 Rank-zero two-isogeny descent for roots zero one and minus two

For $C:y^2=x(x^2+x-2)$, the two-isogenous curve is $D:v^2=u(u^2-2u+9)$. The first-coordinate square-class image on $C$ is $\{1,-2\}$. The prime-support bound permits only $1,-1,2,-2$; square class two would require a primitive integer solution to $N^2=2M^4+M^2{e}^2-e^4$, whose right side is two modulo three whenever $M,e$ are not both divisible by three. Thus that class is excluded, as is minus one by the subgroup property. The identity and the point $(0,0)$ supply the remaining two classes.

The image on $D$ is $\{1,3\}$: the quadratic factor $(u-1{)}^2+8$ is positive, so only positive square classes dividing nine occur, and $(3,6)$ supplies class three. The square-class index formula for two-isogeny descent consequently gives $2^r=(2\cdot 2)/4=1$, so the rank of an elliptic curve is zero. Alternatively, the dual-isogeny kernel lies in the image of the forward isogeny because nine is a square, giving $[C(\mathbb{Q}):2C(\mathbb{Q})]=4$. The four rational two-torsion points already account for that index, again forcing rank zero.