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For $D>0$, a nontorsion rational point on the congruent number elliptic curve $y^2=x^3-D^{2}x$ gives infinitely many positive rational right triangles of area $D$. Apply the formulas $a=|(x^2-D^2)/y|$, $b=|2Dx/y|$, $c=|(x^2+D^2)/y|$ to its multiples. The map is finite-to-one: if $r=c/b$ is fixed, then $|x|$ satisfies $X^2-2DrX+D^2=0$, leaving only finitely many coordinate signs. Thus infinitely many points give infinitely many distinct triangles.

A Diophantine equation is soluble at a place of a number field if it has a solution over the corresponding completion of a valued field. It is locally soluble everywhere if this holds at all finite and infinite places. Over the rational numbers, these are the real field and all p-adic fields. Local solubility is necessary for a rational point, but the local-to-global gap in isogeny descent shows it need not suffice.

Coverings surviving all local solubility tests need not have rational points. The quotient of an isogeny Selmer group by the rational-point image is a kernel in the Tate–Shafarevich group. A nonzero class therefore passes every local test but contributes no rational point to a search. This explains why two-isogeny descent can leave a gap between a certified upper bound and found lower bound for the rank of an elliptic curve. Higher descents address finer obstructions; see Tom Fisher's paper on higher descents with rational two-torsion.

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 $p$.

Let $O=(0:1:0)$ be the point at infinity of the Weierstrass equation of an elliptic curve. It is the identity element for the elliptic curve group law. For an affine rational point $P=(x,y)$, the inverse isIndeed, the two affine intersections with the vertical line have these ordinates, since their sum as roots of the quadratic in $y$ is $-a_{1}x-a_3$. This reflection is valid for a generalized Weierstrass equation of an elliptic curve in every characteristic of a field.

For distinct $P,Q$, draw their line; for $P=Q$, draw the tangent line. Let $R$ be its third intersection with the elliptic curve, counted with intersection multiplicity. Then the sum is the reflection of the third intersection:A vertical line has third intersection $O$, so $P+(-P)=O$. If a tangent has third intersection $O$, then $2P=O$. Define addition with $O$ by $P+O=P$; these conventions include all exceptional cases of the affine construction.

The construction is defined over the coefficient field: after two known intersections, the remaining root is rational over that field. It gives a commutative group operation. To see why it is associative, use the elliptic curve group law from Riemann-Roch: the map $P\mapsto [P-O]$ identifies the elliptic curve with its degree-zero divisor class group. A line has intersection divisor $P+Q+R$ linearly equivalent to $3O$, the divisor of the line at infinity. Thus $[P-O]+[Q-O]+[R-O]=0$. The geometric sum agrees with addition of divisor classes, which is associative.

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.

Choose an integral Minimal Weierstrass equation over ${\mathbb{Z}}_p$. Let $E_0({\mathbb{Q}}_p)$ be the points with nonsingular reduction, and let $E_1({\mathbb{Q}}_p)$ be those reducing to $O$. This distinction matters at bad reduction of an elliptic curve: the whole singular reduced cubic is not a group. The smooth locus is a group under the induced elliptic curve group law, and Hensel's lemma lifts each of its rational points, givingThe formal kernel of a minimal Weierstrass equation identifies $E_1({\mathbb{Q}}_p)$ with $p{\mathbb{Z}}_p$ equipped with $F$. Indeed, the integral series $w(t)$ converges there, and $x=t/w(t)$, $y=-1/w(t)$ give the unique point for each nonzero parameter. If $v_p(t)=r>0$, then $v_p(x)=-2r$ and $v_p(y)=-3r$; the parameter zero gives $O$. Conversely a point reducing to $O$ has these local coordinates of positive valuation, so is recovered by the series.

Define $E_m$ for $m\ge 1$ by $t\in p^m{\mathbb{Z}}_p$. The linear part of the formal group law is addition, and the higher terms have degree at least two. Hence the filtration of elliptic-curve points over a local field hasAt good reduction, $E_0=E$ and $E/E_1\cong \tilde{E}({\mathbb{F}}_p)$. At bad reduction, there is also the finite quotient $E/E_0$: $E({\mathbb{Q}}_p)$ is a compact group because the curve is projective, and $E_0$ is an open subgroup, so its index of a subgroup is finite. Together these quotients and the formal filtration describe the group, including the extra components at bad primes.

The formal logarithm makes the deepest part explicit. Write $L(T)=T+{\sum }_{n\ge 2}{c}_n{T}^n$. Since its derivative is the integral invariant differential of a formal group law, $n{c}_n\in {\mathbb{Z}}_p$. ConsequentlyThe inequality $v_p(n)\le (n-1)/(p-1)$ shows that for an integer $m>1/(p-1)$ every nonlinear term on $p^m{\mathbb{Z}}_p$ has strictly higher valuation than the linear term. After scaling, $L$ is the identity plus a strictly contracting series; solving $L(t)=u$ by successive correction gives its inverse formal group exponential. Thus the deep logarithm subgroup of a formal group givesOne may take $m=1$ for odd $p$ and $m=2$ for $p=2$. In particular, $E({\mathbb{Q}}_p)$ contains an open subgroup isomorphic to the additive p-adic integers, and the remaining information is in finite quotients and their group extensions.

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.

The obstruction is the local-to-global gap in isogeny descent. A two-isogeny descent quartic can have points over $\mathbb{R}$ and every p-adic field without a rational point. Local tests compute an isogeny Selmer group, which can strictly contain the rational-point quotient. For an isogeny of elliptic curves $\varphi :E\to E^{\prime }$, the comparison isThe last group is the kernel of $\varphi :\mathrm{Sha}(E/\mathbb{Q})\to \mathrm{Sha}(E^{\prime }/\mathbb{Q})$, so it records elements of the Tate–Shafarevich group invisible to all local tests. Searching a locally soluble covering which has no rational points never succeeds. Thus a Selmer group of an elliptic curve gives an upper bound and found independent points give a lower bound, and these need not coincide. Higher descent can improve the bounds, but this particular procedure has no unconditional guarantee of closing every gap.

The rational point $P=(0,1)$ lies on $E$ and is neither the identity nor $(-1,0)$. By solution these are all the rational torsion points of an elliptic curve. Hence $P$ has infinite order and $E(\mathbb{Q})$ is infinite: its multiples $nP$ are distinct. No assertion about its being a generator or about the exact rank of an elliptic curve is needed here.

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.

Over ${\mathbb{F}}_2$, both $x=0,1$ give $x^3-x=0$, and both $y=0,1$ give $y^2+y=0$. There are four affine rational points, together with the point $O$ at infinity. Over ${\mathbb{F}}_3$, each $x=0,1,2$ gives $x^3-x=0$, while precisely $y=0,2$ solve $y^2+y=0$. ThusBoth reductions are nonsingular by the discriminant calculation, so these are the point counts of the reduced elliptic curves, with their group laws.

Retain $E:y^2=x(x^2+ax+b)$ and put $b^{\prime }=a^2-4b$. The explicit two-isogeny formula iswith the maps extended at the kernel points. Direct substitution verifies the equations and $\psi \varphi =[2]$.

Here is a concrete kernel test for the descent homomorphisms. A preimage of $(X,Y)\in E^{\prime }$ has $x$ satisfying $x^2+(a-X)x+b=0$, whose discriminant is $(X-a{)}^2-4b$. The target equation says $Y^2=X((X-a{)}^2-4b)$, so for nonexceptional points this quadratic has a rational solution precisely when $X$ is a square. The resulting $y$ follows from the displayed elliptic isogeny formula. For a nonzero target two-torsion point, the preimage quadratic has a double root $x=(X-a)/2$, and the source equation gives $y^2=x^{2}X$, so the same square condition still applies. At $(0,0)\in E^{\prime }$, a rational preimage exists precisely when $b^{\prime }$ is a square, matching the special square-class value ${\alpha }^{\prime }((0,0))=[b^{\prime }]$. The dual argument gives the other kernel. ThusLet $\delta =[\ker \psi :\ker \psi \cap \varphi E(\mathbb{Q})]$. The quotient map induced by $\psi$ has this kernel, givingIf $b^{\prime }$ is a rational square, $E$ has four rational two-torsion points and $\delta =1$; otherwise it has two and $\delta =2$. In either case $\delta \#E(\mathbb{Q})[2]=4$. Combining with finite generation therefore proves the two-isogeny rank formulaThis derivation checks the factor four, including curves with full rational two-torsion.

To compute these images, enumerate the signed square-free divisors $d$ of $b$ and $b^{\prime }$. Substituting $x=d(U/V{)}^2$ yields the quartic covering in a two-isogeny descentThe class $d$ occurs precisely when this covering has a rational point, with the identity and two-torsion handled by the boundary cases. Test at the real place and at primes dividing $2b{b}^{\prime }$ to rule out candidates, and search for rational points to prove that surviving classes occur. The locally soluble classes form a finite elliptic isogeny Selmer group of an elliptic curve, giving an upper rank bound. rational points and independent height pairings give a lower bound. When the bounds agree the rank is proved. Local solubility alone does not prove a rational point exists: if a gap remains, a higher descent or further information about the covering obstructions is needed. This is why the procedure usually computes the rank but is not an unjustified universal termination claim.

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

We can use reduction instead of expanding a very large rational addition formula. Modulo $2$, both $P_1$ and $P_2$ reduce to $(0,0)$, which generates the three-element reduced group. The reduction homomorphism therefore givesThe rational point $P_2$ is not any of the three torsion points just found, so it has infinite order. In particular $R=P_1+5P_2\ne O$: otherwise multiplication by three would give $15P_2=O$.

If $R$ had integral coordinates, its affine integral representative would reduce to an affine point modulo $2$, not to the point at infinity. This contradicts its reduction to $O$. ThusIndeed, it is not even integral at $2$. This is a reduction certificate for a nonintegral elliptic point, with the crucial nonidentity condition verified by the torsion calculation.

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.

The Hessian criterion for flexes of a plane cubic applied to the defining polynomial givesConsequently the inflection points of a plane cubic on this curve are exactly its intersections with the three coordinate lines. Choose $r\in \overline{\mathbb{Q}}$ with $r^3=d$, and a primitive cube root of unity $\zeta$. The nine distinct flexes of a diagonal cubic areWith the chosen flex as identity, these points are precisely $E_d[3]$. A tangent cutting the Weil divisor $3P$ gives $3(P-O)=0$; conversely $3(P-O)=0$ makes the hyperplane Weil divisor equivalent to $3P$, which supplies a tangent with triple contact. This is the three-torsion points are flexes of a plane cubic correspondence.

On $Z=0$ only $O$ is rational, since the other cube roots of unity are not rational. On each other coordinate line a rational point exists exactly when $d$ is a rational cube, and then exactly one does. ThereforeIn particular the rational torsion subgroup cannot contain two independent points of order $3$.

One can also determine the whole rational torsion of a diagonal cubic. First it is killed by $6$. To see this, use the Weierstrass equation of an elliptic curve from (iii). At primes $\ell \equiv 2(\mathrm{mod}3)$ outside the finite set containing $2,3$ and the prime divisors of the numerator and denominator of $d$, the cube map permutes ${\mathbb{F}}_{\ell }$, sowhere the quadratic character is extended by $\chi (0)=0$, and the character sum vanishes after substituting $t=x^3-432d^2$. Reduction of torsion points on an elliptic curve injects the prime-to-$\ell$ part of rational torsion into this finite group. This injection follows because multiplication by an integer prime to $\ell$ is invertible on the formal group of an elliptic curve, and hence its reduction kernel has no such torsion.

For any prime $q>3$, choose a good $\ell \equiv 2(\mathrm{mod}3)$ with $\ell \equiv 1(\mathrm{mod}q)$; the Dirichlet theorem on primes in arithmetic progressions and the Chinese remainder theorem supply such primes outside the finite bad set. Then $q\nmid \ell +1$, excluding $q$-torsion. Similarly good $\ell \equiv 5(\mathrm{mod}12)$ exclude order $4$, and good $\ell \equiv 11(\mathrm{mod}18)$ exclude order $9$. Thus every rational torsion point has order dividing $6$.

The inversion formula in (i) shows that the only possible nonidentity rational point of order $2$ has $X=Y$ and $Z\ne 0$. It exists precisely when $(X/Z{)}^3=-d/2$, that is, when $d/2$ is a rational cube. There is at most one such point. This condition and $d$ being a cube cannot hold together, since $2$ is not a rational cube. Combining it with the three-torsion calculation gives the full answer:

Set $d=-7$. The affine rational pair $(u,v)=(2,-1)$ gives $P=(84,756)$ on the elliptic curve $y^2=x^3-21168$. It is a point of infinite order: in the rational torsion of a diagonal cubic classification, neither $-7$ nor $-7/2$ is a rational cube, so this curve has trivial rational torsion subgroup.

There is also a short direct torsion check specific to this curve. The primes $5$ and $11$ are good and have point counts $6$ and $12$, respectively, because cubing is bijective in their finite fields. Reduction at these two primes bounds the entire rational torsion by order dividing $6$, including its possible $5$- and $11$-primary parts by reduction at the other prime. The flex calculation excludes order $3$, and the inversion calculation excludes order $2$. Hence $P$ is indeed nontorsion.

Every nonzero multiple $nP$ is distinct, and its inverse projective image has $Z\ne 0$: the only rational point at infinity on the diagonal cubic is its identity. Thus each yields a distinct rational pair. More explicitly, if $nP=(x_n,y_n)$ on the Weierstrass model, thenHere $x_n\ne 0$, since $x_n=0$ would require a rational square equal to $-21168$. For example the elliptic-curve addition formula gives $2P=(28,28)$, which produces $(u_2,v_2)=(5/3,4/3)$. This proves there are infinitely many rational pairs of cubes summing to seven.

An isogeny of elliptic curves $\varphi :E_1\to E_2$ over a field $K$ is a nonconstant morphism preserving the identity and addition. It is surjective and finite: its image is a closed connected subgroup, and a nonconstant morphism of smooth projective curves is finite. The degree of an isogeny is the degree of its induced extension of function fields. A separable isogeny has degree equal to the number of geometric kernel points; in general the degree is the length of the finite kernel scheme, incorporating inseparability. The degree of an isogeny multiplies under composition, and degree one means an isomorphism. In characteristic zero all elliptic isogenies are separable. In positive characteristic the Frobenius isogeny illustrates inseparability, while an elliptic isogeny of degree prime to the characteristic is separable.

The homomorphism group of elliptic curves includes the zero homomorphism, the constant identity map, and every nonzero member is an elliptic isogeny. The operations are pointwise addition and negation; for endomorphisms, composition also gives the endomorphism ring of an elliptic curve. We put $q(0)=0$ and $q(\varphi )=\mathrm{deg}\varphi$ otherwise. We now prove the positive definiteness of degree on elliptic-curve homomorphisms in full.

Let $E=E_2$, and choose a Weierstrass model. Its coordinate $x$ has a double pole at $O$, and the generic fibre of $x$ consists of $P$ and $-P$. On $E\times E$ the rational function $x(P)-x(Q)$ therefore has Weil divisorwhere $\Delta =\{P=Q\}$, ${\Delta }_{-}=\{P=-Q\}$, $V=\{O\}\times E$, and $H=E\times \{O\}$. The generic zeros on the two diagonals have multiplicity one, and the generic poles on $V,H$ have multiplicity two; their intersections have codimension two and add no Weil divisor components. This argument also uses the general Weierstrass involution in characteristic $2$, where $-P$ need not mean changing just the sign of $y$.

For the sum and difference maps $a(P,Q)=P+Q$ and $s(P,Q)=P-Q$, the divisors $a^{\ast }(O)$ and $s^{\ast }(O)$ are ${\Delta }_{-}$ and $\Delta$. Consequently, for the degree-one line bundle $L={\mathcal{O}}_E(O)$,Pull back along $P\mapsto (\varphi (P),\psi (P))$ and take degrees of line bundles on $E_1$. A nonconstant map pulls back $L$ to degree equal to its degree, while a constant map pulls it back to a trivial bundle of degree zero. We obtain the degree parallelogram lawUsing the line-bundle identity is important: it still restricts correctly when $\varphi =\psi$ or $\varphi =-\psi$, when restriction of the displayed rational function would be identically zero.

Since $q(-\varphi )=q(\varphi )$, applying the identity to $n\varphi$ and $\varphi$ gives the second-difference recursionStarting from $q(0)=0$ and $q(\varphi )$ yields $q(n\varphi )=n^{2}q(\varphi )$ for all integers $n$. DefineIt is symmetric and odd in each variable. Applying the parallelogram identity to the four terms in its first expression givesInterchanging $u,w$ and using oddness also gives $B(u+w,v)-B(u-w,v)=2B(w,v)$. Adding proves $B(u+w,v)=B(u,v)+B(w,v)$; symmetry proves additivity in the other variable. Thus $q$ is a quadratic form with associated rational bilinear form $B$.

Every nonzero homomorphism has positive degree, so $q(\varphi )>0$ for $\varphi \ne 0$. The scaling identity also shows that the homomorphism group is a torsion-free group. Hence $q$ extends to its rational vector space. On the span of any finitely many rationally independent homomorphisms, the matrix of $B$ is rational. Positivity on nonzero rational vectors implies nonnegativity on real vectors by density. If the real matrix had a nontrivial kernel, its rational linear equations would have a nonzero rational solution, contradicting positivity. Thus its real extension is a positive-definite quadratic form. This establishes the requested positivity as well as the quadratic identities.

An important companion to degree is the dual isogeny. Under $E\cong {\mathrm{Pic}}^0(E)$ via $P\mapsto [P-O]$, the pullback ${\varphi }^{\ast }:{\mathrm{Pic}}^0(E_2)\to {\mathrm{Pic}}^0(E_1)$ defines $\hat{\varphi }:E_2\to E_1$. Divisor pushforward corresponds to $\varphi$, and pushforward after pullback multiplies divisor classes by $n=\mathrm{deg}\varphi$. Therefore $\varphi \hat{\varphi }=[n]$. Composing shows that $\hat{\varphi }\varphi -[n]$ takes values in the finite kernel of $\varphi$; a morphism from a connected smooth curve into this finite scheme must be constant, and its identity value is zero. ThusThe last equality follows from $\mathrm{deg}[n]=n^2$, already proved by applying $q(m\varphi )=m^{2}q(\varphi )$ to the identity map. The dual isogeny is unique, since the difference of two choices composed with $\varphi$ would vanish and $\varphi$ is surjective. It follows that being isogenous is an equivalence relation: duality supplies symmetry, and composition supplies transitivity. For a finite subgroup of geometric points of order prime to the characteristic, the quotient curve and its quotient map give an elliptic isogeny with that kernel; in general one uses finite subgroup schemes. This is the geometric source of explicit descent maps, including the two-isogeny formula in Question 4.

Finally elliptic isogenies preserve the rational rank of an abelian group over a number field. Their dual compositions are multiplication by $n$, so after tensoring the rational point groups with $\mathbb{Q}$ the elliptic isogeny becomes invertible. For a single curve, its endomorphism algebra in characteristic zero is $\mathbb{Q}$ or an imaginary quadratic field, giving ordinary integer multiplication or complex multiplication. In positive characteristic a supersingular curve can instead have a quaternionic endomorphism algebra. The positive degree form controls the sizes of these homomorphisms, while duality makes their arithmetic and geometric properties accessible.

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 $\varphi \hat{\varphi }=[2]$.