The exponent of Ш of $y^2 = x^3 + px$, where $p$ is a Fermat prime

For $d$ a non-zero integer, let $E_d$ be the elliptic curve $$E_d : y^2 = x^3+dx.$$ When we let $d$ be $p = 2^{2^k}+1$, for $k \in \{1,2,3,4\}$, sage tells us that, conditionally on BSD, $$\# Ш(E_p) = 2^{2k-2}.$$ Together with the fact that $\operatorname{Sel}^{(2)}(E_p) \cong (\mathbb{Z}/2\mathbb{Z})^2$ if $k = 1$ and $\operatorname{Sel}^{(2)}(E_p) \cong (\mathbb{Z}/2\mathbb{Z})^3$ if $k \geq 2$ (this can be found in X.6 of Silverman's book) the above implies, still assuming BSD, $$Ш(E_p) \cong (\mathbb{Z}/2^{k-1}\mathbb{Z})^2.$$

Question

Restating the above in a more explicit format, assuming BSD we have: \begin{align} Ш(E_5) & = 0 \\\ Ш(E_{17}) & \cong (\mathbb{Z}/2\mathbb{Z})^2 \\\ Ш(E_{257}) & \cong (\mathbb{Z}/4\mathbb{Z})^2 \\\ Ш(E_{65537}) & \cong (\mathbb{Z}/8\mathbb{Z})^2 \end{align} Is there any "reason" why this should be true? I realize that this is a "soft" question, but the above pattern seems so remarkable that I feel compelled to look for an explanation at some level.

An alternative question would run as follows: if $k \geq 1$ is an integer and $p=2^{2^k}+1$ is prime, do we always have $$Ш(E_p)\cong(\mathbb{Z}/2^{k-1}\mathbb{Z})^2?$$

A weaker question could be: if $k \geq 1$ is an integer and $p=2^{2^k}+1$ is prime, do we always have that the exponent of $Ш(E_p)$ is a multiple of $2^{k-1}$? (And what if we do not insist that $2^{2^k}+1$ be prime?) I have no idea how to prove lower bounds for the exponent of Ш of elliptic curves, other than running an explicit descent, but that doesn't seem the right method to attack the questions above.

How this came up: I was using sage to search for elliptic curves $E_d$ of the form $$E_d : y^2 = x^3 + dx$$ with the additional property that $$Ш(E_d)[2] = 0,$$ the underlying motivation being simply that for the course I am teaching I wanted to compile a list of elliptic curves on which a partial $2$-descent is possible without extending the ground field, and gives a sharp bound on the rank. I noticed then that the smallest $d$ with $\# Ш(E_d)[2^{\infty}] > 1$ is $d=17$, and the smallest $d$ with $\# Ш(E_d)[2^{\infty}] > 4$ is $d=257$.

While $2^{32}+1$ is of course not prime, out of curiosity I did try to compute the analytic rank and conjectural order of Ш of $E_d$ for $d=2^{32}+1$ (both with sage and magma), but the computations seem to be too lengthy to get an answer within a reasonable time.

• There are similar phenomena in connection with class groups of quadratic fields whose discriminants are Mersenne or Fermat primes. These things are not trivial. – Franz Lemmermeyer Dec 10 '13 at 17:21
• Indeed I was wondering about that. Is there a particular reference for this that you would recommend? – RP_ Dec 10 '13 at 17:23
• The value of the $L$-function at $s=1$ for (quartic) twists of $y^2 = x^3+x$ could help. Or maybe the $2$-adic $L$-function. But I don't know enough about CM-curves to help you. – Chris Wuthrich Dec 10 '13 at 22:22
• I think Chris is right: your statement is equivalent to $L(E_d,1)/\Omega=2,8,32$ for $d=17,257,65537$, and this quotient could be perhaps computed with modular symbols? (Or at least proving the congruence $L(E_p,1)/\Omega\cong 0$ mod $2^{...}$ using that $p=2^{2^k}+1$?) – Tim Dokchitser Dec 12 '13 at 17:56
• Magma returns some information about 2^32+1. Try the commands: d:=2^(2^5)+1;E := EllipticCurve([0, 0, 0, d, 0]);MordellWeilShaInformation(E);. Returns (Z/2)^4 <= Sha(E)[4] <= (Z/2)^4 and <4, [ 0, 0 ]> – joro Nov 3 '14 at 13:20