Diagonal cubic elliptic curve ($X^3+Y^3+d{Z}^3=0$)

For $d\in {\mathbb{Q}}^{\times }$, the smooth projective curve $X^3+Y^3+d{Z}^3=0$, with identity $O=(1:-1:0)$, is a genus-one elliptic curve. Its partial derivatives have no common projective zero. The invertible linear projective transformationsends it to ${\mathcal{Y}}^2\mathcal{Z}={\mathcal{X}}^3-432d^2{\mathcal{Z}}^3$, taking $O$ to $(0:1:0)$. Indeed $(X+Y{)}^3+3(X+Y)(X-Y{)}^2=4(X^3+Y^3)=-4d{Z}^3$. In affine coordinates $u=X/Z,v=Y/Z$, this gives $x=-12d/(u+v)$ and $y=-36d(u-v)/(u+v)$, with inverse $u=(y-36d)/(6x)$, $v=(-y-36d)/(6x)$. Thus diagonal sums of two cubes are naturally studied through elliptic-curve addition formula.

There are infinitely many rational pairs $(u,v)$ with $u^3+v^3=7$. The corresponding Weierstrass equation of an elliptic curve is $y^2=x^3-21168$, and $(u,v)=(2,-1)$ maps to $P=(84,756)$. The primes $5$ and $11$ are good, with point counts $6$ and $12$ because cubing is bijective in their residue fields. Reduction of torsion points on an elliptic curve bounds rational torsion by order dividing $6$. The flexes of a diagonal cubic exclude three-torsion for $d=-7$, while $d/2=-7/2$ is not a cube and hence excludes two-torsion. Thus $P$ has infinite order.

All nonzero multiples of $P$ map to distinct rational affine pairs, since the only rational point at infinity of the diagonal cubic is its identity. The inverse map is $u=(y+252)/(6x)$ and $v=(252-y)/(6x)$.

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.

Let $r^3=d\ne 0$ and let $\zeta$ be a primitive cube root of unity. The nine inflection points of a plane cubic on $X^3+Y^3+d{Z}^3=0$ areThe Hessian criterion for flexes of a plane cubic gives the Hessian determinant $216dXYZ$, so every flex lies on one of the three coordinate lines. Their intersections give precisely the displayed points. With $(1:-1:0)$ as identity, these are exactly the three-torsion points are flexes of a plane cubic configurations: the tangent divisor at a flex is $3P$, giving $3(P-O)=0$.

Over $\mathbb{Q}$ only the first point with $j=0$ is rational on $Z=0$. The other two coordinate lines each give one rational point precisely when $d$ is a rational cube. Consequently the rational three-torsion subgroup is $\mathbb{Z}/3\mathbb{Z}$ when $d$ is a cube and is trivial otherwise.