The curve has a <two-isogeny descent>. Its quotient by $\langle T\rangle$ is
$$
E':y^2=x(x^2-2ax+a^2-4b),
$$
and the dual two-isogeny gives a corresponding map $\alpha':E'(\mathbb Q)\to\mathbb Q^\times/(\mathbb Q^\times)^2$. The standard kernel calculation yields
$$
2^{\operatorname{rank}E(\mathbb Q)}
=\frac{|\operatorname{im}\alpha|\,|\operatorname{im}\alpha'|}{4}.
$$

If $d$ represents a class in $\operatorname{im}\alpha$, write
$$
x=d\left(\frac uv\right)^2
$$
with 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 descent>
$$
C_d:\qquad
w^2=du^4+au^2v^2+\frac bdv^4
$$
has 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 $E'$ reduces the rank calculation to finitely many explicit quartic solubility tests.

Solution

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

{title2=$C_d$}

For a square class $d$ dividing $b$, the associated quartic covering is
$$
C_d:\qquad w^2=du^4+au^2v^2+\frac bdv^4.
$$
It has a nontrivial rational point exactly when $d$ occurs in the square-class image attached to $E:y^2=x(x^2+ax+b)$.


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 Incoming links: Quartic covering in a two-isogeny descent