# What is the rational rank of the elliptic curve x^3 + y^3 = 2?

In Unsolved Problems in Number Theory, 3rd edition, section D1, 2004, R. K. Guy says, "Andrew Bremner has computed the rational rank of the elliptic curve x^3 + y^3 = Taxicab(n) as equal to 2, 4, 5, 4 for n = 2, 3, 4, 5, respectively." Since Taxicab(1) = 2, my question can be restated:

What is the rational rank of the elliptic curve x^3 + y^3 = Taxicab(1)?

It's isomorphic to $y^2 = x^3 - 27$, which has rank $0$.

• How do you show those two claims, or what is a reference? Thanks. Oct 25, 2013 at 20:27
• In particular, why is (1,1) a torsion point on x^3 + y^3 = 2 (as your answer implies)? Oct 25, 2013 at 20:34
• Since it has a rational point, it's isomorphic to its Jacobian (for which there are classical formulas - for instance see math.arizona.edu/~wmc/Research/JacobianFinal.pdf) which has the equation I wrote. Then mwrank (which you can call in sage, for instance) tells you it has rank $0$. You can also see that $(1,1)$ is a torsion point by seeing that the line $x + y = 2$ which is tangent to the cubic at $(1,1)$ has as the third point of intersection the point at infinity $(1:-1:0)$ (which you can take to be the origin of the elliptic curve). Oct 25, 2013 at 20:40
• Thanks. I cited your answer at oeis.org/A230564. Oct 27, 2013 at 13:15

To add to Abhinav's answer: the fact that $x^3+y^3=2$ has no solutions other then $x=y=1$ is attributed by Dickson to Euler himself: see Dickson's History of the Theory of Numbers (1920) Vol.II, Chapter XXI "Numbers the Sum of Two Rational Cubes", page 572. The reference (footnote 182) is "Algebra, 2, 170, Art. 247; French transl., 2, 1774, pp. 355–60; Opera Omnia, (1), I, 491". In the next page Dickson also refers to work of Legendre that includes this result (footnote 184: "Théorie des nombres, Paris, 1798, 409; ...").

This result is actually easier than the $n=3$ case of Fermat, because the curve has a $2$-torsion point, so only a $2$-descent is required, and here one soon finds that there are only two rational points (the second being the point at infinity $(x:y:1) = (1:-1:0)$). [It also happens that this curve has conductor $36$, while it's known that the smallest conductor of an elliptic curve of positive rank is $37$; but that's using a huge modern cannon to dispatch a classical fly.]

An easy way to bring the curve $x^3+y^3=2$ into Weierstrass form is to factor $x^3+y^3 = (x+y)(x^2-xy+y^2)$ and substitute $(x,y)=(r+s,r-s)$ to get $r^3+3rs^2=1$. Thus $s^2=(1-r^3)/(3r)$, so $3r(1-r^3)$ is to be a square. Now substitute $r=3/X$ to get $s^2 = (X^3-27)/(3X)^2$, or equivalently $Y^2=X^3-27$ where $y=3sX$. The solution $(x,y)=(1,1)$ corresponds to $(r,s)=(1,0)$ and then $(X,Y)=(3,0)$, which we recognize as a 2-torsion point because $Y=0$.

• Thanks. I cited your comment at oeis.org/A230564. Oct 27, 2013 at 13:18