Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 125 / 5 / b

The curve has a two-isogeny descent. Its quotient by $⟨T⟩$ isand the dual two-isogeny gives a corresponding map ${\alpha }^{\prime }:E^{\prime }(\mathbb{Q})\to {\mathbb{Q}}^{\times }/({\mathbb{Q}}^{\times }{)}^2$. The standard kernel calculation yields

If $d$ represents a class in $\mathrm{im}\alpha$, writewith coprime integers $u,v$. Substitution into the equation and clearing squares shows that the class occurs exactly when the quartic covering in a two-isogeny descenthas a nontrivial rational, equivalently primitive integral, solution. Valuation parity shows that only the finitely many square classes represented by square-free divisors $d$ of $b$ need be considered. Repeating the construction for $E^{\prime }$ reduces the rank calculation to finitely many explicit quartic solubility tests.