How could I calculate the rank of the elliptic curve $y^2 = x^3 - 432$?

Question

The birational change of variables $(u,v) = (\frac{36+y}{6x},\frac{36-y}{6x})$ maps $u^3+v^3=1$ to $y^2 = x^3 - 432$ which has discriminant $-2^{12}\cdot 3^9$.

Using pari/gp we can compute the torsion subgroup:

? elltors(ellinit([0,0,0,0,-432]))
%1 = [3, [3], [[12, 36]]]

This says the torsion subgroup has order 3, is $\mathbf{Z}/3\mathbf{Z}$ and is generated by $(12,36)$ (which corresponds to $1^3+0^3=1^3$). The reason it has order 3 is because this also includes the projective solution $[0:0:1]$ of $X^3+Y^3=Z^3$.

Edit: By Nagell-Lutz one only needs to solve $y^2 = x^3 - 432$ in integers for $y=0$ and $y^2|2^{12}\cdot 3^9$ (which is a simple generate and test) to compute the elements of the torsion subgroup 'on paper'.

The group of rational points for this curve is then (by Mordell's Theorem) of the form $\mathbf{Z}^r \times \mathbf{Z}/3\mathbf{Z}$ where $r$ is the rank of the curve. If we can show the rank is 0 then this would prove fermats last theorem for $n = 3$.

How can it be shown directly the rank of this curve is 0?

Answer 1

I don't know much about this, but there is a theorem due to combined work of Zagier, Kolyvagin, Gross, etc, that says that if the analytic rank of the elliptic curve $E$ is 0 then it's rank is in fact 0. So you can compute the analytic rank for instance using SAGE and then by this theorem you will have shown that the rank is 0.

Of course this is like cheating since I think there ought to be some simpler way to do this, but you'll have to wait for the experts to respond you. For the time being, you can look at Theorem 5.8 and the definition preceding it in the following paper

http://www.ams.org/bull/2002-39-04/S0273-0979-02-00952-7/S0273-0979-02-00952-7.pdf

Answer 2

For some of the programs that output a rank, such as MAGMA, when they do so it is a rigorous certificate that the rank is the one displayed, and not only an apparent rank calculated from the L-function assuming the Birch and Swinnerton-Dyer conjecture. So although for some cases such as $L(E,1) \neq 0$ or using parity etc, the state of knowledge on BSD is enough to conclude the rank from accurate approximate calculation on the L-function and other invariants, these tricks are not needed given the program output.

Answer 3

The approved answer has caused some risibility at mathoverflow, and I'll elaborate on Robin's more reasonable comment (but I'm inclined to attribute the descent argument in this case to Euler--at least he wrote it down). The version I give in an undergrad number theory class is this: First one develops the standard facts about Z[w] where w^2+w+1=0. (It has unique factorization, 2 is prime, the units are 1,-1,w,-w,w^2 and -w^2, any element not 0 or a unit has absolute value >1, and each congruence class mod 2 is represented by 0 or a unit). Then one notes that it's enough to prove:

Theorem--There are no a,b,c in R with a+b+c=0, abc a non-zero cube and a=b=1 mod 2.

The proof of the theorem is a reductio. Let H be max(/a/,/b/,/c/) and choose a solution a,b,c with minimal H. (H^2 is an integer). a, b and c are evidently pairwise prime. Since their product is a non-zero cube, each is (unit)(cube). Since a=b=1 mod 2, a=A^3 and b=B^3, and we may assume that A=B=1 mod 2. Since abc is a cube, c=C^3 for some C in R. Since 2 divides c, 2 divides C and H is at least 8.

Now let S=Aw+Bw^2, T=Aw^2+Bw, and U=A+B. Then S+T+U=0 while STU is A^3+B^3=-C^3. Also S=T=1 mod 2, while max(/S/,/T/,/U/) is at most 2(H^(1/3)). This contradicts the minimality assumption.