= Solution
Put $F=x^3+y^3+13z^3-2xyz$. The <Hessian criterion for flexes of a plane cubic> applies because the projective cubic is smooth. To check smoothness, a singular point with all coordinates nonzero would satisfy $3x^2=2yz$, $3y^2=2xz$, $39z^2=2xy$. Multiplication would force $351=8$, a contradiction. If one coordinate is zero, the three equations force all coordinates to vanish, which is not a projective point.
For completeness, the <Hessian criterion for flexes of a plane cubic> follows locally by putting a smooth point at $(0:0:1)$ and its tangent at $y=0$. After normalization the homogeneous cubic has terms $yz^2+ax^2z+bxyz+cy^2z$ and terms cubic in $x,y$. Its tangent has contact order three precisely when $a=0$. Its <Hessian matrix> at the point has determinant $-8a$, giving exactly the same condition.
For the present cubic the Hessian determinant is
$$
\det\begin{pmatrix}
6x&-2z&-2y\\
-2z&6y&-2x\\
-2y&-2x&78z
\end{pmatrix}
=-24(x^3+y^3+13z^3)+2792xyz.
$$
On $F=0$ this is $2744xyz=8\cdot343\,xyz$. Thus the <plane cubic flexes> are exactly the points on the cubic with $xyz=0$. Let $c=\sqrt[3]{13}$ and let $\omega$ be a primitive cube root of unity. The complete list over the <algebraic closure> is
$$
\boxed{(1:-\omega^j:0),\quad(-c\omega^j:0:1),\quad(0:-c\omega^j:1),\qquad j=0,1,2.}
$$
These nine points are distinct, and the Hessian calculation proves there are no others. They are the <flexes of a Hesse cubic>. The only rational one is $O'=(1:-1:0)$: the other two points with $z=0$ require a nonrational cube root of unity, and a rational point in either remaining triple would give a rational cube root of $-13$. By prime valuations, $13$ is not a rational cube.
Back to article page
