Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2010 / B / 6

Complete the square-class map at the identity by $q(O)=1$; at the finite point $T=(0,0)$ its value is the class of $b$. For another rational point writeThese denominator powers follow because a negative valuation of $x$ forces the leading cubic term to dominate, so that valuation is even and the valuation of $y$ is three halves of it. The equation givesIf a prime $\ell \nmid b$ divides $r$, it does not divide the second factor, since $\ell \nmid s$. Thus $v_{\ell }(r)$ is even. All denominator valuations are even too. Every image class therefore has a representative which is a signed squarefree divisor of $b$, proving that $\mathrm{im}q$ is finite.

Use the standard 2-isogeny with kernel $\{O,T\}$ and dual isogeny satisfying $\hat{\varphi }\varphi =[2]$, $\varphi \hat{\varphi }=[2]$. Away from the exceptional points one may takeApplying the analogous isogeny to $E^{\prime }$ gives a curve with coefficients $4a,16b$; scaling its coordinates by $1/4,1/8$ gives the dual map back to $E$. The analogous descent map $q^{\prime }$ on $D$ has kernel $\varphi (C)$ and finite image. The sequenceis exact at the last two terms. Its outside groups are finite, identified with the two finite descent images, so $C/2C$ is finite.

For $E_p:y^2=x(x^2-p^2)$, the rational 2-torsion is $O,(0,0),(p,0),(-p,0)$, of order four. The first descent image is exactly $\{1,-1,p,-p\}$: these classes occur on the 2-torsion, and the divisor-support bound permits no others. The isogenous curve is $E_p^{\prime }:y^2=x(x^2+4p^2)$. Its real points have $x\ge 0$, so the second image lies in the four positive classes $\{1,2,p,2p\}$.

Here the kernel $\{O,(0,0)\}$ of the dual is contained in $\varphi (C)$, since $\varphi (\pm p,0)=(0,0)$. Thus the displayed map $D/\varphi (C)\to C/2C$ is injective, and exactness givesThe allowed rank formula makes the left side $2^{R+2}$, soNow consider the quartic relation. Normalize a nontrivial integer solution so that $p$ does not divide both $x,y$, by dividing them by $p$ and $z$ by $p^2$ whenever necessary. If $p\nmid x$, reduction modulo $p$ gives $x^4=2z^2$, with $z\ne 0$ modulo $p$, making $2$ a quadratic residue. If $p\mid x$, then $p\nmid y$ and $p\mid z$; write $x=pX$, $z=pZ$, divide by $p^2$, and reduce to get $y^4=2Z^2$ modulo $p$. Again $2$ is a square. ThereforeIf the class $2$ occurred in $\mathrm{im}{q}^{\prime }$, write the corresponding coordinate as $x^{\prime }=2u^2/v^2$. The curve equation then givesHere $u,v$ are nonzero coprime integers and $z$ is rational. Since $2z^2$ is an integer, its reduced denominator must be one, so $z$ is an integer too. For $p\equiv 3,5(\mathrm{mod}8)$ this contradicts the quartic obstruction. Hence class $2$ is absent. A subgroup of $\{1,2,p,2p\}$ omitting $2$ has order at most two, so $2^R\le 2$ and $R\le 1$.

For $p=5$, direct substitution verifies the nonintegral pointA rational torsion point would have integral coordinates by the preceding theorem, so this point has infinite order. Thus $R\ge 1$, and together with the upper bound,