Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 125 (Elliptic Curves)

Put $O=0_E$. Choose rational projective coordinates in which $O=(0:1:0)$ and its tangent is $Z=0$. Because $O$ is an inflection point of a plane cubic, that tangent intersects the cubic in the divisor $3O$. Consequently its cubic equation has the formHere $a\ne 0$, since the restriction to $Z=0$ is a nonzero multiple of $X^3$, and $b\ne 0$, since smoothness at $O$ requires a nonzero $Z$-derivative there. Completing the square in $Y$, by the rational linear change $Y^{\prime }=Y+(cX+eZ)/(2b)$, gives on $Z=1$Set $x^{\prime }=c_{0}x$ and $y^{\prime \prime }=c_0{y}^{\prime }$. ThenThese are invertible rational projective coordinate changes, and $O$ remains the unique point at infinity. Smoothness ensures that the resulting monic cubic has no repeated root. Thus this is a Weierstrass equation of an elliptic curve, without requiring any extraction of a square or cube root in $\mathbb{Q}$.

On this model write $-(x,y)=(x,-y)$ and $-O=O$. To add $P,Q$, take their joining line, or the tangent if $P=Q$, and let $R$ be the third intersection counted with multiplicity. DefineFor a vertical line its third intersection is $O$, so $P+(-P)=O$. The line through $O$ and an affine point is vertical, giving $P+O=P$. The tangent at $O$ has triple contact, giving $O+O=O$. This is the chord-and-tangent group law.

If $P,Q$ are rational, their joining line or tangent is rational. Its residual intersection point is rational: substituting the line in the cubic leaves a cubic whose other two roots, counted with multiplicity, are rational. Thus addition preserves $E(\mathbb{Q})$. The construction is symmetric in $P,Q$, so it is commutative.

For associativity, use divisor classes. A smooth plane cubic has geometric genus $1$, by the genus-degree formula. The Riemann-Roch theorem says that a divisor of degree $1$ on a genus-one curve has a one-dimensional space of sections. Therefore every degree-zero divisor class is uniquely represented by $P-O$: add $O$, take its unique effective degree-one representative, and then subtract $O$. Uniqueness also follows because a nonconstant function with its only pole a simple pole at $Q$ would contradict $\ell (Q)=1$.

Every line section is linearly equivalent to the tangent section $3O$. Thus, whenever $P,Q,R$ are collinear, including tangencies,A vertical section gives $[(-R)-O]=-[R-O]$. Hence the chord-and-tangent group law becomes ordinary addition in the degree-zero Picard group under the injective correspondence $P\mapsto [P-O]$. Associativity follows from associativity in that abelian group. Together with closure, identity and inverses already checked, this proves that $E(\mathbb{Q})$ is an abelian group.

Write $T=(\theta ,0)$, so $f(\theta )=0$, $K=\mathbb{Q}(\theta )$, and $f^{\prime }(\theta )\ne 0$. Let $[a]$ denote the class of $a\in K^{\ast }$ modulo squares. The required two-torsion square-class homomorphism isThe last value is needed in the domain only if $T\in E(\mathbb{Q})$. Every displayed nonexceptional value is nonzero, because an affine point with $x=\theta$ necessarily has $y=0$ and equals $T$.

Consider a nonvertical line $y=mx+c$ whose three intersection points have $x$-coordinates $x_1,x_2,x_3$, counted with multiplicity. Since $f$ is monic,If none of these points is $T$, evaluating at $\theta$ givesand therefore the product of their $\alpha$-values is $1$ in the square-class group. Tangencies are included by repeated factors. Since negation preserves the $x$-coordinate, this identity says $\alpha (P+Q)=\alpha (P)\alpha (Q)$.

If the line passes through $T$, put $f(x)=(x-\theta )g(x)$ and write the line as $y=m(x-\theta )$. The other two intersection abscissae $x_1,x_2$ satisfyEvaluating at $\theta$ now givesThus $\alpha (T)\alpha (P_1)\alpha (P_2)=[f^{\prime }(\theta ){]}^2=1$, as required. This covers a tangent at a different point whose third intersection is $T$ as well.

The tangent at $T$ is vertical. For a vertical chord or tangent the two affine intersections are $P,-P$, so their product of square classes is $\alpha (P{)}^2=1$; in particular $\alpha (T{)}^2=1$ agrees with $2T=O$. Finally, adding $O$ changes neither side. These cases exhaust the chord-and-tangent group law, proving that the displayed map is a group homomorphism.

For an elliptic curve over ${\mathbb{F}}_q$, where $q$ is a prime power, the Hasse theorem for elliptic curves isWe prove it from the degree form on elliptic-curve endomorphisms, including the required degree facts.

A nonzero homomorphism of elliptic curves is a finite surjective morphism; its degree is the degree of the induced extension of function fields. Set the degree of the zero homomorphism to $0$. The degree of an isogeny is also the degree of the pullback of any point divisor: locally, a finite map of smooth curves gives a module of rank equal to the function-field degree, and its fibre length is the sum of the local multiplicities. For a composition, the tower law for function fields givesIf the isogeny of elliptic curves is separable, its pullback of a nonzero invariant differential on an elliptic curve is nonzero. Translation invariance implies that its differential is nonzero at every point, so each fibre point has multiplicity $1$. Every fibre is a translate of the kernel. ThereforeFor an inseparable isogeny, multiplicities must be retained; counting geometric kernel points alone would be incorrect.

Here is a divisor proof of the degree parallelogram law. On $E\times E$, let $\Delta$ be the diagonal and ${\Delta }^{-}$ the locus $P=-Q$. A Weierstrass $x$-function has a double pole at $O$, and the equation $x(P)=x(Q)$ describes $P=Q$ or $P=-Q$. ConsequentlyThis remains valid in characteristic $2$ using a general Weierstrass equation of an elliptic curve: the $x$-map still has degree $2$ and its two points are exchanged by negation. The divisor identity gives an identity of line bundles. Pull it back by $P\mapsto (\alpha (P),\beta (P))$ and take degrees. The diagonal is the inverse image of $O$ under subtraction and the antidiagonal under addition, soUsing line bundles makes this pullback argument valid even when $\alpha =\beta$, $\alpha =-\beta$ or one map is zero, when directly substituting into the rational function would fail.

Let $d_n=\mathrm{deg}[n]$. Applying the degree parallelogram law to $[n]$ and $[1]$ gives $d_{n+1}+d_{n-1}=2d_n+2$, with $d_0=0,d_1=1$. Induction givesMore generally, the same recurrence and polarization show that the degree restricted to the subgroup generated by two endomorphisms is a quadratic form: its mixed coefficient is determined by their sum or difference. This can be checked directly by the second-difference recurrence in each integer variable and the parallelogram identity in the two diagonal directions.

Let $\pi$ be the $q$-power Frobenius isogeny of an elliptic curve. It has degree $q$: it is purely inseparable, has one geometric point in each fibre, and sends a local parameter at the rational point $O$ to its $q$th power, giving fibre multiplicity $q$. Its pullback of any differential is zero. On the other hand, the standard addition rule for an invariant differential on an elliptic curve givesHence $1-\pi$ is separable, and its kernel consists exactly of the points fixed by Frobenius, namely $E({\mathbb{F}}_q)$. Put $N=\#E({\mathbb{F}}_q)$ and $t=q+1-N$. Thenand the quadratic-form calculation gives, for all $m,n\in \mathbb{Z}$,If $t^2>4q$, the real polynomial $X^2-tX+q$ is negative on an open interval. That interval contains a rational $n/m$ with $m\ne 0$, contradicting the displayed nonnegativity after multiplication by $m^2$. Thus $t^2\le 4q$, which is precisely the Hasse theorem for elliptic curves.

Let $\pi ,{\pi }^{\prime }$ be the $p$-power Frobenius isogenies of $E,E^{\prime }$. Since $\psi$ is defined over ${\mathbb{F}}_p$, it commutes with Frobenius:Taking degrees, using multiplicativity and cancelling the nonzero degree of an isogeny $\mathrm{deg}\psi$, givesBoth differences are separable, so the preceding proof identifies their degrees with rational point counts. Therefore

Isogenous elliptic curves can have different rational point groups. Here is an explicit example. Over ${\mathbb{F}}_7$, takeThe mapextends across $O$ and $(0,0)$ to a degree-$2$ isogeny of elliptic curves with those two points as its kernel. Substitution verifies the target equation, or this follows from the two-isogeny formula with $a=0,b=-1$. Both curves are smooth modulo $7$.

For $x=0,1,\dots ,6$, the numbers of affine points with that abscissa on $E$ are $(1,1,0,0,2,2,1)$, and on $E^{\prime }$ they are $(1,0,2,2,0,0,2)$. Adding the point at infinity gives $8$ on each. The first curve has four rational points of order dividing $2$, from $O$ and the three roots $0,1,-1$. The second has only two: its quadratic factor $X^2+4$ has no root modulo $7$, since $3$ is not a square. By the classification of finite abelian groups,They are thus isogenous with equal orders and nonisomorphic groups.

Here a formal group means a one-dimensional commutative formal group law. Over a ring $R$, it is a formal power series $F(X,Y)\in R[[X,Y]]$ satisfyingIn particular, $F(X,Y)=X+Y+$ terms of total degree at least $2$. A formal inverse $\iota (T)=-T+\cdots$ with $F(T,\iota (T))=0$ is obtained recursively. An isomorphism of formal group laws from $F$ to $G$ is a series $h(T)\in TR[[T]]$ with invertible linear coefficient andIts compositional inverse exists over $R$ by coefficient recursion.

For $R={\mathcal{O}}_K$, all these series converge on $\pi {\mathcal{O}}_K$; the operations preserve that set and make it the group $F(\pi {\mathcal{O}}_K)$. Use the uniformizer $\pi$ and normalize the valuation by $v_K(\pi )=1$, and write $e=v_K(p)$.

We explicitly construct its formal logarithm. DefineThen putThis is a formal series over $K$ with linear coefficient $1$. Differentiating associativity in its last variable at $Z=0$ givesIt follows that the $Y$-derivative of ${\log }_F(F(X,Y))$ is $1/f(Y)$, the same as that of ${\log }_F(Y)$. Evaluating at $Y=0$ therefore provesThe differential $dT/f(T)$ is the invariant differential of a formal group law.

Choose an integer $m>e/(p-1)$. In degree $n\ge 2$, the coefficient of ${\log }_F(T)$ has valuation at least $-e{v}_p(n)$. Thus the scaled serieshasHere $v_p(n)\le (n-1)/(p-1)$, and the coefficient valuations tend to infinity. Consequently $H(U)=U+\pi R(U)$ with $R$ a convergent integral power series on ${\mathcal{O}}_K$. Such a series is $1$-Lipschitz. For any $a\in {\mathcal{O}}_K$, the equation $H(U)=a$ is equivalent to $U=a-\pi R(U)$, a strict contraction, to which the contraction mapping theorem applies on the complete ring ${\mathcal{O}}_K$. It has exactly one solution.

The deep logarithm subgroup of a formal group is obtained as follows. The formal group exponential is the formal compositional inverse ${\exp }_F(T)$ of ${\log }_F(T)$, obtained recursively; equivalently it solves ${\exp }_F^{\prime }(T)=f({\exp }_F(T))$, with ${\exp }_F(0)=0$. The preceding contraction, applied to the scaled series, constructs its convergent inverse on ${\pi }^m{\mathcal{O}}_K$ and agrees with that formal inverse. This supplies the required convergence sketch for ${\exp }_F$ as well as a bijectivity proof for the logarithm.

The integral formal law and inverse preserve ${\pi }^m{\mathcal{O}}_K$, so this is a subgroup. The logarithm identity and bijectivity giveReduction of the formal law modulo ${\pi }^m$ has kernel $F({\pi }^m{\mathcal{O}}_K)$. Its underlying image set is $\pi {\mathcal{O}}_K/{\pi }^m{\mathcal{O}}_K$, which has size $q^{m-1}$, where $q=\#({\mathcal{O}}_K/\pi )$. Thus this subgroup has finite index, explicitly $q^{m-1}$.

In fact, $2,3,5,7$ must all be primes of bad reduction.

At a prime of good reduction of an elliptic curve, the kernel $E_1({\mathbb{Q}}_p)$ of reduction is the formal group of an elliptic curve evaluated on $p{\mathbb{Z}}_p$, using its parameter at the identity. This kernel has no prime-to-$p$ torsion: for an integer $\ell$ prime to $p$, the multiplication series is $[\ell {]}_F(T)=\ell T+$ higher-degree integral terms, so for nonzero $T\in p{\mathbb{Z}}_p$ its first term has strictly smaller valuation than the others and $[\ell {]}_F(T)\ne 0$.

For each odd good prime, the entire given torsion subgroup, of order $16$, would therefore inject into $E({\mathbb{F}}_p)$. But the Hasse theorem for elliptic curves givesNone can contain a subgroup of order $16$. Hence $3,5,7$ are bad.

For the prime-two torsion bound for good reduction, use the deeper formal subgroup corresponding to $4{\mathbb{Z}}_2$. Part (i), with $m=2>1/(2-1)$, shows that $E_2({\mathbb{Q}}_2)$ is torsion-free and has index $2$ in $E_1({\mathbb{Q}}_2)$. Therefore any finite torsion subgroup of $E_1({\mathbb{Q}}_2)$ injects into $E_1/E_2$ and has order at most $2$.

If $E$ had good reduction at $2$, its rational torsion subgroup of order $16$ would consequently have reduction image of order at least $8$. The Hasse theorem for elliptic curves instead gives $\#E({\mathbb{F}}_2)\le 5$. This contradiction proves bad reduction at $2$ as well, and thereforeThe special formal-group argument at $2$ is necessary; prime-to-$p$ injectivity alone says nothing about the given $2$-primary torsion there.

Heights measure arithmetic size, and their finiteness and growth properties turn finite divisibility information into finite generation. For an elliptic curve $E$ over a number field $K$, the resulting Mordell-Weil theorem isThe distinction between this statement and the Weak Mordell-Weil theorem, which only asserts that $E(K)/mE(K)$ is finite, is exactly where heights enter.

For a rational point of the projective line, choose coprime integers $a,b$ and define the logarithmic projective heightFor a fixed number field, its absolute logarithmic version iswith the usual normalized absolute values. The product formula makes this independent of scaling $(a,b)$, and the local degree formula makes it independent of enlarging the number field. Heights are nonnegative and are useful precisely because they control numerator and denominator together, unlike an ordinary real absolute value.

The Northcott theorem states that algebraic points of bounded degree and bounded height form a finite set. Over $\mathbb{Q}$ this is immediate from the coprime numerator-denominator description. In general, bounded degree and height bound the coefficients of the primitive minimal polynomial: their elementary symmetric functions are bounded by its Mahler measure, which is the appropriate power of the multiplicative height. There are only finitely many such integer polynomials.

On a Weierstrass equation of an elliptic curve, define the naive logarithmic height $h_x(P)=h(x(P))$ and $h_x(O)=0$. Since an $x$-coordinate has at most two points above it, Northcott theorem implies that only finitely many points of $E(K)$ have bounded $h_x$.

Duplication descends to a rational map of the projective $x$-line of degree $4$. More generally, if a morphism $\varphi :{\mathbb{P}}^1\to {\mathbb{P}}^1$ has degree $d$, thenuniformly in $z$. Write $\varphi$ as two homogeneous degree-$d$ polynomials without common zeros. Bounding their coefficients gives the upper estimate at each place. A nonzero resultant, or homogeneous Bezout identities, prevents simultaneous cancellation and gives the lower estimate; only finitely many places contribute a nonzero constant. Applied to duplication, this gives the height growth under a morphism of the projective line estimate

Define the canonical height of an elliptic curve byThe factor $1/2$ is a convention making it the height for the degree-one symmetric divisor $O$. If $a_n=4^{-n}{h}_x(2^{n}P)$, then $|a_{n+1}-a_n|\le C_E{4}^{-(n+1)}$, so the limit exists by a geometric-series estimate. Summing that estimate also provesThus bounded canonical height still gives only finitely many $K$-points, and $\hat{h}(2P)=4\hat{h}(P)$ exactly.

The addition formula on the quotient by negation gives the uniform approximate parallelogram identityThis height parallelogram identity follows as follows. One way to obtain it is to express the unordered pair of their $x$-coordinates as a morphism ${\mathbb{P}}^1\times {\mathbb{P}}^1\to {\mathrm{Sym}}^2{\mathbb{P}}^1$ of bidegree $(2,2)$; the height of that unordered pair is the sum of the two heights, up to a bounded constant. It is also the height identity for the symmetric divisor $O$. Apply this identity to $2^{n}P,2^{n}Q$, divide by $2\cdot 4^n$, and take limits. The errors disappear, givingTogether with evenness and $\hat{h}(O)=0$, the recurrence implies $\hat{h}(mP)=m^2\hat{h}(P)$ for every integer $m$. The associated canonical height pairingis bilinear. Nonnegativity of the height implies its Cauchy-Schwarz inequality, by applying it to integer combinations and then rational approximations. In particular,The height vanishes precisely at torsion points: torsion gives a finite set of multiples, whereas if $\hat{h}(P)=0$, every multiple of $P$ has bounded naive height and therefore the set of multiples is finite by Northcott theorem.

Now apply the Weak Mordell-Weil theorem with $m=2$, proved cohomologically in the other essay. Choose a finite set $R$ of representatives for $E(K)/2E(K)$, and put $H={\max }_{R_i\in R}\hat{h}(R_i)$. Write an arbitrary point as $P=2Q+R_i$. ThenIf $\hat{h}(P)>H+1$, this decreases the height by more than $1/2$. Repeat until the current point has height at most $H+1$. There are only finitely many such points. Unwinding the equations $P=2Q+R_i$ shows that $E(K)$ is generated by this finite bounded-height set together with $R$. This is the height descent lemma, and proves the Mordell-Weil theorem.

Once finite generation is established, the structure theorem for finitely generated abelian groups gives the displayed decomposition. The canonical height pairing is positive definite on the real vector space of the free part: a null direction would, by approximation with integer combinations of generators, give infinitely many lattice points of bounded height, contradicting Northcott theorem. Its determinant is the regulator of an elliptic curve, and its geometry gives practical tools for searching for generators and comparing independence. Thus heights supply both the finiteness mechanism in the proof and the quadratic size function used in arithmetic computations.

Galois cohomology records the obstruction to choosing Galois-invariant division points. Let $G_K=\mathrm{Gal}(\overline{K}/K)$ be the absolute Galois group and give its algebraic-point modules the discrete topology. A continuous $1$-cocycle with values in a Galois module $M$ is a function $c:G_K\to M$ satisfyingA group coboundary has the form $c_{\sigma }=\sigma a-a$. Quotienting cocycles by coboundaries defines $H^1(K,M)$. For the finite module $E[m]$ these cocycles have finite image and factor through finite data; continuity is essential because $G_K$ is a profinite group.

In characteristic zero, multiplication by $m\ge 2$ on the algebraic points of an elliptic curve is surjective, with kernel $E[m]\cong (\mathbb{Z}/m\mathbb{Z}{)}^2$. The short exact sequence of Galois modulesgives the Kummer exact sequence of an elliptic curveExplicitly, choose $Q$ with $mQ=P$ and set $\delta (P{)}_{\sigma }=\sigma Q-Q$. Changing $Q$ by an $m$-torsion point changes the cocycle by a coboundary. The class vanishes exactly when a suitable choice of $Q$ is Galois-fixed, that is, when $P\in mE(K)$. This proves the injection directly and identifies its arithmetic meaning.

The entire group $H^1(K,E[m])$ need not be finite. For example, Kummer theory gives $H^1(K,{\mu }_m)=K^{\ast }/K^{\ast m}$, which has classes supported on arbitrarily many different primes. The finite part needed for the Weak Mordell-Weil theorem comes from a ramification restriction on the image of $\delta$.

Choose a finite set $S$ of places containing the archimedean places, the primes dividing $m$ and all primes of bad reduction of an elliptic curve. At a finite place $v\notin S$, the curve has good reduction and $m$ is invertible in its valuation ring. The elliptic curve extends to a smooth proper group scheme, and $[m]$ on that model is finite etale. A point $P\in E(K_v)$ extends to an integral section by properness. Its division-point fibre is consequently finite etale over the valuation ring. Over the maximal unramified extension it has a point, so the Kummer cocycle restricts trivially to the inertia group. The module $E[m]$ itself is unramified there for the same reason. Thuswhere $H_S^1$ denotes classes unramified outside $S$.

Here is a proof that this restricted group is finite. Take a finite Galois extension $L/K$ containing all coordinates of $E[m]$ and the $m$th roots of unity, and enlarge $S$ by its ramified primes. Over $L$ the module is trivial and, after choosing a basis, is isomorphic to ${\mu }_m^2$. Hilbert theorem 90 and the multiplicative Kummer sequence therefore giveFor unramified Kummer classes with bounded prime support, a class unramified outside $S_L$ has valuations divisible by $m$ at every prime outside $S_L$: the valuation of an $m$th root in an unramified extension is integral. Consequently both coordinates lie in

To show this set finite, write the outside-$S_L$ divisor of $a$ as $mD$. The ideal class of $D$ belongs to the $m$-torsion of the ideal class group of the ring of $S_L$-integers. This gives the exact sequenceThe last map is onto: if $mD$ is principal in the $S_L$-ideal group, a generator represents a class with outside valuations divisible by $m$. The kernel consists exactly of S-units modulo $m$th powers. The Dirichlet unit theorem, with the finitely many inverted primes adjoined, makes the S-unit group finitely generated; the ideal class group is finite, and localization only quotients it. Both ends of the sequence are therefore finite.

Finally, the inflation-restriction exact sequence bounds the kernel of restriction from $H^1(K,E[m])$ to $H^1(L,E[m])$ by the finite group $H^1(\mathrm{Gal}(L/K),E[m])$. The unramified subgroup has finite image, contained in $L(S_L,m{)}^2$, and finite kernel, so it is finite. The Kummer injection now provesThis is the Weak Mordell-Weil theorem, proved without first assuming finite generation of $E(K)$.

For computation one refines the unramified group by local solvability. The Selmer group of an elliptic curve consists of classes whose restriction at every completion lies in the corresponding local Kummer image. It is finite and sits inwhere the Tate–Shafarevich group $\mathrm{Sha}$ measures classes in $H^1(K,E)$ that become trivial at every completion. Thus locally soluble descent equations can give an upper bound without every class coming from a rational point. No finiteness assumption on the whole Tate-Shafarevich group is needed for the weak theorem. Combining the finite quotient with the height descent lemma from the height essay yields the full Mordell-Weil theorem. Replacing multiplication by $m$ with a smaller isogeny gives the same cohomological framework for two-isogeny descent.

Start with an elliptic curve having a rational point $T$ of order $2$. Move that point to $(0,0)$ and take an integral modelPut $b^{\prime }=a^2-4b$ andThe two-isogeny formula and its dual areThey extend to the projective curves, with kernels $\{O,T\}$ and $\{O,T^{\prime }\}$, where $T^{\prime }=(0,0)\in E^{\prime }$. Substitution verifies their target equations, and the elliptic-curve addition formula verifies $\hat{\varphi }\varphi =[2]$ and $\varphi \hat{\varphi }=[2]$. The division by $4,8$ in the dual identifies the twice-transformed curve, with coefficients $4a,16b$, with $E$.

Define the two-torsion square-class homomorphismsQuestion 1(ii) proves these are homomorphisms, since $f^{\prime }(0)=b$. Their kernels areFor completeness, the first coordinate of the dual is $Y^2/(4X^2)$, so a nonexceptional dual image has square $x$-coordinate. Conversely, if $x=u^2\ne 0$, the preimage equation iswhose discriminant is $16(x^2+ax+b)=16(y/u{)}^2$. The roots $X=2x+a\pm 2y/u$ are rational and nonzero. Choosing $Y=\pm 2uX$ gives a point on $E^{\prime }$, and a choice of sign makes its dual image exactly $(x,y)$. The point $T$ has a rational dual preimage precisely when $b$ is a square, as seen from the roots $a\pm 2\sqrt{b}$ of the nonzero two-torsion polynomial on $E^{\prime }$. The identity is already a dual image. Applying this argument to the transformed curve proves the second kernel assertion as well, because scaling an $x$-coordinate by $4$ does not change its square class.

The two-isogeny descent is finite because an image square class has a signed square-free integer representative dividing $b$. Indeed, if $\ell \nmid b$ and $v_{\ell }(x)>0$, the other factor $x^2+ax+b$ is a unit, so $v_{\ell }(y^2)=v_{\ell }(x)$ forces an even valuation. If $v_{\ell }(x)<0$, the term $x^2$ dominates that factor and $v_{\ell }(y^2)=3v_{\ell }(x)$, again forcing even valuation. Thus odd valuations can occur only at primes dividing $b$. The same argument applies to $b^{\prime }$ on $E^{\prime }$.

For each candidate signed squarefree divisor $d\mid b$, put $x=d(U/V{)}^2$, with coprime integers $U,V$. The curve equation becomes the two-isogeny descent quarticA solution with $UV\ne 0$ yields $x=d{U}^2/V^2$ and $y=dUN/V^3$. Conversely, every point in that class gives such a primitive integer solution, since a rational square root of an integer is integral. The boundary solutions $U=0$ and $V=0$ account respectively for the classes of $T$ and $O$. Rational solutions prove that a class occurs; a real or congruence obstruction excludes it. Merely finding local solutions everywhere does not automatically prove a rational solution.

To extract the rank, write $A=\alpha (E(\mathbb{Q}))$ and $A^{\prime }={\alpha }^{\prime }(E^{\prime }(\mathbb{Q}))$. The Mordell-Weil theorem givesThe isogeny factorization gives the index $|A|$ for the first quotient by $\hat{\varphi }{E}^{\prime }$. For the remaining index, apply $\hat{\varphi }$ to $E^{\prime }/\varphi E$. Its kernel has size $2/e$, whereThe equality follows because $\hat{\varphi }(\varphi P)=2P$. Consequentlyand cancellation yields the two-isogeny rank formulaThis formula is valid whether there is just one rational nonzero two-torsion point or all three.

For the first curve, $a=-1,b=1$, and the isogenous curve isThe only candidate square classes on $E$ are $1,-1$. Since $x^2-x+1>0$ for every real $x$, a real affine point has $x\ge 0$; the exceptional torsion class is $[b]=1$. ThusOn $E^{\prime }$, the candidates are $1,-1,3,-3$. They all occur: the identity gives $1$, $T^{\prime }$ gives $-3$, $(3,6)$ gives $3$, and $(-1,2)$ gives $-1$. Hence $|A^{\prime }|=4$, andThe product in the rank formula is $1\cdot 4=4$, not $1\cdot 2$; all four classes on the companion curve are essential.

For the second curve, $a=5,b=-6$, andThe candidate classes on $E$ are $\{\pm 1,\pm 2,\pm 3,\pm 6\}$. The identity, $T$, $(2,4)$ and $(-3,6)$ show thatThis is a subgroup of order $4$. Its other coset is $\{3,6,-1,-2\}$, so it suffices to exclude the representative $3$.

The modulo-eight obstruction to a two-isogeny descent class uses the corresponding two-isogeny descent quartic,If both $U,V$ are odd, its right side is $6$ modulo $8$. If $U$ is odd and $V$ even, it is $3$ or $7$ modulo $8$. If $U$ is even and $V$ odd, it is $6$ or $2$ modulo $8$. These are all primitive parity cases, and none is a square modulo $8$. Hence the class $3$ is impossible. Since $A$ is a subgroup, every class in its coset is impossible, and thereforeOn the companion curve, the only candidates are $\pm 1,\pm 7$. Its quadratic factor is $(X-5{)}^2+24>0$, so every nonzero real affine $X$ is positive. The exceptional value $[49]$ is $1$, and the point $(7,14)$ supplies the class $7$. ThusThe rank formula gives $2^r=4\cdot 2/4=2$, soEvery included class has an explicit rational representative and every excluded coset has a proved real or congruence obstruction, so these are exact ranks rather than bounds obtained from a point search.