# Making an elliptic curve out of a cubic polynomial made a cube, or $ax^3+bx^2+cx+d = y^3$

What is the transformation such that a general cubic polynomial to be made a cube,

$$ax^3+bx^2+cx+d = y^3\tag{1}$$

can be transformed to Weierstrass form,

$$x^3+Ax+B = t^2\tag{2}$$

(The special case $b = c = 0$ is easier.) Given an initial rational point, I know a method how to find subsequent ones, but it would be nice to know the general transformation. For example,

$$3x^3+9x^2+15x+9 = y^3\tag{3}$$

I find that,

$$x_1 = 3$$

$$x_2 = -1839/1871$$

$$x_3 = -13898941449153/12222218425537$$

and so on. (I may have skipped some points.) But how do you transform $(3)$ to $(2)$?

For those interested, the cubic $(3)$ in two-variable form is equivalent to,

$$3p^3+9p^2q+15pq^2+9q^3 =\; p^3 + (p+q)^3+(p+2q)^3 = t^3$$

or three cubes (not necessarily positive) in arithmetic progrees. Thus, $p,q = 3,1$ gives the well-known,

$$3^3+4^3+5^3 = 6^3$$

and $p,q = -1839, 1871$ yields,

$$(-1839)^3+(-1839+1871)^3+(-1839+2\cdot1871)^3 = 876^3$$

• The general case of cubics is dealed with in this article. – Cantlog Sep 10 '13 at 7:29

The point $P=(x,y)=(-1,0)$ is a point of inflection in the sense that the line $x=-1$ makes 3-fold contact there. This makes it a prime candidate to be moved to the point at infinity. Recalling that in Weierstrass form the two coordinates have poles of orders $2$ and $3$ respectively at the point of infinity (following the known process outlined e.g. in the proof of Proposition 3.1. of chapter 3 in The Arithmetic of Elliptic Curves, GTM#106 by J.Silverman) this suggests the pair $$u=\frac1{x+1},\qquad v=\frac{y}{x+1}$$ as new coordinates. These are chosen to have poles of orders $3$ and $2$ at $P$, and no other poles. This is clear in the case of $u$. To see that $v$ has no other poles we observe that your curve has three points at infinity with homogeneous coordinates $$[X_0:X_1:X_2]=[1,\omega^k\root3\of3:0],$$ $\omega=e^{2\pi i/3}, k=0,1,2$, where $v$ might potentially go to infinity. But we see that the ratio $v=y/(x+1)=X_1/(X_0+X_2)$ is well defined and finite at all those points, so none of these points are poles of $v$. Furthermore, $y$ is a local parameter at $P$, so the pole of $v$ at $P$ is of order $2$.
Indeed, the calculation $$v^3=\frac{y^3}{(x+1)^3}=\frac{3(x+1)[(x+1)^2+2]}{(x+1)^3}=3+\frac{6}{(x+1)^2}=3+6u^2$$ then shows that you can write your curve in the form $$6u^2=v^3-3.$$ To get to the Weierstrass form we should multiply this equation by $6^3=216$. This gives us $$(36u)^2=6^4u^2=216v^3-348=(6v)^3-3\cdot216.$$ The final coordinates should thus be $X=6v$ and $Y=36u$, and the Weierstrass form $$Y^2=X^3-648.$$ Altogether the transformation was $$X=\frac{6y}{x+1},\qquad Y=\frac{36}{x+1}.$$ So for example the point $Q=(x,y)=(3,6)$ in the original coordinates becomes $(X,Y)=(9,9).$ As the curve is now in Weierstrass form, $-Q=(X,Y)=(9,-9)$ is another rational point with original coordinates $(x,y)=(-5,-6).$ Of course, you could equally well do the curve's group operation in the original coordinates using the point $P$ as the neutral element. The three points at infinity in the original coordinates have now become the three points with $Y=0$.
• Thanks, Jyrki. I was hoping for the general transformation, but this detailed answer and references will do. As far back as Dickson's History of the Theory of Numbers he gives a transformation, but it is only for the general monic cubic (with leading coefficient $a=1$) as $x^3+bx^2+cx+d = y^3$. – Tito Piezas III Aug 28 '13 at 23:56