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.
Entering the curve's coefficient vector [0,0,0,0,-2] into mwrank (which must be what MAGMA uses), we find that the rank is $1$ and the group of rational points is generated by $(3,5)$ (up to torsion, but gp quickly reports that the torsion is trivial). Now mwrank uses $2$-descent, which means that you need either the arithmetic of the pure cubic field ${\bf Q}(\root 3 \of 2)$ or some reduction theory of binary quartic forms. It can surely be done also using descent via the $3$-isogenies between $E$ and the curve $y^2 = x^3 + 54$, which will require only the arithmetic of ${\bf Q}(\sqrt{-2}\,)$ and ${\bf Q}(\sqrt6\,)$; but though that's in principle easier than $2$-descent it would still take much longer to do by hand than to press the mwrank button. But perhaps it's done somewhere as an exercise in a textbook or lecture notes.
You could have a look at this paper:
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.
