Let $F(x,y) = a\_3 x^3 + a\_2x^2 y + a\_1 xy^2 + a\_0 y^3$ be a binary cubic form, say with real coefficients. Put $H(x,y) = H\_F(x,y)$ for the *Hessian covariant* of $F$, defined by


$$\displaystyle H\_F(x,y) = \frac{1}{4} \begin{vmatrix} F\_{xx} & F\_{xy} \\ F\_{xy} & F\_{yy} \end{vmatrix},$$


and put


$$\displaystyle G(x,y) = G\_F(x,y) = \begin{vmatrix} F\_x & F\_y \\ H\_x & H\_y \end{vmatrix}.$$


$H,G$ are both covariants of $F$ under the substitution action of $\operatorname{GL}\_2$, and in particular, $H, G, F$ and the invariant $\Delta(F)$, the discriminant of $F$, generate the ring of polynomial covariants. They are connected by a single syzygy, given by


$$\displaystyle 4H(x,y)^3 + G(x,y)^2 = -27 \Delta(F) F(x,y)^2.$$


As can be verified by immediate calculation, we have


$$\displaystyle \Delta(G) = 729 \Delta(F)^3,$$


which is a perfect cube. 


My question is, suppose that $G$ is a binary cubic form with integer coefficients satisfying $\Delta(G) = 729 n^3$ for some non-zero integer $n$. Does it follow that $G$ is the $G\_F$-covariant for some binary cubic form $F$ with integer (rational) coefficients? If not, what is a counter-example, and what would be a stricter condition that guarantees $G$ is such a covariant?


On a related note, if $G'$ is the non-trivial cubic covariant of $G\_F$, then $G' = -729 \Delta(F)^2 F(x,y)$.



A curious identity involving a covariant of binary cubic forms
As Abdelmalek notes, if $G$ is to be the cubic covariant $G\_F$ of another form $F$, and has a nonzero discriminant of the expected shape $\Delta(G)=3^6n^3$, then $F$ must divide $G'=G\_G$ - indeed the only candidate (comparing discriminants and exploiting homogeneity) is $F=-G\_G/(3^6n^2)$, which will have rational coefficients when $n$ is rational, and integral coefficients when (and only when) those of $G\_G$ are integral multiples of $3^6n^2$.


And when this is the case, this $F$ does indeed work:
$$G\_F = -G\_{G\_G}/(3^6n^2)^3 = 3^6\Delta(G)^2G/(3^{18}n^6) = 3^6(3^6n^3)^2G/(3^{18}n^6) = G\,.$$


