alpha =2x+yt$ with integers $x,y$. Taking norms gives $4x^2+30y^2=2p$, proving the reverse implication. This is prime representation via an ideal class, and it proves sufficiency as well as necessity.

Identify the required class in $\mathrm{Gal}(H/K)$ through the Artin map. We can use the explicit representation $17=2+15$: $2+t$ has norm $34$, so the prime above $17$ obtained from $(2+t){\mathfrak{p}}_2^{-1}$ has class $[{\mathfrak{p}}_2]$. Its Frobenius in the multiquadratic extension acts on the three radicals with signsTheir product is one, so this automorphism fixes $K$. Because $H/\mathbb{Q}$ is abelian, for any unramified rational prime $p$ that splits in $K$, the Frobenius at either prime of $K$ of norm $p$ is its rational Frobenius restricted to $H/K$. The Artin isomorphism therefore makes the required class condition equivalent toConversely, these signs already imply splitting in $K$, since their product is $(\frac{-30}{p})=1$. No extra splitting condition is missing.

The supplementary law for $2$, quadratic reciprocity for $-3$ and $5$, and then the Chinese remainder theorem turn these sign conditions intoTheir four simultaneous residue classes modulo $120$ are $17,23,47,113$. Finally, $2$ is represented by $(x,y)=(\pm 1,0)$, whereas $3$ and $5$ cannot be represented: $y\ne 0$ gives a value at least $15$, and $y=0$ gives twice a square. We have proved the complete criterion

Translate the rational point of order two to $(0,0)$ and complete the square. After clearing denominators by admissible scaling, writeThe companion curve is $E^{\prime }:Y^2=X(X^2-2aX+a^2-4b)$. The two-isogeny formula gives maps $\varphi :E\to E^{\prime }$ and $\psi :E^{\prime }\to E$ satisfying $\psi \varphi =[2]$.

For the Weak Mordell-Weil theorem in this setting, use the two-torsion square-class homomorphismand its counterpart ${\alpha }^{\prime }$ on $E^{\prime }$. These are homomorphisms: a nonvertical line intersecting the cubic in three points has the product of their $x$-coordinates a square, by comparison of the constant term of the intersection cubic. For a line through $(0,0)$, the product of the other two coordinates is $b$, explaining the special value there. The kernels are $\psi E^{\prime }(\mathbb{Q})$ and $\varphi E(\mathbb{Q})$, respectively, as follows by solving the elliptic isogeny preimage quadratic and testing its discriminant for a square.

The prime-support bound in two-isogeny descent makes both images finite. Indeed at a prime not dividing $b$, a negative valuation of $x$ makes the leading cubic term dominate, giving $2v(y)=3v(x)$; a positive valuation makes $x^2+ax+b$ a unit, giving $2v(y)=v(x)$. In either case $v(x)$ is even. Thus the square class of $x$ is supported only on primes dividing $b$, together with a possible sign. The same argument uses $b^{\prime }=a^2-4b$ on $E^{\prime }$. Since $2E=\psi \varphi E$, the two finite elliptic isogeny quotients imply that $E(\mathbb{Q})/2E(\mathbb{Q})$ is finite.

Finite generation still needs a descent on height. Let $h_x$ be the logarithmic height of the $x$-coordinate. The duplication formula has a degree-four rational $x$-coordinate map with coprime numerator and denominator. Polynomial height estimates, with the resultant supplying the lower bound, give the height growth under a morphism of the projective line relation $h_x(2P)=4h_x(P)+O(1)$. Thus the convergent telescoping limitdefines the canonical height of an elliptic curve, without assuming finite generation. It is nonnegative, satisfies $\hat{h}(2P)=4\hat{h}(P)$ and the height parallelogram identity, and differs from $h_x/2$ by a bounded amount. Consequently its bounded subsets of rational points are finite by the Northcott theorem.

Choose the finite list $R_i$ of representatives modulo $2E(\mathbb{Q})$ and let $H={\max }_i\hat{h}(R_i)$. For $P=2Q+R_i$, the canonical height pairing inequality givesRepeated halving brings the height below $H+1$. There are only finitely many such terminal points, and recovering $P$ expresses it as an integer combination of those points and the $R_i$. This proves the canonical-height proof of Mordell-Weil finite generation, so

To compute the rank of an elliptic curve, translate the rational point of order two to $(0,0)$ and put the curve in the form $y^2=x^3+a{x}^2+bx$, with $a,b\in \mathbb{Z}$ after clearing denominators and $b(a^2-4b)\ne 0$. Its two-isogenous partner and the elliptic isogeny areThe maps extend over their exceptional points. The dual isogeny isThe two-torsion square-class homomorphism is $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\times }/{\mathbb{Q}}^{\times 2}$, defined by $\alpha (O)=1$, $\alpha ((0,0))=[b]$, and $\alpha ((x,y))=[x]$ otherwise. Define ${\alpha }^{\prime }$ similarly on $E^{\prime }$ with $b^{\prime }=a^2-4b$.

Here are the ingredients behind the descent procedure. For a generic chord $y=mx+c$, the cubic of intersection coordinates has root product $c^2$. As the third intersection is the negative of the group sum and negation leaves $x$ unchanged, this proves the square-class homomorphism property. For a line through $(0,0)$, the two nonzero intersection coordinates have product $b$, so the assigned class $[b]$ gives the same multiplicativity. A vertical line has two equal coordinates and contributes their square, and $O$ has identity class. These observations also cover tangencies and the exceptional sums. Its kernel is $\hat{\varphi }(E^{\prime }(\mathbb{Q}))$. Indeed, the displayed formula gives $x(\hat{\varphi }(X,Y))=(Y/(2X){)}^2$. Conversely, if a point on $E$ has $x=r^2\ne 0$, the possible preimage coordinates satisfyIts elliptic-curve discriminant is $16(r^4+a{r}^2+b)=16y^2/r^2$, a rational square. A root $X$ and $Y=2rX$, with the sign chosen appropriately, give a preimage. The identity is always in the image. The point $(0,0)$ is in the image exactly when $b$ is a square: its nonzero preimage coordinates are $X=a\pm 2\sqrt{b}$, with $Y=0$. Thus the exceptional points satisfy precisely the same kernel statement. Interchanging the curves similarly gives $\ker {\alpha }^{\prime }=\varphi (E(\mathbb{Q}))$.

A prime not dividing $b$ can occur only with even valuation in an $x$-coordinate. If its valuation is positive, the factor $x^2+ax+b$ is a unit and $2v(y)=v(x)$; if negative, $x^2$ dominates that factor and $2v(y)=3v(x)$. Thus square classes in $\mathrm{im}\alpha$ have signed square-free representatives dividing $b$, giving finitely many candidates. To test a candidate $d$, put $x=d(u/v{)}^2$ and $y=duN/v^3$. The curve equation becomesA nontrivial rational solution, including the appropriate limiting points, is exactly a realization of the class. Clear denominators and arrange $\gcd (u,v)=1$ for congruence tests. Do the same for $E^{\prime }$ with $b^{\prime }$.

The factor in the two-isogeny rank formula can also be checked directly. Let $A=E(\mathbb{Q})$, $B=E^{\prime }(\mathbb{Q})$ and $t=\#A[2]$. The index of $\hat{\varphi }(B)$ in $A$ is $\#\mathrm{im}\alpha$, while the map from $B/\varphi (A)$ onto $\hat{\varphi }(B)/2A$ has kernelThe numerator has order two. Moreover $\ker \hat{\varphi }\cap \varphi (A)=\varphi (A[2])$, since $\hat{\varphi }\varphi P=2P$, and the latter has order $t/2$. Hence this quotient has order $4/t$, andThe Mordell-Weil theorem gives $[A:2A]=2^{r}t$, where $r$ is the rank. Cancelling $t$ yieldsIn practice, local congruence and real-solubility tests remove many candidate quartics, giving an upper bound, while rational point searches realize classes and give a lower bound. These often coincide. Local solubility alone is not a proof of a rational point: an everywhere locally soluble covering can still fail globally, so a remaining gap must be reported rather than treated as an exact rank.

For the curve at hand, $a=8$, $b=-7$, $b^{\prime }=92$, andThe first image is contained in $\{1,-1,7,-7\}$ and contains $1,-7$, the latter from the point $(0,0)$. To exclude $-1$, its quartic would require coprime $u,v$ withModulo $16$, an odd fourth power is $1$ and an even fourth power is $0$. If $u$ is odd and $v$ even the right side is $15$; if $u$ is even and $v$ odd it is $7$; if both are odd it is $14$. None is among the square residues $0,1,4,9$. Thus $-1$ is impossible. The image is a subgroup, so $7$ is also impossible: its product with the known class $-7$ would be $-1$. We obtain $\mathrm{im}\alpha =\{1,-7\}$.

For $E^{\prime }$, its quadratic factor is $X^2-16X+92=(X-8{)}^2+28>0$. Every nonzero rational $X$-coordinate is therefore positive. Prime support leaves only $1,2,23,46$ as possible classes; $1$ and $23=[92]$ are supplied by $O$ and the two-torsion point. The explicit point $(18,48)$ is on $E^{\prime }$ becauseIts class is $[18]=2$. Multiplying classes supplies $46$, so $\mathrm{im}{\alpha }^{\prime }=\{1,2,23,46\}$. The two-isogeny rank formula now provesAs an additional concrete point on the original curve, the dual isogeny sends $(18,48)$ to $(16/9,116/27)$. The descent calculation proves the rank; it does not by itself certify that this point generates the full free quotient.

A rational point of order $2$ can be moved to $(0,0)$, giving an integral modelIts two-isogeny descent partner and the two-isogeny formula areThe maps extend over the identity and $(0,0)$, have these two points as their respective kernels, and direct substitution and the doubling formula give $\hat{\varphi }\varphi =[2]$ and $\varphi \hat{\varphi }=[2]$.

Define the two-torsion square-class homomorphisms byand similarly ${\alpha }^{\prime }$ for $E^{\prime }$, using $b^{\prime }$. To see the homomorphism property, intersect a line $y=mx+c$ with $E$. Its three $x$-coordinates have product $c^2$, since the resulting monic cubic has constant term $-c^2$. The third point is the negative of the sum of the first two, and negation preserves $x$, so the square classes multiply as required. If the line passes through $(0,0)$, the other two coordinates have product $b$, giving exactly the prescribed exceptional value. Vertical lines and repeated intersections give the identity and tangent cases.

We also verify the kernel needed for descent:For a point on $E^{\prime }$, the first coordinate of its image under $\hat{\varphi }$ is $Y^2/(4X^2)$, a square. Conversely, if $P=(s^2,y)$ has square $x$-coordinate, solving $x(\hat{\varphi }(X,Y))=s^2$ givesIts discriminant is $16(s^4+a{s}^2+b)=(4y/s{)}^2$, so it has rational nonzero roots $X=a+2s^2\pm 2y/s$. Taking $Y=\pm 2sX$ gives a point on $E^{\prime }$; choose the signs so its image has second coordinate $y$. The exceptional point $(0,0)$ lies in the image exactly when $b$ is a square, as seen from the points with $Y=0$, $X=a\pm 2\sqrt{b}$. This proves the first kernel formula. Applying the same argument to $E^{\prime }$ and then rescaling the second partner by $(X,Y)\mapsto (X/4,Y/8)$ proves the second.

There are only finitely many candidate square classes. For a rational point write $x=M/e^2$ and $y=N/e^3$, with $\gcd (M,e)=1$. These denominator forms follow because a negative prime valuation of $x$ makes the leading cubic term the unique lowest-valuation term, forcing that valuation to be even. ThenIf a prime divides $M$ but not $b$, the second factor is a unit at that prime, so its valuation in $M$ is even. Thus every class in $\alpha (E)$ has a signed squarefree representative $d\mid b$. This is the prime-support bound in two-isogeny descent; the analogous candidates for ${\alpha }^{\prime }(E^{\prime })$ divide $b^{\prime }$.

The class $d$ is realized by a nonexceptional rational point precisely when its quartic covering in a two-isogeny descenthas a rational point with $UV\ne 0$. Indeed substitution $x=d(U/V{)}^2$, $y=dUN/V^3$ gives the original equation, and conversely the square-class condition gives these parameters. Clearing denominators permits $U,V,N$ integral with $\gcd (U,V)=1$. The zero-coordinate cases account for the already known identity and two-torsion classes.

For clarity, the two-isogeny rank formula contains a factor $4$, even when $E$ has full rational two-torsion. Put $t=\#E(\mathbb{Q})[2]$, which is $2$ or $4$, and let $r$ be the rank. From the two kernel formulas and $2E=\hat{\varphi }\varphi E$,Indeed the kernel of the induced map $E^{\prime }/\varphi E\to \hat{\varphi }{E}^{\prime }/2E$ is $\ker \hat{\varphi }/(\ker \hat{\varphi }\cap \varphi E)$. The intersection is $\varphi (E[2])$, of size $t/2$, so $\delta =4/t$. The Mordell-Weil theorem gives $[E:2E]=2^{r}t$. Therefore

This gives a practical rank procedure. Enumerate the finitely many signed squarefree divisors of $b,b^{\prime }$, and test their quartics over the reals and local fields at $2$ and primes dividing the relevant coefficients or elliptic-curve discriminant. These necessary local conditions give finite Selmer group of an elliptic curve upper bounds. Search the remaining coverings for rational points, or search the curves directly, to realize square classes and obtain lower bounds. One can also exhibit independent points and check independence by the canonical height pairing. When the upper and lower bounds agree, the rank is determined. Local solubility alone does not guarantee a rational point, so a persistent gap can reflect a nontrivial Tate–Shafarevich group rather than a missing congruence test.

Now take $a=0$, $b=-p^2$ for an odd prime $p$. The curve has four rational two-torsion points $O,(0,0),(p,0),(-p,0)$. Their images under $\alpha$ are $1,-1,p,-p$. The prime-support bound allows no other classes, soIts partner is $E_p^{\prime }:Y^2=X^3+4p^{2}X$. Any nonidentity rational point with $X\ne 0$ has $X>0$, because $X^2+4p^2>0$. Also the exceptional point has square class $4p^2=1$. Thus its image is contained in the four positive candidate classes $\{1,2,p,2p\}$. The rank bound for a prime congruent-number elliptic curve follows:

For an exact rank-one example, take $p=5$. The point $(X,Y)=(5,25)$ on $E_5^{\prime }:Y^2=X^3+100X$ realizes the class $5$. The five-adic obstructions for the congruent-number isogeny covers exclude the other two possible nontrivial classes. For $d=2$, the quartic isModulo $5$, if $5\nmid U$ the right side is $2$, a nonsquare. Hence $U=5U_1$ and $N=5N_1$; primitivity gives $5\nmid V$. Dividing by $25$ then gives $N_1^2=50U_1^4+2V^4\equiv 2(\mathrm{mod}5)$, again impossible. For $d=10$, the quartic is $N^2=10(U^4+V^4)$. At least one of $U,V$ is nonzero modulo $5$, and their fourth powers are $0$ or $1$, so their sum is $1$ or $2$ modulo $5$. The right side has five-adic valuation $1$, impossible for a square. Thus ${\alpha }^{\prime }(E_5^{\prime })=\{1,5\}$ exactly, andThe dual isogeny maps $(5,25)$ to $(25/4,-75/8)$ on $E_5$, giving an explicit rational point associated with this rank-one computation.

For each finite place $v\in S$, choose $d_v\in K_v^{\times }$ such that $K_v(\sqrt{d_v})/K_v$ is the unramified quadratic extension. Such a choice exists even at residue characteristic two. At odd residue characteristic use a unit with nonsquare residue. At residue characteristic two, choose $\overline{c}$ for which $X^2-X-\overline{c}$ is irreducible over the finite residue field, lift $c$ to ${\mathcal{O}}_v$, and set $d_v=1+4c$. The lifted polynomial has unit derivative at every residue point and gives the unramified degree-two extension; its root is $(1+\sqrt{d_v})/2$.