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() print([rank1,rank2])`

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 $a=\pm p^2$.

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