TITLE: What is the rank of the Mordell equation $y^2 = x^3 - 2$?
QUESTION [12 upvotes]: The mordell equation $E$ defined by $y^2 = x^3 - 2$ over $\mathbb{Q}$ is known to have only one non-trivial integer solution $P = (3,5)$ from here. However, the rank of Mordell-Weil group $E(\mathbb{Q})$ is not zero because formulas known since the time of Bachet show that $P$ has infinite order.
How can we compute the rank of $E$ without using MAGMA or the BSD conjecture? I'd be happy to see an argument by 3-descent.
The torsion subgroup of $E(\mathbb{Q})$ is known to be trivial.

REPLY [12 votes]: You could have a look at this paper:

M. Stoll, On the arithmetic of the curves  $y^2 = x^\ell + A$, II;
J. Number Theory 93, 183-206 (2002).
Corollary 2.1 says that for $A = -2$, one gets a rank bound of
$1$ plus twice the $3$-rank of the class group of ${\mathbb Q}(\sqrt{6})$.
So it only remains to verify that the class number of this field is not
divisible by $3$ (in fact, it is $1$).
The method is $(1-\zeta_3)$-descent, which is pretty close to a $3$-isogeny
descent. Apart from the class number determination, no computation is
necessary; it all follows from theoretical considerations.