I was playing around with sage, when I found that the Mordell-Weil ranks (over $\mathbb{Q}$) of the elliptic curves $y^2=x^3+p^3$ and $y^2=x^3-p^3 $ almost always agree, for $p$ prime. The first few exceptions occur at $p=37$, $p=61$, $p=157$, $p=193$, $\ldots$. This pattern struck me as odd, since the two curves are non-isogenous over the ground field, so why would their ranks be correlated?

After some reflection and further experimentation, I found out that if one looks instead at the $2$-Selmer ranks, there is even a stronger pattern: they seem to agree for all primes $p>2$.

I verified this using the following code, written in sage:

for p in primes(100):
    E1 = EllipticCurve(QQ,[0,p^3])
    E2 = EllipticCurve(QQ,[0,-p^3])
    print("p = "+QQ(p).str()+":"),
    rank1 = E1.selmer_rank()
    rank2 = E2.selmer_rank()

which gives

p = 2: [2, 1] p = 3: [1, 1] p = 5: [1, 1] p = 7: [2, 2] p = 11: [2, 2]
p = 13: [1, 1] p = 17: [1, 1] p = 19: [2, 2] p = 23: [2, 2] p = 29: [1, 1]
p = 31: [2, 2] p = 37: [3, 3] p = 41: [1, 1] p = 43: [2, 2] p = 47: [2, 2]
p = 53: [1, 1] p = 59: [2, 2] p = 61: [3, 3] p = 67: [2, 2] p = 71: [2, 2]
p = 73: [1, 1] p = 79: [2, 2] p = 83: [2, 2] p = 89: [1, 1] p = 97: [1, 1]

I have been trying to prove this by following a case distinction according to the residue class of $p$ modulo $12$, and performing a partial $2$-descent for each case, but I keep getting distracted by the thought that there must be a neater explanation that I'm missing. Hence my question:

Is there?

Edit: It might be useful to note that similar Sage experiments suggest that also (a) the $2$-Selmer ranks of the elliptic curves $y^2=x^3 \pm p$ and $y^2=x^3 \mp p^5$ agree for all $p>2$ and (b) the $2$-Selmer ranks of the elliptic curves $y^2=x^3 \pm p^2$ and $y^2=x^3 \mp p^4$ agree tot all $p>2$.

In fact, here's a conjecture, also born out by computer experiments, which goes even further and subsumes all cases mentioned before:

Conjecture. Let $a$ be an odd jnteger. Then the $2$-Selmer ranks of the elliptic curves $y^2=x^3 + a$ and $y^2=x^3-a^{-1}$ (which of course is isomorphic to $y^2=x^3 - a^5$) are equal.

We get the original statement with $a=p^3$, we get (a) with $a=\pm p$, and (b) with $a=\pm p^2$.

  • 2
    $\begingroup$ Both groups live in ${\mathbb Q}(\sqrt{-3})^\times$ mod squares. Did you check whether the groups are actually equal as subgroups of this quotient? $\endgroup$ Jan 8, 2016 at 20:45
  • $\begingroup$ No, but that's a good idea. What I did do was checking whether a certain more easily checkable consequence of (both groups being equal as subgroups of $\mathbb{Q}(\sqrt{-3})^\times/\mathbb{Q}(\sqrt{-3})^{\times 2}$) held, and it did. $\endgroup$
    – Milton
    Jan 8, 2016 at 21:05
  • 5
    $\begingroup$ Since the two elliptic curves are quadratic twists of each other (by the quadratic character cut out by $\mathbb{Q}(\sqrt{-1})$), their 2-torsion subgroups are canonically isomorphic as Galois modules, so one can identify $H^1(\mathbb{Q},E_1[2])$ with $H^1(\mathbb{Q},E_2[2])$. The respective 2-Selmer groups are therefore subgroups of this common overgroup, given by (potentially different) local conditions. By Mazur-Rubin, Ranks of twists of elliptic curves and Hilbert’s tenth problem, Lemma 2.10, the local conditions are equal at all places except possibly 2, 3, and $p$. contd... $\endgroup$
    – Alex B.
    Jan 8, 2016 at 23:37
  • 1
    $\begingroup$ ... One could try to analyse carefully what happens to the local conditions at those remaining places under twisting, which should lead to a comparison of the 2-Selmer ranks. $\endgroup$
    – Alex B.
    Jan 8, 2016 at 23:40
  • 1
    $\begingroup$ No, I would not necessarily expect the local conditions at 2, 3, and $p$ to be the same for the two curves. But if you get lucky, then the twisting changes these conditions in "opposite directions", so that the Selmer rank does not change in the end. As I say, this would actually require careful analysis. $\endgroup$
    – Alex B.
    Jan 9, 2016 at 11:15

2 Answers 2


Your curves seem to be quadratic twists of each other. There are many results in the literature about ranks of such curves. E.g. Ranks of quadratic twists of elliptic curves by Donnelly and others, using Stoll's formula for the size of a Selmer group. Chang (in Note on the rank of quadratic twists of Mordell equations) studied Mordell curves of certain type (your curves are Mordell curves).

  • 7
    $\begingroup$ Yes, I noticed they are quadratic twists; in particular, all the pairs become isomorphic over fields containing $\sqrt{-1}$. However, I do not see how that helps out a lot... $\endgroup$
    – Milton
    Jan 8, 2016 at 20:59

The (Selmer) ranks of the isogenous twists $y^2=x^3+D$ and $y^2=x^3-27D$ satisfy various relations that often mean the ranks are equal (or differ by 1). I realize these are not the pairs you're studying, but the theory, and especially the techniques, might well be applicable. (Actually, if you work over $\mathbb Q(\sqrt3)$, then $y^2=x^3-27D$ is isomorphic to $y^2=x^3-D$, so that would be your situation.) Here are a couple of references to get you started, they'll contain information about how the problem has been approached.

  • MR1451686 Chen, Yen-Mei J. The Selmer groups of elliptic curves and the ideal class groups of quadratic fields. Comm. Algebra 25 (1997), no. 7, 2157–2167.
  • MR0846960 Satgé, Philippe. Groupes de Selmer et corps cubiques. J. Number Theory 23 (1986), no. 3, 294–317.
  • $\begingroup$ +1. Thank you, I appreciate this. $\endgroup$
    – Milton
    Jan 10, 2021 at 21:59

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