$ax^3+by^3+cz^3=0$ and Elliptic curves

Question

What is relation between $ax^3+by^3+cz^3=0$ and Elliptic curves?

Answer 1

Consider the map $$(x,y,z)\quad\mapsto\quad (X,Y) = \left( \frac{-4bcyz}{x^2}, \frac{-4bc(by^3-cz^3)}{x^3} \right)$$ By brute force expansion, one can verify $$Y^2 - X^3 - (4abc)^2 = \frac{16 b^2c^2(-ax^3+by^3+cz^3)(ax^3+by^3+cz^3)}{x^6} $$ This means every non-zero integer solution of $\quad\displaystyle a x^3 + b y^3 + c z^3 = 0\quad$ corresponds to a rational point on the elliptic curve $$Y^2 = X^3 + (4abc)^2$$

Answer 2

Furthermore, the correspondence point out by Hui implies that if $ax^3+by^3+cz^3 = N$ has a rational solution, then one can generally find an infinite more. This can be proven by the identity, $$ap^3+bq^3+cr^3 = (ax^3+by^3+cz^3)(ax^3-by^3)^3$$ where, $$p,\,q,\,r = x(ax^3+2by^3),\; -y(2ax^3+by^3),\; z(ax^3-by^3) $$ For example, this shows that, $$x^3+y^3 = 9$$ has an infinite number of rational solutions.

For more, see Form 19 at http://sites.google.com/site/tpiezas/011.

Answer 3

Actually, the map given by achille hui gives a 3-to-1 map from your curve $C$ to the elliptic curve $E$ that he lists, so although it is true that every point in $C(\mathbb{Q})$ gives a rational point in $E(\mathbb{Q})$, the converse is not true, at least, not via the map that he gives. The way to really understand your curve is that it is a homogeneous space of degree 3 for the elliptic curve $E$. You'll find homogeneous spaces described in Chapter X of my book The Arithmetic of Elliptic Curves (Springer). There is a long, influential article of Selmer [1] in which he makes a detailed study of your curves, and in particular finds a large number of them that have $p$-adic points for all $p$, but no $\mathbb{Q}$-rational points. In modern terminology, they correspond to elements of order 3 in the Shafarevich-Tate group of $E$.

[1] E. Selmer, The {D}iophantine equation $ax^3+by^3+cz^3=0$, Acta Math. 85 (1951), 203-362.