 Solution

The discriminant is $-64D^3$, which is nonzero modulo three because $D\equiv 2(\mathrm{mod}3)$. Thus there is good reduction at three. For every $x\in {\mathbb{F}}_3$, the reduced right side is $x^3-x=0$, so its only affine points are $(0,0),(1,0),(2,0)$. Together with the identity they form $C_2\times C_2$.

By part (ii), the rational torsion subgroup injects into this four-element group and hence has exponent two. Over $\mathbb{Q}$, the two-torsion points have $Y=0$, requiring $X(X^2+D)=0$. Since $D>0$, the only rational root is zero. Therefore