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.

  • $\begingroup$ it seems the answer is "no" --- mathoverflow.net/a/202080/11260 $\endgroup$ Nov 26, 2016 at 19:49
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    $\begingroup$ That's about a criterion for a curve in the family $y^2 = x^3 + k$ to have rank zero. It can be much sipmler for a single curve in this family, which is what this question is asking (with $k=-2$). $\endgroup$ Nov 26, 2016 at 21:35
  • $\begingroup$ What might be a trivial integer solution? $\endgroup$ Nov 27, 2016 at 11:40
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    $\begingroup$ Hm, good point. Often for such equations there are trivial solutions with one variable equal zero; e.g. $y^2 = x^3 + 1$ has trivial solutions $(x,y) = (-1,0)$ (2-torsion) and $(0,\pm 1)$ (3-torsion), and also the nontrivial solutions $(2,\pm 3)$ (6-torsion) and no others. But $y^2 = x^3 - 2$ has no integer solutions with $xy=0$. There is the trivial element ("point at infinity") of the group of rational points, but that point is not integral. $\endgroup$ Nov 27, 2016 at 16:50

2 Answers 2


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.

  • $\begingroup$ I have looked everywhere but can't seem to find anything for 3-descent of this particular elliptic curve. I don't know why one would do 2-descent as it's a prime of bad reduction. $\endgroup$ Nov 27, 2016 at 0:19
  • $\begingroup$ 3 is also a prime of bad reduction . . . $$ $$ In general 2-descents are much easier than 3-descents; bad reduction at 2 (or 3) doesn't really affect this. The reason that this curve is a natural candidate for 3-descent is that it has a rational 3-isogeny, so you need only work over quadratic number fields, not octic which would be the usual situation. A quick skim through Dickson's History finds nothing on this particular curve, so possibly Mordell might have been the first to work it out, and you might try searching through his works. $\endgroup$ Nov 27, 2016 at 1:44
  • $\begingroup$ Also, the curve is simple enough that you can also look it up in LMFDB lmfdb.org/EllipticCurve/Q/1728/o/3 $-$ but of course the information there was also computed by 2-descent, and thus ultimately using mwrank. $\endgroup$ Nov 27, 2016 at 2:39

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.


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