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Move the nonidentity rational 2-torsion point to $T=(0,0)$ and complete the square. After a rational change of coordinates, the elliptic curve has an integral modelThe two-isogeny formula gives a partnerand the isogeny of elliptic curves and its dual isogeny areBoth extend at the exceptional points; their kernels are the identities and the distinguished 2-torsion points, and $\hat{\varphi }\varphi =[2]$.

The two-torsion square-class homomorphism is $\alpha (O)=1$, $\alpha (T)=[b]$, and $\alpha (x,y)=[x]$ otherwise, with values in ${\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$. Its kernel is $\hat{\varphi }{E}^{\prime }(\mathbb{Q})$. Define ${\alpha }^{\prime }$ on the partner in the same way. By the prime-support bound in two-isogeny descent, representatives are signed squarefree divisors $d$ of $b$, or of $b^{\prime }$ for the partner. A class is realized precisely when its quartic covering in a two-isogeny descent has a rational solution:For $UV\ne 0$, this comes from $x=d(U/V{)}^2$ and $y=dUN/V^3$; the cases $U=0$ or $V=0$ account for the distinguished point and the identity. Test this finite list using signs and local congruences, and exhibit rational solutions for the surviving classes. The local-to-global gap in isogeny descent matters: local solubility gives an upper bound from the isogeny Selmer group, so a rank determination requires matching actual classes or further descent, not merely declaring every locally soluble cover to have a rational point.

Once the two actual images are determined, the two-isogeny rank formula isFor completeness, put $A=E(\mathbb{Q})$ and $B=E^{\prime }(\mathbb{Q})$. The Mordell-Weil theorem gives $[A:2A]=2^r|A[2]|$. The index of a subgroup formula through $\hat{\varphi }B$ givesHere $\ker \hat{\varphi }\cap \varphi A=\varphi (A[2])$, since $\hat{\varphi }\varphi =[2]$, and this image has size $|A[2]|/2$. The denominator is therefore $4/|A[2]|$, proving the displayed rank of an elliptic curve formula.

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.

The Mordell-Weil theorem gives $E(\mathbb{Q})\cong E(\mathbb{Q}{)}_{\mathrm{tors}}\oplus {\mathbb{Z}}^r$. Move the rational point of order two to $(0,0)$ and clear denominators to obtain an integral Weierstrass equation of an elliptic curveThe two-isogeny formula gives $E^{\prime }:Y^2=X(X^2-2aX+a^2-4b)$ and dual isogenies $\varphi :E\to E^{\prime }$, $\hat{\varphi }:E^{\prime }\to E$ with $\hat{\varphi }\varphi =[2]$. Compute the torsion subgroup using reduction of torsion points on an elliptic curve at several good primes, followed by division polynomials of an elliptic curve to identify the possible points.

For the free part, use two-isogeny descent. The two-torsion square-class homomorphism isin ${\mathbb{Q}}^{\ast }/{\mathbb{Q}}^{\ast 2}$, with kernel $\hat{\varphi }{E}^{\prime }(\mathbb{Q})$; define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly. The prime-support bound in two-isogeny descent restricts the first image to signed square-free divisors of $b$ and the second to those of $a^2-4b$. For each candidate $d$, solve its two-isogeny descent quarticTest local solubility at the real place and the relevant p-adic fields, discard impossible classes, and search the surviving coverings for rational points. A point with $UV\ne 0$ gives $x=d(U/V{)}^2$, $y=dUN/V^3$. Include the identity and rational 2-torsion cases separately. Once both rational images are determined, the two-isogeny rank formula givesThe point searches give explicit candidates for a basis. Their canonical height pairing can certify independence; Mordell-Weil saturation and effective height bounds then certify that they generate the full free part. Rank alone does not certify generators of the whole Mordell-Weil group. If the rational images are not fully determined, the procedure instead supplies lower and upper bounds on the rank of an elliptic curve.

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.

For a rank-zero example, take $E_0:y^2=x^3+x$, with rational two-torsion $(0,0)$. Its two-isogeny formula companion is $E_0^{\prime }:y^2=x^3-4x$. The prime-support bound in two-isogeny descent allows only $\pm 1$ on $E_0$; all nonzero affine $x$ are positive because $x(x^2+1)$ is a square. Hence $A_0=\{1\}$. For $E_0^{\prime }$ the possible classes are $\{1,-1,2,-2\}$, all realized: $O$ gives $1$, $(0,0)$ gives $[-4]=-1$, and $(2,0),(-2,0)$ give $2,-2$. ThereforeThis is the rank-zero elliptic curve with cubic x cubed plus x certificate; no unresolved locally soluble covering is being counted as soluble.

For a positive-rank example, take $E_1:y^2=x^3-2x$. Its companion is $E_1^{\prime }:y^2=x^3+8x$. The first image is contained in $\{1,-1,2,-2\}$. The identity gives $1$, $(0,0)$ gives $-2$, and $(2,2)$ gives $2$; multiplicativity of the square-class group homomorphism also gives $-1$. Thus $A_1$ has all four classes. On $E_1^{\prime }$, the same prime-support bound allows $\pm 1,\pm 2$, but $x(x^2+8)$ is nonnegative only for $x\ge 0$, eliminating the negative classes. The identity gives $1$ and $(0,0)$ gives $[8]=2$, so $B_1=\{1,2\}$. ConsequentlyThe rank-one elliptic curve with cubic x cubed minus twice x therefore supplies the required positive-rank example. The exhibited point $(2,2)$ is itself nontorsion: doubling gives $(9/4,-21/8)$, whose nonintegral coordinates contradict the Nagell–Lutz theorem if it were torsion. No assertion that this point generates the full free part is needed for the rank calculation.

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

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.

We first determine the torsion. The rational 2-torsion points are $O,(0,0),(5,0),(-5,0)$. At the good primes three and seven, part (b) gives group orders four and eight. Prime-to-residue-characteristic torsion injects under reduction: an element in the kernel has a formal parameter $t$ of positive valuation, and multiplication by an integer prime to the residue characteristic has series $mt+O(t^2)$ with unit leading coefficient, so cannot kill a nonzero such parameter. At seven this excludes rational 3-primary torsion; at three it excludes 7-primary torsion and every odd-primary component other than the already excluded 3-primary component. Thus all torsion is 2-primary and injects at three, bounding its order by four. Consequently

For the rank, use two-isogeny descent withThe two-isogeny formula givesThey extend at the exceptional points, have kernels $\{O,(0,0)\}$ on their respective curves, and satisfy $\psi \varphi =[2]$. These assertions can be checked by substitution; for example the composed x-coordinate is $(x^2+25{)}^2/[4x(x^2-25)]$, the duplication x-coordinate. The maps respect addition because they are morphisms fixing $O$: pushforward of divisor classes is additive, takes principal divisors to principal divisors by the norm of a rational function, and under the identification in part 1(a) sends $[(P)-(O)]$ to $[(\varphi P)-(O)]$, and likewise for $\psi$.

Define the two-torsion square-class homomorphism $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\times }/{\mathbb{Q}}^{\times 2}$ by $\alpha (x,y)=[x]$, $\alpha (O)=1$, and $\alpha ((0,0))=[-25]=[-1]$. Define ${\alpha }^{\prime }$ on $E^{\prime }$ similarly, with ${\alpha }^{\prime }((0,0))=[100]=1$. To see the homomorphism property, a nonvertical line $y=mx+c$ cuts either curve in three points whose x-coordinates have product $c^2$. The chord-and-tangent law therefore makes their square classes multiply to one. When the line passes through $(0,0)$, the two remaining x-coordinates have product equal to the coefficient of $x$, exactly explaining the assigned value there. Vertical lines give the inverse-point relation. The same argument includes repeated tangent intersections.

The kernels are $\ker \alpha =\psi E^{\prime }(\mathbb{Q})$ and $\ker {\alpha }^{\prime }=\varphi E(\mathbb{Q})$. Here is an explicit check, not just an invocation of the descent correspondence. The x-coordinate of every nonexceptional $\psi$-image is a square. Conversely, if $(x,y)\in E$ has $x=s^2\ne 0$, put $X=2x+2y/s$ and $Y=2sX$. Then $X^2-4xX+100=0$, so $(X,Y)\in E^{\prime }$ and the displayed $\psi$ maps it to $(x,y)$. For $E^{\prime }$, every nonexceptional $\varphi$-image has $X=(y/x{)}^2$; if $X=s^2\ne 0$, put $x=(X+Y/s)/2$, $y=sx$. Substitution gives a point of $E$ mapping to $(X,Y)$. The point $(0,0)\in E^{\prime }$ has square class one and equals $\varphi (5,0)$; the point $(0,0)\in E$ has nonsquare class $-1$. Together with the origins these checks handle every exception.

The prime-support bound in two-isogeny descent can be read directly from $y^2=x(x^2+B)$. At a prime not dividing the integral coefficient $B$, positive valuation of $x$ gives $2v(y)=v(x)$, while negative valuation gives $2v(y)=3v(x)$; either way the valuation of $x$ is even. Thus the first image has only the candidate classes $1,-1,5,-5$. All four occur, already at the four torsion points. Hence $\#\alpha (E)=4$.

On $E^{\prime }$, nonzero real x-coordinates are positive since $X(X^2+100)$ has the sign of $X$. Prime support leaves only $1,2,5,10$. The point $(5,25)$ supplies class five. Class two would give the quartic covering in a two-isogeny descentIndeed write $X=2(U/V{)}^2$ and clear the square factor in the curve equation. If $5\nmid U$, the right side is two modulo five, impossible for a square. If $5\mid U$, then $5\nmid V$ and $5\mid W$; division by twenty-five gives $(W/5{)}^2\equiv 2V^4\equiv 2(\mathrm{mod}5)$, again impossible. This is the five-adic obstruction to square class two on the curve with coefficient one hundred. Because the image is a subgroup and contains five, class ten is excluded as well. Therefore ${\alpha }^{\prime }(E^{\prime })=\{1,5\}$ has size two.

Now $[E:\psi E^{\prime }]=4$. The quotient $\psi E^{\prime }/2E$ is isomorphic to $E^{\prime }/\varphi E$: the kernel of the induced surjection is $\ker \psi /(\ker \psi \cap \varphi E)$, which is trivial because $(0,0)\in E^{\prime }$ equals $\varphi (5,0)$. Thus $[E:2E]=4\cdot 2=8$. Finite generation follows from the Mordell-Weil theorem, whose height-descent proof is supplied in the essay below. A finitely generated group with the torsion already found has $[E:2E]=2^{r+2}$, so $r=1$. HenceThe point $(-4,6)$ has infinite order, since it is not one of the torsion points. No assertion that this particular point generates the free factor is needed.

The elliptic-curve discriminant of $y^2=x^3-12x^2+20x$ is $16\cdot 20^2\cdot (144-80)=409600=2^{14}{5}^2$. Thus three and eleven are primes of good reduction. Counting the reduced points gives four points over ${\mathbb{F}}_3$ and sixteen over ${\mathbb{F}}_{11}$. For example, the numbers of affine points above successive $x=0,1,2$ modulo three are $1,1,1$. Modulo eleven, for $x=0,\dots ,10$, they are $1,2,1,2,0,0,2,2,2,2,1$; add the point at infinity in each case.

The reduction of torsion points on an elliptic curve is injective on torsion prime to the residue characteristic. The three-primary rational torsion injects into the group of order sixteen at eleven, so is trivial; likewise the eleven-primary torsion injects into the group of order four at three. Every other primary component injects at both primes. Therefore the rational torsion subgroup has order dividing four. All four rational 2-torsion points are visible, so

The rational point $P=(1,3)$ lies on the curve. The tangent slope in the elliptic-curve addition formula is $(3-24+20)/6=-1/6$, giving $2P=(361/36,-323/216)$. The integral substitution $X=x-4$ changes the equation to $y^2=X^3-28X-48$; the doubled point has $X=217/36$. The Nagell–Lutz theorem says that a rational torsion point on this integral short equation has integral coordinates, so $2P$, and hence $P$, is nontorsion. An infinite-order point is $(1,3)$.

We prove the rank bound by two-isogeny descent. For $a=-12,b=20$, the two-isogeny formula gives the partnerand maps $\varphi :E\to E^{\prime }$, $\psi :E^{\prime }\to E$ with $\psi \varphi =[2]$. The two-torsion square-class homomorphism $\alpha :E(\mathbb{Q})\to {\mathbb{Q}}^{\ast }/({\mathbb{Q}}^{\ast }{)}^2$ is $\alpha (x,y)=[x]$ away from $O,(0,0)$, with values $1$ and $[20]=[5]$ at those exceptional points. Its kernel is $\psi E^{\prime }(\mathbb{Q})$. The corresponding map ${\alpha }^{\prime }$ on $E^{\prime }$ has kernel $\varphi E(\mathbb{Q})$.

The prime-support bound in two-isogeny descent can be checked directly here. For a prime $\ell \nmid 20$, if $v_{\ell }(x)>0$ then $x^2-12x+20$ is a unit, so $2v_{\ell }(y)=v_{\ell }(x)$ is even; if $v_{\ell }(x)<0$, the $x^3$ term has strictly least valuation and $2v_{\ell }(y)=3v_{\ell }(x)$ is even. Thus the square class of $x$ is supported only on two and five. Also $x(x-2)(x-10)\ge 0$ forces $x\in [0,2]\cup [10,\infty )$, so nonzero $x$ is positive. The possible classes are $1,2,5,10$, and the four torsion points provide all of them:Consequently $\#\alpha (E(\mathbb{Q}))=4$.

The same valuation argument on $E^{\prime }$ gives possible signed classes $1,-1,2,-2$, since its coefficient $b^{\prime }=64$ has only the prime two. The identity gives class $1$, and $(-4,8)\in E^{\prime }(\mathbb{Q})$ gives class $-1$. For $d=\pm 2$, the two-isogeny descent quartic iswith integral coprime $U,V$ after scaling a rational solution. If $U$ is odd, its right side is $2$ or $10$ modulo sixteen for $d=2$, and $14$ or $6$ for $d=-2$, according to the parity of $V$. None is a square. If $U=2u$ is even, $V$ is odd and the right side is $32(\pm u^4+3u^2{V}^2\pm V^4)$. The bracket is odd for either parity of $u$, so its 2-adic valuation is five, again impossible for a square. The signed divisor-two obstruction for an isogeny covering excludes both classes. Hence ${\alpha }^{\prime }(E^{\prime }(\mathbb{Q}))=\{1,-1\}$.

For completeness, the index correction in the two-isogeny index formula over a number field is one: $\ker \psi =\{O,(0,0)\}$ and $\varphi ((2,0))=(0,0)$, so the whole kernel lies in $\varphi E(\mathbb{Q})$. ThusThe Mordell-Weil theorem and the four rational 2-torsion points identify this index as $2^{r+2}$. Therefore $2^{r+2}=8$ andThis is the rank-one elliptic curve with roots zero two and ten. We have exhibited a point of infinite order but have not needed to prove that it generates the free part.

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 $O,(0,0),(p,0),(-p,0)$.