TITLE: How could I calculate the rank of the elliptic curve $y^2 = x^3 - 432$?
QUESTION [13 upvotes]: 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?

REPLY [7 votes]: 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.