Rational torsion of a diagonal cubic ($E_d(\mathbb{Q}{)}_{\mathrm{tors}}$)

For $d\in {\mathbb{Q}}^{\times }$ the rational torsion subgroup of $X^3+Y^3+d{Z}^3=0$ is cyclic of order $3$ if $d$ is a cube, cyclic of order $2$ if $d/2$ is a cube, and trivial otherwise.

To bound torsion, use the Weierstrass equation of an elliptic curve $y^2=x^3-432d^2$. At every good prime $\ell \equiv 2(\mathrm{mod}3)$, cubing is a bijection on ${\mathbb{F}}_{\ell }$, so the number of points is $\ell +1$. Reduction injects the prime-to-$\ell$ torsion, since multiplication by an integer prime to $\ell$ is invertible in the formal group of an elliptic curve. For any prime $q>3$, the Dirichlet theorem on primes in arithmetic progressions supplies a good $\ell \equiv 2(\mathrm{mod}3)$, $\ell \equiv 1(\mathrm{mod}q)$, excluding $q$-torsion. Good primes $\ell \equiv 5(\mathrm{mod}12)$ exclude order $4$, and good primes $\ell \equiv 11(\mathrm{mod}18)$ exclude order $9$. Thus every rational torsion point is killed by $6$.

The flexes of a diagonal cubic give rational three-torsion exactly when $d$ is a cube, and then it has order $3$. Rational two-torsion requires $x^3=432d^2$, whose sole possible rational root exists exactly when $d/2$ is a cube. This equivalence follows by comparing prime valuations modulo $3$, including the valuation at $2$. The two conditions cannot hold together because $2$ is not a rational cube. The asserted classification follows.