Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2009 / B / 6 / b / Solution

For $E:y^2=x^3+px$ with $p>0$, the factor $x^2+p$ is strictly positive on the real line. Thus every nonzero rational $x$-coordinate is positive. In the two-isogeny descent map, prime support restricts its square class to $1$ or $p$. Both are realized by the identity and $(0,0)$, although only the upper bound is needed:The two-isogenous curve is $E^{\prime }:Y^2=X^3-4pX$. Its two-torsion square-class homomorphism has signed squarefree representatives supported on the primes dividing $4p$. For odd $p$ these are amongHence the second image has at most eight elements. At $p=2$ the list reduces to $1,-1,2,-2$, so the bound by eight remains valid. The two-isogeny rank formula therefore givesIn fact the same simple count yields $r\le 1$ when $p=2$. The positivity restriction on the first descent image is essential for the stated bound of two for all primes.