# Quadratic twists of genus one curves and Diophantine quintuples

Matija Kazalicki  Department of Mathematics
University of Zagreb
Bijenička cesta 30
10000 Zagreb
Croatia<EMAIL_ADDRESS>
###### Abstract.

Motivated by the theory of Diophantine $m$-tuples, we study rational points on
quadratic twists $H^{d}:dy^{2}=(x^{2}+6x-18)(-x^{2}+2x+2)$, where $|d|$ is a
prime. If we denote by
$S(X)=\\{d\in\mathbb{Z}:H^{d}(\mathbb{Q})\neq\emptyset,|d|\textrm{ is a
prime}\textrm{ and }|d|<X\\},$ then, by assuming some standard conjectures
about the ranks of elliptic curves in the family of quadratic twists, we prove
that as $X\rightarrow\infty$

$\frac{43}{256}+o(1)\leq\frac{\\#S(X)}{2\pi(X)}\leq\frac{46}{256}+o(1).$

###### Key words and phrases:

elliptic curves, quadratic twists, Selmer groups

###### 2010 Mathematics Subject Classification:

14H52, 11G05

##  1\. Introduction

For an integer $d$, a set of $m$ distinct nonzero rational numbers with the
property that the product of any two of its distinct elements plus $d$ is a
square is called a rational Diophantine $m$-tuple with the property $D(d)$ or
$D(d)$-$m$-tuple. The $D(1)$-$m$-tuples (with rational elements) are called
simply rational Diophantine $m$-tuples and have been studied since ancient
times, starting with Diophantus, Fermat, and Euler.

It is not known how large can a rational Diophantine tuple be. Dujella,
Kazalicki, Mikić, and Szikszai [DKMS17] proved that there are infinitely many
rational Diophantine sextuples, while no example of a rational Diophantine
septuple is known. Also, no example of rational $D(d)$-sextuple is known if
$d$ is not a perfect square. For more information on Diophantine $m$-tuples
see the survey article [Duj16].

We are interested in the following question.

######  Question.

Does there exist a rational $D(d)$-quintuple for every $d\in\mathbb{Z}$?

Dujella and Fuchs [DF12] proved that there are infinitely many squarefree
integers $d$’s for which there are infinitely many rational $D(d)$-quintuples,
and Dražić [Dra22] (improving the similar result from [DF12]) proved, assuming
the Parity conjecture for the quadratic twists of several explicitly given
elliptic curves, that for at least $99.5\%$ of squarefree integers $d$ there
are infinitely many rational $D(d)$-quintuples.

Following an idea from [DF12], we start with a
$D(\frac{16}{9}x^{2}(x^{2}-x-3)(x^{2}+2x-12))$-quintuple in $\mathbb{Z}[x]$

${\left\\{\frac{1}{3}(x^{2}+6x-18)(-x^{2}+2x+2),\frac{1}{3}x^{2}(x+5)(-x+3),(x-2)(5x+6),\frac{1}{3}(x^{2}+4x-6)(-x^{2}+4x+6),4x^{2}\right\\}}$

found by Dujella [Duj99] (and used to prove that there are infinitely many
$D(-1)$-quintuples in [Duj02]). Note that for rational $u\neq 0$, if
$\\{a,b,c,d,e\\}$ is $D(qu^{2})$-quintuple, then
$\\{\frac{a}{u},\frac{b}{u},\frac{c}{u},\frac{d}{u},\frac{e}{u}\\}$ is
$D(q)$-quintuple. In particular, for squarefree integer $d$, if

$dy^{2}=(x^{2}-x-3)(x^{2}+2x-12)$

for some $x,y\in\mathbb{Q}$ then by dividing the elements of quintuple above
with $\frac{4}{3}xy$ we obtain $D(d)$-quintuple. Thus, if the equation above
has infinitely many solution, we may conclude that there are infinitely many
$D(d)$-quintuples.

Consider the genus one quartic

$H:\quad y^{2}=(x^{2}-x-3)(x^{2}+2x-12).$

For a squarefree integer $d$, we denote by $H^{d}:dy^{2}=f(x)$ the quadratic
twist of $H$ with respect to $\mathbb{Q}(\sqrt{d})$. Quartic $H$, as a
(singular) genus one curve with a rational point at infinity, is birationally
equivalent to the elliptic curve $E/\mathbb{Q}$

$E:y^{2}=(x-9)(x-8)(x+18).$

Likewise, we denote by $E^{d}$ the quadratic twist of $E$ by
$\mathbb{Q}(\sqrt{d})$. Thus $H^{d}(\mathbb{Q})\neq\emptyset$ implies that
$H^{d}$ is birationally equivalent to $E^{d}$. Since, by Proposition 3.4,
$H^{d}(\mathbb{Q})\neq\emptyset$ implies that $H^{d}(\mathbb{Q})$ is infinite
and consequently that there are infinitely many $D(d)$-quintuples, we are led
to the study of squarefree integers $d$ for which
$H^{d}(\mathbb{Q})\neq\emptyset$.

In this paper we will focus on twists by $\mathbb{Q}(\sqrt{d})$ where $|d|$ is
prime. Let

$S=\\{d\in\mathbb{Z}:H^{d}(\mathbb{Q})\neq\emptyset\textrm{ and }|d|\textrm{
is a prime}\\}.$

######  Question.

What is asymptotically the size of set $S(X)=\\{d\in S:|d|<X\\}$ as
$X\rightarrow\infty$?

Surprisingly, and in contrast with the analogous problem for the quadratic
twists of elliptic curves, not much is known about this question.

Çiperiani and Ozman gave a criterion for the set of rational points of the
quadratic twist of quartic to be non-empty in terms of the image of the global
trace map $tr_{\mathbb{Q}(\sqrt{d})/\mathbb{Q}}$ on an elliptic curve (see
Section 2 of [cO15]), but in general, no estimates for the size of set $S(X)$
are known.

For a squarefree $d$, the quartic $H^{d}$, as a $2$-covering of $E^{d}$,
represents an element of $\operatorname{Sel^{(2)}}(E^{d})$, the $2$-Selmer
group of $E^{d}$, provided that $H^{d}$ is everywhere locally solvable (i.e.
$H^{d}(\mathbb{Q}_{v})\neq\emptyset$ for all places $v$ – we write ELS for
short). For the interpretation of Selmer group elements as $2$-covers of $E$
see Section 1.2 of [Sto12].

If $|d|=p$ is a prime, then Proposition 2.1 implies that $H^{d}$ is ELS if and
only if $\genfrac{(}{)}{}{}{p}{13}=1$ or $p=13$. Thus, for such $d$,
$H^{d}(\mathbb{Q})=\emptyset$ if and only if $H^{d}$ represents a nontrivial
element in $\Sha(E^{d})[2]$ (where $\Sha(E^{d})$ denotes the Tate-Shafarevich
group of $E^{d}$), or more precisely, if and only if the image of $H^{d}$
under the map $\iota:\operatorname{Sel^{(2)}}(E^{d})\rightarrow\Sha(E^{d})[2]$
from the exact sequence

(1.1) $0\longrightarrow
E^{d}(\mathbb{Q})/2E^{d}(\mathbb{Q})\longrightarrow\operatorname{Sel^{(2)}}(E^{d})\longrightarrow\Sha(E^{d})[2]\longrightarrow
0$

is nonzero. In this case, we say that $H^{d}$ represents the element of order
two in $\Sha(E^{d})$.

Our main tool for studying the image of $H^{d}$ in $\Sha(E^{d})[2]$ is the
Cassels-Tate pairing on $\Sha(E^{d})$ with values in $\mathbb{Q}/\mathbb{Z}$,
or more precisely, its extension to a pairing on $2$-Selmer group by (1.1)

$\langle\cdot,\cdot\rangle_{CT}:\operatorname{Sel^{(2)}}(E^{d})\times\operatorname{Sel^{(2)}}(E^{d})\rightarrow\mathbb{Z}/2\mathbb{Z}=\\{0,1\\}.$

This pairing is bilinear, alternating, and non-degenerate on
$\Sha(E^{d})[2]/2\Sha(E^{d})[4]$, or equivalently, on
$\operatorname{Sel^{(2)}}(E^{d})/2\,\textrm{Sel}^{(4)}(E^{d})$ (see Section
4). In particular, $\dim_{\mathbb{F}_{2}}\Sha(E^{d})[2]/2\Sha(E^{d})[4]$ is
even, thus equal to $0$ or $2$ if $|d|$ is a prime (see Proposition 3.1).
Thus, if we find a class $L\in\operatorname{Sel^{(2)}}(E^{d})$ such that
$\langle H^{d},L\rangle_{CT}=1$, we can conclude that $\iota(H^{d})\neq 0$,
and, hence, that $H^{d}$ represents the element of order two in $\Sha(E^{d})$.
If $\Sha(E^{d})[2]$ is nontrivial and $\Sha(E^{d})[2]=2\Sha(E^{d})[4]$ (see
Proposition 3.9), then we can not obtain any information about $H^{d}$ using
this method.

For estimating the asymptotic behaviour of $\\#S(X)$ as $X\rightarrow\infty$
we will assume the following “standard” conjectures.

######  Conjecture 1.

$100\%$ of quadratic twists $E^{d}$ where $|d|$ is a prime have rank $0$ or
$1$.

Note that this conjecture is now a theorem under the BSD conjecture if we let
$d$ range over all squarefree integers (see Smith [Smi22a, Smi22b]).

######  Conjecture 2 (The parity conjecture).

For all $d\in\mathbb{Z}$ where $|d|$ is prime,

$(-1)^{\operatorname{rank}(E^{d})}=w(E^{d}),$

where $w(E^{d})$ is the root number of the elliptic curve $E^{d}$.

It follows from Proposition 3.4 that the contribution of $d$’s ($|d|$ is a
prime) for which the root number $w(E^{d})$ is equal to $1$ to the $\\#S(X)$
is negligible since by Conjecture 1 $100\%$ of the curves $E^{d}$ will have
rank $0$ or $1$ and by Conjecture 2 that rank is even, hence zero.

On the other hand, in the case $w(E^{d})=-1$, if
$\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})=3$ (see Proposition 3.4
for the description of $\operatorname{Sel^{(2)}}(E^{d})$) then by Conjecture 2
$\operatorname{rank}(E^{d})=1$ so $\Sha(E^{d})[2]$ is trivial (note that
$E^{d}$ has full rational two torsion, hence
$\dim_{\mathbb{F}_{2}}\Sha(E^{d})[2]=\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})-\operatorname{rank}(E^{d})-2=0$).

Hence the only interesting case (in which we expect $\Sha(E^{d})[2]$
generically to be nontrivial) is when
$\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})=5$ or equivalently (see
Proposition 3.1) when $d\in T=T^{+}\cup T^{-}$ where

$\displaystyle T^{+}$ $\displaystyle=\\{d>0:|d|\textrm{ is
prime},\genfrac{(}{)}{}{}{d}{13}=1,\genfrac{(}{)}{}{}{d}{3}=1,d\equiv
1\pmod{8}\\},$ $\displaystyle T^{-}$ $\displaystyle=\\{d<0:|d|\textrm{ is
prime},\genfrac{(}{)}{}{}{d}{13}=1,\genfrac{(}{)}{}{}{d}{2}\cdot\genfrac{(}{)}{}{}{d}{3}=-1,d\equiv
5,7\pmod{8}\\}.$

Define

(1.2)
$\displaystyle\begin{split}H_{1}&:y^{2}=4x^{4}-56x^{2}+169\in\operatorname{Sel^{(2)}}(E),\\\
H_{2}&:y^{2}=18x^{4}-24x^{3}-32x^{2}+40x+34\in\operatorname{Sel^{(2)}}(E),\\\
F_{1}&:y^{2}=11x^{4}+12x^{3}+56x^{2}+24x+68\in\operatorname{Sel^{(2)}}(E^{-1}),\\\
F_{2}&:y^{2}=x^{4}+56x^{2}+676\in\operatorname{Sel^{(2)}}(E^{-1}).\\\
\end{split}$

We show in Proposition 3.1 that if $d\in T$, $\operatorname{Sel^{(2)}}(E^{d})$
is generated by the image of the two torsion $E^{d}[2]$ under the Kummer map,
$H^{d}$, and by the quadratic twists of those classes in (1.2) which land in
$\operatorname{Sel^{(2)}}(E^{d})$. Hence for such $d$’s
$\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})=5$. Proposition 2.3
describes when these twists of quartics in (1.2) are ELS. Note that this
simple explicit description of $\operatorname{Sel^{(2)}}(E^{d})$ (see
Proposition 3.1) is the main reason why we considered only quadratic twists by
$d$ where $|d|$ is prime. In general, for squarefree $d$,
$\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})$ is unbounded.

Assuming the parity conjecture for $E^{d}$, where $d\in T$, we can deduce that
$\dim_{\mathbb{F}_{2}}\Sha(E^{d})[2]=0$ or $2$. Assume further that
$\Sha(E^{d})[2]\neq 2\Sha(E^{d})[4]$. The non-degeneracy of the Cassels-Tate
pairing implies that for $d$ such that $\iota(H^{d})\neq 0$ there exists class
$L\in\operatorname{Sel^{(2)}}(E^{d})$ (also with $\iota(L)\neq 0$) for which
$\langle H^{d},L\rangle_{CT}=1$. The following theorem then follows easily
from Section 4, Proposition 3.1 and the previous discussion.

######  Theorem 1.1.

Let $d\in T$ such that $\Sha(E^{d})[2]\neq 2\Sha(E^{d})[4]$. Assuming the
parity conjecture for $E^{d}$, the following is true.

  1. a)

If $d<0$ and $d\equiv 1\pmod{4}$ then $\langle
H^{d},F_{1}^{-d}\rangle_{CT}=1$. In particular, $\iota(H^{d})\neq
0\in\Sha(E^{d})[2]$.

  2. b)

If $d<0$ and $d\equiv 3\pmod{4}$ then $\iota(H^{d})\neq 0$ if and only if
$\langle H^{d},F_{2}^{-d}\rangle_{CT}=1$.

  3. c)

If $d>0$ then $\iota(H^{d})\neq 0$ if and only if $\langle
H^{d},H_{1}^{d}\rangle_{CT}=1$ or $\langle H^{d},H_{2}^{d}\rangle_{CT}=1$.

It remains to explain how to compute the Cassels-Tate pairing of the quadratic
twists of quartics. To each pair $(A,B)$ of quartics from Table 1 (see (1.2)),
by the work of Smith (see Theorem 3.2. in [Smi16]), we can associate the
governing field $L_{A,B}$ such that the value of pairing $\langle
A^{d},B^{d}\rangle_{CT}$ is determined by $\langle A,B\rangle_{CT}$ and the
splitting behaviour of $d$ in $L_{A,B}$. For example, for $d\in T$, it follows
that $\langle H^{d},H^{d}_{2}\rangle_{CT}=0$ if and only if $d$ splits
completely in
$L=\mathbb{Q}(\sqrt{-1},\sqrt{2},\sqrt{13})(\sqrt{4+2\sqrt{13}})$. For
complete description of governing fields see Table 1 and Section 4. Section 4
and Proposition 3.9 imply the following corollary of Theorem 1.1.

$\langle A^{d},B^{d}\rangle_{CT}$ | $K_{A,B}$ | $\alpha_{A,B}$
---|---|---
$\langle H^{d},H_{1}^{d}\rangle_{CT}$ | $\mathbb{Q}(\sqrt{3},\sqrt{13})$ | $4+\sqrt{13}$
$\langle H^{d},H_{2}^{d}\rangle_{CT}$ | $\mathbb{Q}(\sqrt{-1},\sqrt{2},\sqrt{13})$ | $4+2\sqrt{13}$
$\langle H^{-d},F_{1}^{d}\rangle_{CT}$ | $\mathbb{Q}(\sqrt{-2},\sqrt{13})$ | $-1$
$\langle H^{-d},F_{2}^{d}\rangle_{CT}$ | $\mathbb{Q}(\sqrt{13},\sqrt{-1},\sqrt{-3})$ | $3(1+\sqrt{13})(3+\sqrt{13})$
$\langle H_{1}^{d},H_{2}^{d}\rangle_{CT}$ | $\mathbb{Q}(\sqrt{3},\sqrt{-1},\sqrt{2})$ | $8(1+\sqrt{3})(4+2\sqrt{3})$
$\langle F_{1}^{d},F_{2}^{d}\rangle_{CT}$ | $\mathbb{Q}(\sqrt{3},\sqrt{-1},\sqrt{2})$ | $8(1+\sqrt{3})(4+2\sqrt{3})$
Table 1.  For $d=p>0$ which splits completely in $K_{A,B}$ (and in the case
$\langle H^{d},H_{1}^{d}\rangle_{CT}$ we in addition require $p\equiv
1\pmod{4}$), we have $\langle A^{d},B^{d}\rangle_{CT}=0$ if and only if $d$
splits completely in a governing field $L_{A,B}=K_{A,B}(\sqrt{\alpha_{A,B}})$.

######  Corollary 1.2.

Let $d\in T$. Assuming the parity conjecture for $E^{d}$, if $d$ does not
split completely in $L_{H_{1},H_{2}}=L_{F_{1},F_{2}}$ and

  1. a)

$d=-p<0$ with $p\equiv 1\bmod{4}$ and $p$ splits completely in
$L_{H^{-1},F_{2}}$, or

  2. b)

$d=p>0$ and $p$ splits completely in $L_{H,H_{1}}$ and $L_{H,H_{2}}$,

then $H^{d}(\mathbb{Q})\neq\emptyset$. Hence, for such $d$ there exists
infinitely many $D(d)$-quintuples.

######  Remark 1.3.

As we already noted, if for $d=\pm p$ we have that
$\genfrac{(}{)}{}{}{p}{13}=1$ and
$\genfrac{(}{)}{}{}{p}{2}\cdot\genfrac{(}{)}{}{}{p}{3}\cdot\genfrac{(}{)}{}{}{p}{13}=1$
(hence $w(E^{d})=-1$ by Proposition 2.4), but if $d\notin T$ (hence
$\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})=3$), then by Conjecture
2 $\Sha(E^{d})[2]$ is trivial, and $H^{d}(\mathbb{Q})\neq\emptyset$, so there
exists infinitely many $D(d)$-quintuples.

######  Example 1.4.

The set of $d\in T$, $|d|<3000$, for which Corollary 1.2 implies that
$H^{d}(\mathbb{Q})\neq\emptyset$ is equal to

$\\{-2857,-2833,-1993,-601,-337,-313,1993,2833,2857\\}.$

For $d=-313$, we find a point $(-2107/1202,389073/1444804)\in
H^{-313}(\mathbb{Q})$ which produces a $D(-313)$-quintuple

$\left\\{\frac{81062614477261}{1313828969096},\frac{15660515591}{623554328},\frac{9009021853}{546517874},\frac{28246175292437}{1313828969096},\frac{2532614}{129691}\right\\}.$

######  Remark 1.5.

Results about infinite number of $D(d)$-quintuples obtained as above from
$d\in T$ where $d<0$ are new, they are not covered in [Dra22].

Using Chebotarev’s density theorem to determine the factorization of primes in
governing fields, we obtain the following bounds for $S(X)$.

######  Corollary 1.6.

Assuming Conjecture 1, we have that as $X\rightarrow\infty$

$C_{1}+o(1)\leq\frac{\\#S(X)}{2\pi(X)}\leq C_{2}+o(1),$

where $C_{1}=\frac{43}{256}$ and $C_{2}=\frac{46}{256}$.

######  Remark 1.7.

We can rephrase the result above by saying that the classes
$H^{d}\in\operatorname{Sel^{(2)}}(E^{d})$, for $d\in T$, are “equidistributed”
in the quotient $\operatorname{Sel^{(2)}}(E^{d})/\kappa({E^{d}[2]})-\\{0\\}$
with respect to the image of rational points $E^{d}(\mathbb{Q})$ (rank is
generically $1$) under the Kummer map
$\kappa:E^{d}(\mathbb{Q})/2E^{d}(\mathbb{Q})\rightarrow\operatorname{Sel^{(2)}}(E^{d})$,
since we have that the probability for
$H^{d}\in\kappa({E^{d}(\mathbb{Q})})/\kappa({E^{d}[2]})\subset\operatorname{Sel^{(2)}}(E^{d})/\kappa({E^{d}[2]})$
is $1/7$. Recall that by Proposition 3.4 $H^{d}$ is never an element of
$\kappa({E^{d}[2]})$ (thus $1/7$ and not $1/8$ is the “right” answer).

By Proposition 3.9 and Conjecture 1 the density of $d$’s ($|d|$ is a prime)
for which $\Sha(E^{d})[2]$ is nontrivial and $\Sha(E^{d})[2]=2\Sha(E^{d})[4]$
is $\frac{3}{256}$, so with this method we can not bridge the gap between
$C_{1}$ and $C_{2}$.

##  2\. Local properties

######  Proposition 2.1.

For a square-free $d\in\mathbb{Z}$, the quartic $H^{d}$ is everywhere locally
solvable if and only if for all primes $p|d$ we have
$\genfrac{(}{)}{}{}{p}{13}=1$ or $p=13$.

###### Proof.

Assume that $H^{d}$ is ELS. It follows that for every prime $p|d$, $p\neq 13$,
the equation $(x^{2}-x-3)(x^{2}+2x-12)=0$ has a solution in $\mathbb{F}_{p}$,
which implies that $\genfrac{(}{)}{}{}{p}{13}=1$ since the discriminant of
quadratic factors is $13$ and $4\cdot 13$ respectively.

Conversely assume that for all primes $p|d$ we have
$\genfrac{(}{)}{}{}{p}{13}=0$ or $1$. Obviously,
$H^{d}(\mathbb{R})\neq\emptyset$. If $p|d$, then by assumption there is a
solution $(x^{2}-x-3)(x^{2}+2x-12)=0$ in $\mathbb{F}_{p}$ which lifts by
Hensel lemma to $H^{d}(\mathbb{Q}_{p})$. If $p\nmid 2\cdot 3\cdot 13d$, then
$H^{d}$ has a good mod $p$ reduction since $2,3$ and $13$ are only primes
dividing discriminant of $(x^{2}-x-3)(x^{2}+2x-12)$. It follows that
$H^{d}/\mathbb{F}_{p}$ is a genus one curve, hence
$H^{d}(\mathbb{F}_{p})\neq\emptyset$, thus by Hensel’s lemma
$H^{d}(\mathbb{Q}_{p})\neq\emptyset$. It remains to consider cases $p=2,3,13$
and $p\nmid d$. Here reductions mod $2$, $3$ and $13$ of $H^{d}$ are
geometrically irreducible genus zero curve, so it follows that
$H^{d}(\mathbb{F}_{p})\neq\emptyset$, and consequently
$H^{d}(\mathbb{Q}_{p})\neq\emptyset$ for $p=2,3,13$. ∎

######  Remark 2.2.

Novak [Nov22] showed (assuming GRH) that asymptotically the number of
squarefree $d$’s, $0<d<x$, for which $H^{d}$ is ELS is equal to

$\frac{2\sqrt{273}}{13}\pi^{-3/2}\prod_{p}\left(1+\frac{1}{p}\right)^{\genfrac{(}{)}{}{}{p}{13}/2}\frac{x}{\sqrt{\log{x}}}.$

Similarly, the following proposition describes local solvability of quartics
from (1.2).

######  Proposition 2.3.

Let $p$ be a prime and $d=\pm p$.

  1. a)

$H_{1}^{d}$ is everywhere locally solvable if and only if $d\equiv 1\pmod{12}$
or $d=-3$.

  2. b)

$H_{2}^{d}$ is everywhere locally solvable if and only if $d>0$ and $d\equiv
1\pmod{8}$.

  3. c)

$F_{1}^{d}$ is everywhere locally solvable if and only if $d>0$ and $d\equiv
1,3\pmod{8}$.

  4. d)

$F_{2}^{d}$ is everywhere locally solvable if and only if $d>0$ and $d\equiv
1\pmod{12}$.

The following proposition computes the root number of $E^{d}$.

######  Proposition 2.4.

For $d=\pm p$ where $p\neq 2,3,13$ is a prime, the root number $w(E^{d})$ is
equal to $-1$ if and only if

$\genfrac{(}{)}{}{}{p}{2}\cdot\genfrac{(}{)}{}{}{p}{3}\cdot\genfrac{(}{)}{}{}{p}{13}=1.$

Here $\genfrac{(}{)}{}{}{\cdot}{2}$ is the Kronecker symbol for odd $d$
defined by

$\genfrac{(}{)}{}{}{d}{2}=\begin{cases}1,&\textrm{if }|d|\equiv 1,7\bmod(8)\\\
-1,&\textrm{if }|d|\equiv 3,5\bmod(8).\end{cases}$

###### Proof.

Theorem 1.1. in [Des20] implies that

(2.1)
$w(E^{d})=-w_{2}(E^{d})w_{3}(E^{d})w_{13}(E^{d})\genfrac{(}{)}{}{}{-1}{p},$

where $w_{p}(E^{d})$ is a local root number at $p$ of $E^{d}$. Since $E^{d}$
has multiplicative reduction at $13$, Proposition 2 in [Roh93] implies that
$w_{13}(E^{d})=-\genfrac{(}{)}{}{}{6b}{13}$ where $b=64108800d^{3}$, thus
$w_{13}=-\genfrac{(}{)}{}{}{p}{13}$ (since $w_{13}(E^{d})=-1$ if and only if
the reduction is split multiplicative). Likewise, for $p\neq 3$, $E^{d}$ has
multiplicative reduction at $3$, hence
$w_{3}(E^{d})=-\genfrac{(}{)}{}{}{d}{3}=-sgn(d)\genfrac{(}{)}{}{}{p}{3}$.
Moreover, since $j$-invariant $j(E^{d})=\frac{22235451328}{123201}$ is
integral at $2$, $E^{d}$ has additive, potentially good reduction at $2$. One
can check that $v_{2}(c_{4}(E^{d}))=4$, $c_{4}(E^{d})/2^{4}\equiv 3\pmod{4}$,
$v_{2}(c_{6}(E^{d}))=7$ and $v_{2}(\Delta(E^{d}))=6$, hence it follows from
Table 1. in [Hal98] that $w_{2}(E^{d})=1$ if and only if
$c_{4}/2^{4}-4c_{6}/2^{7}\equiv 7,11\pmod{16}$. Thus, one can check that
$w_{2}(E^{d})=1$ if and only if $d\equiv 1,3\pmod{8}$, or equivalently
$w_{2}(E^{d})=\genfrac{(}{)}{}{}{-1}{d}\genfrac{(}{)}{}{}{d}{2}=sgn(d)\genfrac{(}{)}{}{}{-1}{p}\genfrac{(}{)}{}{}{p}{2}$.
Claim now follows from (2.1). ∎

##  3\. Structure of $\operatorname{Sel^{(2)}}(E^{d})$

In this section we describe the structure of $\operatorname{Sel^{(2)}}(E^{d})$
in the case when $|d|$ is prime. We prove the following proposition.

######  Proposition 3.1.

For prime $p\neq 2,3,13$, let $d=\pm p$ be such that
$\genfrac{(}{)}{}{}{d}{13}=1$ and $w(E^{d})=-1$.

  1. a)

If $d\in T$ (i.e. $d\equiv 1\pmod{8}$ if $d>0$ or $d\equiv 5,7\pmod{8}$ if
$d<0$), then $\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})=5$. More
precisely, if $d>0$, then $\operatorname{Sel^{(2)}}(E^{d})$ is generated by
torsion classes, $H^{d}$, $H_{1}^{d}$ and $H_{2}^{d}$. If $d<0$, then
$\operatorname{Sel^{(2)}}(E^{d})$ is generated by torsion classes, $H^{d}$,
$F_{1}^{-d}$, and $F_{2}^{-d}$ if $d\equiv 7\pmod{8}$ or $H_{1}^{d}$ if
$d\equiv 5\pmod{8}$.

  2. b)

If $d\notin T$, then we have that
$\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})=3$.

Since $E$ has full $2$-torsion over $\mathbb{Q}$, each class in
$H^{1}(\mathbb{Q},E[2])$ can be identified with an element of
$(\mathbb{Q}^{\times}/\mathbb{Q}^{\times 2})^{3}$ in the following way. Denote
by $P_{1}=(8,0),P_{2}=(-18,0)$ and $P_{3}=(9,0)$ nontrivial elements in
$E[2]$, by $e_{2}:E[2]\times E[2]\rightarrow\mu_{2}$ the Weil pairing (hence
$e_{2}(P_{i},P_{j})=-1$ if and only if $i\neq j$), and by
$\omega:E[2]\rightarrow\operatorname{Hom}(E[2],\mu_{2}^{3})$,
$T\mapsto(P_{i}\mapsto e_{2}(T,P_{i}))$ the group homomorphism induces by
$e_{2}$. For each class $F\in H^{1}(\mathbb{Q},E[2])$, we denote by
$\omega_{*}(F)$ the pushforward of $\omega$ from $H^{1}(\mathbb{Q},E[2])$ to
$H^{1}(\mathbb{Q},\mu_{2}^{3})\cong
H^{1}(\mathbb{Q},\mu_{2})^{3}\cong(\mathbb{Q}^{\times}/\mathbb{Q}^{\times
2})^{3}$ where the last isomorphism is given by the Kummer map sending
$\alpha\in\mathbb{Q}^{\times}/\mathbb{Q}^{\times 2}$ to $\xi\in
H^{1}(\mathbb{Q},\mu_{2})$ such that
$\xi(\sigma)=\frac{\sqrt{\alpha}^{\sigma}}{\sqrt{\alpha}}$ for every
$\sigma\in\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})}$. One has that
$\omega_{*}(F)=(a_{1},a_{2},a_{3})$ is equivalent to
$F(\sigma)=\chi_{a_{1}}(\sigma)P_{1}+\chi_{a_{2}}(\sigma)P_{2}$, for all
$\sigma\in\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})}$, and
$a_{1}a_{2}a_{3}\in\mathbb{Q}^{\times 2}$, where, for $a\in\mathbb{Q}$. Here,
we denote by $\chi_{a}$ the nontrivial character of $\mathbb{Q}(\sqrt{a}))$
with values in $\mathbb{Z}/2\mathbb{Z}$ (if $\mathbb{Q}(\sqrt{a})=\mathbb{Q}$
then $\chi_{a}$ is trivial). It follows that $F$ is defined over
$\mathbb{Q}(\sqrt{a_{1}},\sqrt{a_{2}})$.

We start with the following standard lemma.

######  Lemma 3.2.

For elliptic curve $\tilde{E}:y^{2}=(x-a_{1})(x-a_{2})(x-a_{3})$, where
$a_{1},a_{2},a_{3}\in\mathbb{Q}$, let $F$ be a quartic $y^{2}=g(x)$,
$g(x)\in\mathbb{Z}[x]$, isomorphic (over $\overline{\mathbb{Q}}$) to
$\tilde{E}$, which represents an element in $H^{1}(\mathbb{Q},\tilde{E}[2])$
(the quartic is not necessarily everywhere solvable, i.e. the element of the
$2$-Selmer group). For $d\in\mathbb{Z}$ let $F^{d}$ be the quadratic twist of
$F$, thus representing the element in $H^{1}(\mathbb{Q},\tilde{E}^{d}[2])$.
After identifying $H^{1}(\mathbb{Q},\tilde{E}[2])\cong
H^{1}(\mathbb{Q},\tilde{E}^{d}[2])$, we have

$\omega_{*}(F)=\omega_{*}(F^{d}).$

###### Proof.

The claim follows directly from the interpretation of the map $\omega_{*}$ in
terms of two-descent theory. If $\omega_{*}(F)=(q_{1},q_{2},q_{3})$, then $F$
is isomorphic (over $\mathbb{Q}$) to the curve

$\displaystyle q_{1}y^{2}$ $\displaystyle=x-a_{1},$ $\displaystyle q_{2}y^{2}$
$\displaystyle=x-a_{2},$ $\displaystyle q_{3}y^{2}$ $\displaystyle=x-a_{3},$

while its twist over $\mathbb{Q}(\sqrt{d})$ is given by

$q_{1}y^{2}=x-da_{1},\quad q_{2}y^{2}=x-da_{2},\quad q_{3}y^{2}=x-da_{3}.$

where the isomorphism $F\rightarrow F^{d}$ maps
$(x,y_{1},y_{2},y_{3})\mapsto(dx,\sqrt{d}y_{1},\sqrt{d}y_{2},\sqrt{d}y_{3})$.
Since $\tilde{E}^{d}$ is isomorphic to $y^{2}=(x-da_{1})(x-da_{2})(x-da_{3})$,
we recognize from the above that $\omega_{*}(F^{d})=(q_{1},q_{2},q_{3})$ (we
identified $(a_{i},0)$ with $(da_{i},0)$), and the claim follows. ∎

For the proof of Proposition 3.1, we need to introduce three more quartics.

$\displaystyle H_{3}$
$\displaystyle:y^{2}=25x^{4}+48x^{3}-114x^{2}-144x+225\in Sel(E^{3}),$
$\displaystyle F_{3}$
$\displaystyle:y^{2}=-71x^{4}-336x^{3}-538x^{2}-336x-71\in\operatorname{Sel^{(2)}}(E^{-1}),$
$\displaystyle F_{4}$
$\displaystyle:y^{2}=-5x^{4}+76x^{3}-168x^{2}-296x-92\in\operatorname{Sel^{(2)}}(E^{-3}).$

Recall that quadratic twist $E^{d}$ has the Weierstrass model
$E^{d}:y^{2}=(x-8d)(x-9d)(x+18d)$.

Next, we prove linear independence of classes needed for the proof of
Proposition 3.1.

######  Lemma 3.3.

For $d\in\mathbb{Z}$, $|d|$ prime, and $|d|\notin\\{2,3,13\\}$, denote by
$Q_{1}=(8d,0)$ and $Q_{2}=(-18d,0)$ elements in $E^{d}[2]$ which correspond to
$P_{1}$ and $P_{2}$ under the natural isomorphism $E[2]\cong E^{d}[2]$, and by
$\kappa:E^{d}(\mathbb{Q})/2E^{d}(\mathbb{Q})\rightarrow\operatorname{Sel^{(2)}}(E^{d})\subset
H^{1}(\mathbb{Q},E^{d}[2])$ the Kummer map. We have that

$\displaystyle\omega_{*}(H^{d})$
$\displaystyle=(13,13,1),\omega_{*}(\kappa(Q_{1}))=(26d,-26,-d),\omega_{*}(\kappa(Q_{2}))=(78,-26d,-3d),$
$\displaystyle\omega_{*}(H_{1}^{d})$
$\displaystyle=(3,1,3),\omega_{*}(H_{2}^{d})=(2,-2,-1),\omega_{*}(H_{3}^{d})=(6,-6,1),$
$\displaystyle\omega_{*}(F_{1}^{d})$
$\displaystyle=(-2,-2,1),\omega_{*}(F_{2}^{d})=(-3,-1,3),\omega_{*}(F_{3}^{d})=(6,2,3),\omega_{*}(F_{4}^{d})=(6,6,1).$

Moreover,

  1. a)

if $d>0$, the the classes $\omega_{*}(F)$, for
$F\in\\{\kappa(Q_{1}),\kappa(Q_{2}),H^{d},H_{1}^{d},H_{2}^{d},H_{3}^{d},F_{3}^{-d}\\}$
are (multiplicatively) independent in $(\mathbb{Q}^{\times}/\mathbb{Q}^{\times
2})^{3}$ and locally solvable at infinity,

  2. b)

if $d<0$, the classes $\omega_{*}(F)$, for
$F\in\\{\kappa(Q_{1}),\kappa(Q_{2}),H^{d},H_{1}^{d},F_{1}^{-d},F_{3}^{-d},F_{4}^{-3d}\\}$
are (multiplicatively) independent in $(\mathbb{Q}^{\times}/\mathbb{Q}^{\times
2})^{3}$, and locally solvable at infinity.

###### Proof.

Using Magma [BCP97], we can easily compute the values of $\omega_{*}(F^{d})$
for quartics $F$ from 1.2 as they don’t depend on $d$ by Lemma 3.2.

We can also compute classes of torsion points explicitly. For example, for
$Q_{1}=(8d,0)\in E^{d}(\mathbb{Q})$, one can check that $2R_{1}=Q_{1}$, where
$R_{1}=(\frac{1}{2}r^{2}-\frac{9d}{2},\frac{1}{2}r^{3}-\frac{25d}{2})$, with
$r^{4}-50dr^{2}+3^{6}d^{2}=0$. Here
$\mathbb{Q}(r)=\mathbb{Q}(\sqrt{-d},\sqrt{-26})$, and by inspection one
obtains that
$R_{1}^{\sigma}-R_{1}=\chi_{26d}(\sigma)Q_{1}+\chi_{-26}(\sigma)Q_{2}$, thus
$\omega_{*}(\kappa(Q_{1}))=(26d,-26,-d)$. Similarly, one computes
$\omega_{*}(\kappa(Q_{2}))$.

The existence of real points on quartic (which determine local solvability at
infinity) can be checked for each quartic separately.

If $d>0$, it is not hard to see that the classes will be independent unless
$d$ is divisible only by $2,3$ and $13$. In particular, for squarefree $d$, we
compute that this happens for $\\{1,2,3,6,13,26,39,78\\}$, thus the claim in
$a)$ follows. The claim in b) is proved in a similar way.

∎

We have the following proposition as a consequence of the previous lemma.

######  Proposition 3.4.

If $d\in\mathbb{Z}$ is square free integer such that
$H^{d}(\mathbb{Q})\neq\emptyset$, then $H^{d}(\mathbb{Q})$ is infinite.

###### Proof.

Assume that for some $d\in\mathbb{Z}$, $H^{d}(\mathbb{Q})\neq\emptyset$ and
$H^{d}(\mathbb{Q})$ is finite. It follows that the rank of Mordell-Weil group
of $E^{d}(\mathbb{Q})$ is zero, hence $H^{d}$ as an element of $2$-Selmer
group $\operatorname{Sel^{(2)}}(E^{d})$ is in the image of the two torsion
$E^{d}[2]$ under the map
$E^{d}(\mathbb{Q})/2E^{d}(\mathbb{Q})\hookrightarrow\operatorname{Sel^{(2)}}(E^{d})$
from (1.1). More precisely, there is a point of order $4$, $Q\in E^{d}[4]$,
such that $H^{d}$ corresponds to the cocycle $\sigma\mapsto Q^{\sigma}-Q$. It
follows from Lemma 3.3 that the image of this cocycle is of order $2$ which
implies that $Q$ is defined over quadratic field. There are only finitely many
$d^{\prime}s$ that have a point of order $4$ defined over quadratic field.
Note that if $x_{0}$ is an $x$-coordinate of point of order $4$ on $E$ (it is
defined over quadratic field), then $d\cdot x_{0}$ is an $x$-coordinate of
point of order $4$ on $E^{d}$. Moreover, if
$E_{d}:y^{2}=f_{d}(x)=(x-8d)(x-9d)(x+18d)$, then $f_{d}(dx_{0})=d^{3}\cdot
f_{1}(x_{0})$ is a square in $\mathbb{Q}(x_{0})$ if and only if $d\cdot
f_{1}(x_{0})$ is a square. One can check that this is the case if and only if
$d=\\{-26,-3,-1,1,3,26\\}$. The proposition follows after verifying the claim
for these special cases. ∎

To obtain an upper bound for the size of $2$-Selmer group, we will use the
method and terminology from the paper of Mazur and Rubin [MR10][Section 3]
(see also [Kra81, BD10]).

######  Definition 3.5.

Suppose $\tilde{E}$ is an elliptic curve over $\mathbb{Q}$. For every place
$v$ of $\mathbb{Q}$, let $H_{f}(\mathbb{Q}_{v},\tilde{E}[2])$ denote the image
of the Kummer map

$\tilde{E}(\mathbb{Q}_{v})/2\tilde{E}(\mathbb{Q}_{v})\rightarrow
H^{1}(\mathbb{Q}_{v},\tilde{E}[2]).$

The $2$-Selmer group $\operatorname{Sel^{(2)}}(\tilde{E})$ is the
$\mathbb{F}_{2}$-vector space defined by the exactness of the sequence

$0\rightarrow\operatorname{Sel^{(2)}}(\tilde{E})\rightarrow
H^{1}(\mathbb{Q},\tilde{E}[2])\rightarrow\bigoplus_{v}H^{1}(\mathbb{Q}_{v},\tilde{E}[2])/H^{1}_{f}(\mathbb{Q}_{v},\tilde{E}[2]).$

We say that $2$-Selmer group $\operatorname{Sel^{(2)}}(\tilde{E})$ is cut out
by the local conditions $H_{f}(\mathbb{Q}_{v},\tilde{E}[2])$.

The following lemma describes the size of local conditions.

######  Lemma 3.6.

Let $v$ be a finite rational place and $d$ an odd squarefree integer. We have

$\dim_{\mathbb{F}_{2}}H_{f}^{1}(\mathbb{Q}_{v},E^{d}[2])=\begin{cases}2&\textrm{
if }v\neq 2\\\ 3&\textrm{ if }v=2.\\\ \end{cases}$

###### Proof.

By Lemma 2.2 in [MR10], if $v\nmid\,2\infty$, then
$\dim_{\mathbb{F}_{2}}H_{f}^{1}(\mathbb{Q}_{v},E^{d}[2])=\dim_{\mathbb{F}_{2}}E^{d}(\mathbb{Q}_{v})[2]=2$.

Following [Sil09, Chapter 4.], denote by $\mathcal{F}$ the formal group
associated to the elliptic curve $E^{d}/\mathbb{Q}_{2}$, and by
$\mathcal{F}(2\mathbb{Z}_{2})$ the group associated to that formal group.
Theorem 6.4. b) in [Sil09] implies that $\mathcal{F}(4\mathbb{Z}_{2})$ is
isomorphic (via formal logarithm map) to the additive group
$\hat{\mathbb{G}}_{a}(4\mathbb{Z}_{2})$ which implies that
$\mathcal{F}(4\mathbb{Z}_{2})/2\mathcal{F}(4\mathbb{Z}_{2})\cong\mathbb{Z}/2\mathbb{Z}$.
On the other hand, since $\mathcal{F}(x,y)=x+y-a_{1}xy-
a_{2}(x^{2}y+xy^{2})+\cdots$, where $a_{1}$ and $a_{2}$ are the usual
Weierstrass coefficients of $E^{d}$, it follows that $[2](x)=2x+O(x^{3})$ (as
$a_{1}=0$), thus $2\mathcal{F}(2\mathbb{Z}_{2})=\mathcal{F}(4\mathbb{Z}_{2})$.
In particular,
$\mathcal{F}(2\mathbb{Z}_{2})/2\mathcal{F}(2\mathbb{Z}_{2})\cong\mathbb{Z}/2\mathbb{Z}$.

If we denote by $E^{d}_{1}(\mathbb{Q}_{2})$ the subgroup of points in
$E^{d}(\mathbb{Q}_{2})$ which reduce to the point at infinity modulo two, then
it is well known that
$E^{d}_{1}(\mathbb{Q}_{2})\cong\mathcal{F}(2\mathbb{Z}_{2})$. Moreover,
$E^{d}_{0}(\mathbb{Q}_{2})/E^{d}_{1}(\mathbb{Q}_{2})$, where
$E^{d}_{0}(\mathbb{Q}_{2})$ is the subgroup of points of nonsingular
reduction, is generated by two torsion point with odd $x$ coordinate. Finally,
$E^{d}(\mathbb{Q}_{2})/E^{d}_{0}(\mathbb{Q}_{2})$ is generated by the point of
order two with even $x$ coordinate (Tamagawa number of $E^{d}$ is two), and we
have that
$E^{d}(\mathbb{Q}_{2})/2E^{d}(\mathbb{Q}_{2})\cong(\mathbb{Z}/2\mathbb{Z})^{3}$,
so the claim follows. ∎

There is a natural identification of Galois modules $E[2]\cong E^{d}[2]$ \-
which is crucial for our argument. We identify point $(a,0)\in E(\mathbb{Q})$
with $(8a,0)\in E^{d}(\mathbb{Q})$ for $a\in\\{8,9,-18\\}$. It allows us to
view $\operatorname{Sel^{(2)}}(E^{d})$ as a subspace of the
$H^{1}(\mathbb{Q},E[2])$, but defined by the different sets of local
conditions $H^{1}_{f}(\mathbb{Q}_{v},E^{d}[2])\subset
H^{1}(\mathbb{Q}_{v},E[2])$.

######  Definition 3.7.

If $\tilde{T}$ is a finite set of places of $\mathbb{Q}$, define relaxed
$2$-Selmer group $\mathcal{S}^{\tilde{T}}$ by the exactness of

$0\rightarrow\mathcal{S}^{\tilde{T}}\rightarrow
H^{1}(\mathbb{Q},E[2])\rightarrow\bigoplus_{v\notin\tilde{T}}H^{1}(\mathbb{Q}_{v},E[2])/H^{1}_{f}(\mathbb{Q}_{v},E[2]),$

where the second arrow is induced by the sum of localization maps
$H^{1}(\mathbb{Q},E[2])\rightarrow H^{1}(\mathbb{Q}_{v},E[2])$.

By definition $\operatorname{Sel^{(2)}}(E)\subset\mathcal{S}^{\tilde{T}}$ for
any $\tilde{T}$. We will choose $\tilde{T}$ such that
$\operatorname{Sel^{(2)}}(E^{d})\subset\mathcal{S}^{\tilde{T}}$ holds as well.
For that we will need the following criteria for equality of local conditions
after twist (see Lemma 2.10 and Lemma 2.11 in [MR10]).

######  Lemma 3.8.

Let $\tilde{E}/\mathbb{Q}$ be an elliptic curve. Let $v$ be a place of
$\mathbb{Q}$ and $d$ a squarefree integer. If at least one of the following
conditions holds

  1. a)

$v$ splits in $\mathbb{Q}(\sqrt{d})$,

  2. b)

$v$ is a prime of good reduction of $\tilde{E}$ and $v$ is unramified in
$\mathbb{Q}(\sqrt{d})/\mathbb{Q}$,

then
$H^{1}_{f}(\mathbb{Q}_{v},\tilde{E}[2])=H^{1}_{f}(\mathbb{Q}_{v},\tilde{E}^{d}[2])$.
Moreover, if $\tilde{E}$ has good reduction at $v$, and $v$ is ramified in
$\mathbb{Q}(\sqrt{d})/\mathbb{Q}$, then

$H^{1}_{f}(\mathbb{Q}_{v},\tilde{E}[2])\cap
H^{1}_{f}(\mathbb{Q}_{v},\tilde{E}^{d}[2])=0.$

Since primes of bad reduction of $E^{d}$ are $\\{2,3,13,p\\}$, and since $13$
splits in $\mathbb{Q}(\sqrt{d})$, it follows from Lemma 3.8 that local
conditions $H^{1}_{f}(\mathbb{Q}_{v},E^{d}[2])$ and
$H^{1}_{f}(\mathbb{Q}_{v},E[2])$ are equal outside the set
$\tilde{T}=\\{2,3,p,\infty\\}$.

###### Proof of Proposition 3.1.

Lower bound for the $\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})$ in
both cases follows from Lemma 3.3 and Proposition 2.3. Note that if $d\in T$,
the classes $H_{1}^{d}$ and $H_{2}^{d}$ are ELS if $d>0$, classes $F_{1}^{-d}$
and $F_{2}^{-d}$ are ELS if $d<0$ and $d\equiv 7\pmod{8}$, and classes
$F_{1}^{-d}$ and $H_{1}^{d}$ are ELS if $d<0$ and $d\equiv 5\pmod{8}$.

For the upper bound we first consider the case $d\in T$. From the definition
of $T$ it follows that for $|d|>3$ primes $2$ and $3$ split in
$\mathbb{Q}(\sqrt{d})$, thus Lemma 3.8 implies that local conditions
$H^{1}_{f}(\mathbb{Q}_{v},E^{d}[2])$ and $H^{1}_{f}(\mathbb{Q}_{v},E[2])$
differ only at $v=p$ (and possibly at $v=\infty$ if $d<0$ \- note that if
$d>0$ elliptic curves $E$ and $E^{d}$ are isomorphic over $\mathbb{R}$).

Assume that $d>0$ and set $\tilde{T}=\\{p\\}$. Define a strict $2$-Selmer
group $\mathcal{S}_{\tilde{T}}:=\mathcal{S}_{\tilde{T}}(E)$ by the exactness
of

$0\rightarrow\mathcal{S}_{\tilde{T}}\rightarrow\mathcal{S}^{\tilde{T}}\rightarrow\bigoplus_{v\in\tilde{T}}H^{1}(\mathbb{Q}_{v},E[2]),$

where the second arrow is the sum of the localization maps.

From the construction, it follows that
$\mathcal{S}_{\tilde{T}}\subset\operatorname{Sel^{(2)}}(E^{d})\subset\mathcal{S}^{\tilde{T}}$,
and
$\mathcal{S}_{\tilde{T}}\subset\operatorname{Sel^{(2)}}(E)\subset\mathcal{S}^{\tilde{T}}$.
We will show that $\mathcal{S}_{\tilde{T}}=\operatorname{Sel^{(2)}}(E)$. One
can compute that $E(\mathbb{Q})$ is generated by $2$-torsion points
$S_{1}=(-18,0)$, $S_{2}=(8,0)$ and point $S_{3}=(45/4,-117/8)$ of infinite
order, and that $\operatorname{Sel^{(2)}}(E)$ is generated by the
$\kappa(S_{i})$, $i=1,2,3$, where
$\kappa:E(\mathbb{Q})/2E(\mathbb{Q})\rightarrow\operatorname{Sel^{(2)}}(E)$ is
the Kummer map - thus $\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E)=3$. It
is enough to show that the image of the $\kappa(S_{i})$ in
$H^{1}(\mathbb{Q}_{p},E[2])$ is trivial. Choose $Q_{i}\in
E(\overline{\mathbb{Q}})$ such that $2Q_{i}=S_{i}$. The fields of definitions
$K_{i}$ of points $Q_{i}$ are $K_{1}=\mathbb{Q}(\alpha_{1})$ where
$\alpha_{1}^{4}+106\alpha_{1}^{2}+1=0$, $K_{2}=\mathbb{Q}(\alpha_{2})$ where
$\alpha_{2}^{4}-50\alpha_{2}^{2}+729=0$ and $K_{3}=\mathbb{Q}(\alpha_{3})$
where $\alpha_{3}^{2}-3\alpha_{3}-43/4=0$. It happens that $p$ splits
completely in all the fields, hence the claim follows.

Lemma 3.2 in [MR10] implies that
$\dim_{\mathbb{F}_{2}}\mathcal{S}^{\tilde{T}}-\dim_{\mathbb{F}_{2}}\mathcal{S}_{\tilde{T}}=\dim_{\mathbb{F}_{2}}H^{1}_{f}(\mathbb{Q}_{p},E[2])$.
By Lemma 3.6 and inclusion
$\operatorname{Sel^{(2)}}(E^{d})\subset\mathcal{S}^{T}$, it follows
$\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})\leq 3+2=5$, and the
claim follows.

The case $d<0$ is analogous - to get the equality of local conditions at
$v=\infty$ one replaces $E$ with $E^{-1}$, and then proceeds as in the $d>0$
case.

Now assume that $d\notin T$. Consider the case $d<0$. In the case $d>0$ one
repeats the same argument with $E^{-1}$ replaced by $E$. Primes $2$ and $3$ do
not need to split in $\mathbb{Q}(\sqrt{d})$ any more, hence we set
$\tilde{T}=\\{2,3,p\\}$ and
$\mathcal{S}^{\tilde{T}}:=\mathcal{S}^{\tilde{T}}(E^{-1})$ and
$\mathcal{S}_{\tilde{T}}:=\mathcal{S}_{\tilde{T}}(E^{-1})$ (we replaced $E$
with $E^{-1}$ in definitions to ensure the equality of local conditions at
$v=\infty$). Lemma 3.2 in [MR10] and Lemma 3.6 imply that
$\dim_{\mathbb{F}_{2}}\mathcal{S}^{\tilde{T}}-\dim_{\mathbb{F}_{2}}\mathcal{S}_{\tilde{T}}=\dim_{\mathbb{F}_{2}}H^{1}_{f}(\mathbb{Q}_{2},E^{-1}[2])+\dim_{\mathbb{F}_{2}}H^{1}_{f}(\mathbb{Q}_{3},E^{-1}[2])+\dim_{\mathbb{F}_{2}}H^{1}_{f}(\mathbb{Q}_{p},E^{-1}[2])=3+2+2=7$.
Since $\mathcal{S}_{\tilde{T}}\subset\operatorname{Sel^{(2)}}(E^{-1})$, if we
show that the image of each class in $\operatorname{Sel^{(2)}}(E^{-1})$ (which
is generated by $H^{-1},F_{1}$ and $F_{3}$) under the localization
$\operatorname{\textrm{loc}}_{2}:\operatorname{Sel^{(2)}}(E^{-1})\rightarrow
H^{1}(\mathbb{Q}_{2},E^{-1}[2])$ is different than zero, then it follows that
$\mathcal{S}_{\tilde{T}}={0}$. One can check that, for any $P\in
E^{-1}(\mathbb{Q})/2E^{-1}(\mathbb{Q})$ and $Q\in
E^{-1}(\overline{\mathbb{Q}})$ such that $2Q=P$, $2$ is ramified in the field
of definition of $Q$, hence the localization of $\kappa(P)$ at $v=2$ is
nontrivial, and $\mathcal{S}_{\tilde{T}}={0}$. It follows that
$\dim_{\mathbb{F}_{2}}\mathcal{S}^{\tilde{T}}=7$.

Lemma 3.3 b) provides us with the generators of $\mathcal{S}^{\tilde{T}}$ once
we show that the torsion classes together with classes
$H,F_{1},H_{1},F_{2},F_{4}\in H^{1}(\mathbb{Q},E)$ satisfy local conditions
$H^{1}_{f}(\mathbb{Q}_{v},E^{-1}[2])$ for $v$ outside the set $\tilde{T}$.
Equivalently, one can check that the quartics $H^{-1},F_{1},H_{1}^{-1},F_{2}$
and $F_{4}^{3}$ (as two covers of $E^{-1}$) are locally solvable outside the
set $\tilde{T}$. Local solvability at the finite places outside the set
$\\{2,3,13\\}$ follows immediately from Hensel lemma argument (as in the proof
of Proposition 2.1) since these are the bad primes of $E^{-1}$, while
solvability at $v=\infty$ (i.e. existence of the real points on quadratic
twists) follows from the observation that polynomials of degree $4$ defining
$H$ and $H_{1}$ have real roots. The local solvability at $v=13$ follows from
the fact that $\genfrac{(}{)}{}{}{p}{13}=1$, which implies that $p$ is a
square in $\mathbb{Q}_{13}$, thus quadratic twist by $\mathbb{Q}(\sqrt{d})$ or
$\mathbb{Q}(\sqrt{-d})$ of any quartic from Lemma 3.3 b) is isomorphic over
$\mathbb{Q}_{13}$ to that quartic. Hence, we only need to check that
$F_{4}^{3}$ is locally solvable at $v=13$ which is checked readily.

We will prove that $\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})\leq
4$, which will imply that
$\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})=3$ since
$\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})$ is odd (by [DD10]
$(-1)^{\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})}=w(E^{d})=-1$) and
greater or equal to $3$ (since $H^{d}$ and the torsion classes of $E^{d}$ are
linearly independent in $\operatorname{Sel^{(2)}}(E^{d})$). Essentially, for
each class in $\mathcal{S}^{\tilde{T}}$ (generators are given by Lemma 3.3
b)), we will check if it satisfies the local conditions
$H_{f}^{1}(\mathbb{Q}_{v},E^{d}[2])$.

Observe that the local condition at $v=p$,
$H_{f}^{1}(\mathbb{Q}_{p},E^{d}[2])$, for $p\neq\\{2,3,13\\}$ is determined
with the image of $2$-torsion
$\kappa(P_{1})(\sigma)=\chi_{3}(\sigma)P_{1}+\chi_{d}(\sigma)P_{3}$,
$\kappa(P_{2})=\chi_{-13d}(\sigma)P_{1}+\chi_{-2}(\sigma)P_{3}$ (since the
elements are independent and dimension of the local condition is $2$). As the
remaining generators of $\mathcal{S}^{\tilde{T}}$,
$H:\sigma\mapsto\chi_{13}(\sigma)P_{3}$,
$F_{1}:\sigma\mapsto\chi_{-2}(\sigma)P_{3}$,
$H_{1}:\sigma\mapsto\chi_{3}(\sigma)P_{1}$,
$F_{4}:\sigma\mapsto\chi_{6}(\sigma)P_{3}$ and
$F_{2}+H_{1}=\chi_{-1}(\sigma)P_{3}$ do not depend on $d$ (here $\chi_{q}$
denotes the nontrivial character of $\mathbb{Q}(\sqrt{q}$)), the local
condition at $v=p$ can be satisfied by some class from the subspace generated
by $H,H_{1},F_{1},F_{4}$ and $F_{2}$ only if the localization of that class at
$v=p$ is trivial. If $p\equiv 5\pmod{8}$, then $-1,13$ are squares in
$\mathbb{Q}_{p}$ while $2$ and $3$ are not, thus $H$, $F_{2}+H_{1}$ and
$F_{4}$ generate the subspace of $\mathcal{S}^{\tilde{T}}$ with required
property, while if $p\equiv 7\pmod{8}$, then $13,2$ are squares in
$\mathbb{Q}_{p}$ while $-1$ and $3$ are not, thus $H,F_{1}+F_{4}$ and
$F_{1}+F_{2}+H_{1}$ generate the subspace of $\mathcal{S}^{\tilde{T}}$
consisting of elements whose localization at $v=p$ is trivial.

Next, to rule out remaining classes, we focus on the local condition at $v=3$.
If $p\equiv 5\pmod{8}$, then $d$ is a square in $\mathbb{Q}_{3}$, and the
classes
$\operatorname{\textrm{loc}}_{3}\kappa(P_{1})(\sigma)=\chi_{3}(\sigma)P_{1}$
and
$\operatorname{\textrm{loc}}_{3}\kappa(P_{2})(\sigma)=\chi_{-1}(\sigma)P_{1}$
linearly independent, thus they generate $2$-dimensional
$\mathbb{F}_{2}$-vector space $H_{f}^{1}(\mathbb{Q}_{3},E^{d}[2])$. Since,
$\operatorname{\textrm{loc}}_{3}(F_{4})(\sigma)=\chi_{6}(\sigma)P_{3}=\chi_{-3}(\sigma)P_{3}\notin
H_{f}^{1}(\mathbb{Q}_{3},E^{d}[2])$, we conclude that in this case
$\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})\leq 4$, hence equal to
$3$.

If $p\equiv 7\pmod{8}$, then
$\operatorname{\textrm{loc}}_{3}\kappa(P_{1})(\sigma)=\chi_{3}(\sigma)P_{1}+\chi_{-1}(\sigma)P_{3}$
and $\operatorname{\textrm{loc}}_{3}\kappa(P_{2})(\sigma)=0$ generate a
$1$-dimensional subspace of the $2$-dimensional vector space
$H_{f}^{1}(\mathbb{Q}_{3},E^{d}[2])$. Note that not all the localisations of
the classes of interest
$\operatorname{\textrm{loc}}_{3}(F_{1}+F_{4})(\sigma)=\chi_{-3}(\sigma)P_{3}$
and
$\operatorname{\textrm{loc}}_{3}(F_{1}+F_{2}+H_{1})(\sigma)=\chi_{-1}(\sigma)P_{3}$
can lie in $H_{f}^{1}(\mathbb{Q}_{3},E^{d}[2])$ (since the subspace they
generated does not contain
$\operatorname{\textrm{loc}}_{3}\kappa(P_{1})(\sigma)$), hence
$\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})\leq 4$, and the claim
follows. ∎

The following proposition follows immediately from the explicit description of
$\operatorname{Sel^{(2)}}(E^{d})$ given in Proposition 3.1.

######  Proposition 3.9.

Let $d\in T$ (hence $\dim_{\mathbb{F}_{2}}\operatorname{Sel^{(2)}}(E^{d})=5$).
We have that $\Sha(E^{d})[2]=2\Sha(E^{d})[4]$ if and only if

  1. a)

$\langle H_{1}^{d},H_{2}^{d}\rangle_{CT}=0$ and $\langle
H^{d},H_{i}^{d}\rangle_{CT}=0$ for $i=1,2$ if $d>0$,

  2. b)

$\langle F_{1}^{-d},F_{2}^{-d}\rangle_{CT}=0$ and $\langle
H^{d},F_{i}^{-d}\rangle_{CT}=0$ for $i=1,2$ if $d<0$ and $d\equiv 7\pmod{8}$,

###### Proof.

If $\Sha(E^{d})[2]=2\Sha(E^{d})[4]$, then the Cassels-Tate pairing on
$\operatorname{Sel^{(2)}}(E^{d})$ is trivial (since it is non-degenerate on
$\Sha(E^{d})[2]/2\Sha(E^{d})[4]$), hence the claim follows. Similarly, if
a),b) o holds, then Proposition 3.1 implies the Cassels-Tate pairing on
$\operatorname{Sel^{(2)}}(E^{d})$ is trivial, hence
$\Sha(E^{d})[2]=2\Sha(E^{d})[4]$. Note that in the case $d<0$ and $d\equiv
5\pmod{8}$, we always have $\langle H^{d},F_{1}^{-d}\rangle_{CT}=1$ (see
Theorem 1.1a)), hence $\Sha(E^{d})[2]\neq 2\Sha(E^{d})[4]$. ∎

##  4\. Cassels-Tate pairing and governing fields

Our main tool for studying Cassels-Tate pairing of quadratic twists of
elements of $2$-Selmer groups is the following specialisation of the theorem
of Smith (see Section $3$ in [Smi16]).

######  Theorem 4.1 (Smith).

Let $\tilde{E}$ be an elliptic curve over $\mathbb{Q}$ with full $2$-torsion
over $\mathbb{Q}$. Let

$F,F^{\prime}\in H^{1}(\mathbb{Q},\tilde{E}[2]),$

and let $K$ be the minimal field over which $F$ and $F^{\prime}$ are trivial.
Next, let $S$ be any set of places of $\mathbb{Q}$ which contains all places
of bad reduction of $\tilde{E}$, the archimedean place and $2$. Take
$\mathcal{D}$ to be the set of pairs $(d_{1},d_{2})$ of elements in
$\mathbb{Q}^{\times}$ such that $d_{1}/d_{2}$ is square at all places of $S$,
and $F^{d_{1}}$ and $F^{\prime d_{2}}$ are elements of $2$-Selmer group of
$\tilde{E}^{d_{1}}$ and $\tilde{E}^{d_{2}}$ respectively.

If $F\cup F^{\prime}$ is alternating (as defined in Section $3$ of [Smi16]),
then $\langle F^{d_{1}},F^{\prime d_{1}}\rangle_{CT}=\langle
F^{d_{2}},F^{\prime d_{2}}\rangle_{CT}$ for all $(d_{1},d_{2})\in\mathcal{D}$.
Otherwise, there is a quadratic extension $L$ of $K$ that is ramified only at
primes in $S$ such that

$\langle F^{d_{1}},F^{\prime d_{1}}\rangle_{CT}=\langle F^{d_{2}},F^{\prime
d_{2}}\rangle_{CT}+\left[\frac{L/K}{\mathbf{d}}\right],$

for all $(d_{1},d_{2})\in\mathcal{D}$, where the Galois group
$\operatorname{Gal}(L/K)$ is identified with
$\frac{1}{2}\mathbb{Z}/\mathbb{Z}$. Here $\mathbf{d}$ is any ideal of $K$
coprime to the conductor of $L/K$ that has norm in
$\mathbb{Q}^{\times}/{\mathbb{Q}^{\times}}^{2}$ equal to $(d_{1}/d_{2})$. Such
$\mathbf{d}$ exists for all $(d_{1},d_{2})\in\mathcal{D}$. We denote by
$\left[\frac{\ \cdot\ }{\cdot}\right]$ the Artin symbol.

######  Remark 4.2.

We will call field $L$ from the statement of Theorem 4.1 a governing field of
$F$ and $F^{\prime}$. It needs not to be unique.

Next, we compute the governing fields of some pairs of classes defined by
quartics from (1.2) (see Table 1).

In general, following Section 3.1. in [Smi16], for $F,F^{\prime}\in
H^{1}(\mathbb{Q},E[2])$ let $\omega_{*}(F)=(a_{1},a_{2},a_{3})$ and
$\omega_{*}(F^{\prime})=(a_{1}^{\prime},a_{2}^{\prime},a_{3}^{\prime})$. For
every place $v$ we have the following relation of Hilbert symbols
$(a_{1},a_{1}^{\prime})_{v}(a_{2},a_{2}^{\prime})_{v}(a_{3},a_{3}^{\prime})_{v}=1$.
We can choose $b\in\mathbb{Q}^{\times}$ such that
$(a_{1},ba_{1}^{\prime})_{v}=(a_{2},ba_{2}^{\prime})_{v}=(a_{3},ba_{3}^{\prime})_{v}=1$
which implies that we can find $x_{i},y_{i},z_{i}\in\mathbb{Q}^{\times}$ such
that $x_{i}^{2}-a_{i}y_{i}^{2}=ba_{i}^{\prime}z_{i}^{2}$ for $i=1,2,3$. We can
further scale $x_{i},y_{i}$ and $z_{i}$ by a common factor so that the field

$L_{F_{,}F^{\prime}}=K_{F,F^{\prime}}\left(\sqrt{(x_{1}+y_{1}\sqrt{a_{1}})(x_{2}+y_{2}\sqrt{a_{2}})(x_{3}+y_{3}\sqrt{a_{3}})}\right)$

avoids ramification at places unramified in the common field of definition

$K_{F,F^{\prime}}:=\mathbb{Q}(\sqrt{a_{1}},\sqrt{a_{2}},\sqrt{a_{1}^{\prime}},\sqrt{a_{2}^{\prime}}).$

######  Lemma 4.3 (Smith).

If $F\cup F^{\prime}$ is not alternating and $\deg
K_{F,F^{\prime}}/\mathbb{Q}=16$, then $L_{F,F^{\prime}}$ is a governing field
of $F$ and $F^{\prime}$.

Although in our case $\deg K_{F,F^{\prime}}/\mathbb{Q}$ is either four or
eight, we can still compute governing fields using the following lemma which
follows from the proof of Proposition 2.1. in [Smi16].

######  Lemma 4.4.

For integers $a$ and $b$ such that $ab$ is not a perfect square let
$L_{a,b}/\mathbb{Q}(\sqrt{a},\sqrt{b})$ be quadratic extension such that
$L_{a,b}/\mathbb{Q}$ is Galois with Galois group isomorphic to dihedral group
$D_{8}$. There exist a map

$\gamma_{a,b}:\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})}\xtwoheadrightarrow{res}\operatorname{Gal}(L_{a,b}/\mathbb{Q})\rightarrow\mu_{2}$

which satisfies $d\gamma_{a,b}=\chi_{a}\cup\chi_{b}\in
H^{2}(\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})},\mu_{2})$. Here
$\mu_{2}=\\{\pm 1\\}$ and the cup product $\chi_{a}\cup\chi_{b}$ is induced by
the natural bilinear map
$\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}\rightarrow\mathbb{Z}/2\mathbb{Z}$
(hence for
$\sigma,\tau\in\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})}$ we have
that $(\chi_{a}\cup\chi_{b})(\sigma,\tau)=-1$ if and only if
$\sqrt{a}^{\,\sigma}=-\sqrt{a}$ and $\sqrt{b}^{\,\tau}=-\sqrt{b}$).

###  4.1.
$L_{H^{-1},F_{2}}=\mathbb{Q}(\sqrt{13},\sqrt{-1},\sqrt{-3})(\sqrt{3(1+\sqrt{13})(3+\sqrt{13})})$

It follows from Lemma 3.3 that
$H^{-1}(\sigma)=\chi_{13}(\sigma)P_{1}+\chi_{13}(\sigma)P_{2}$ and
$F_{2}(\sigma)=\chi_{-3}(\sigma)P_{1}+\chi_{-1}(\sigma)P_{2}$ for all
$\sigma\in\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})}$. If we define
the cup product
$\cup:H^{1}(\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})},E[2])\times
H^{1}(\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})},E[2])\rightarrow
H^{2}(\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})},\mu_{2})$ using the
Weil pairing $e_{2}:E[2]\times E[2]\rightarrow\mu_{2}$, it follows that
$H^{-1}\cup
F_{2}=\chi_{13}\cup\chi_{-1}\cdot\chi_{13}\cup\chi_{-3}=\chi_{13}\cup\chi_{3}$.
The field $L_{H,F_{2}}$ has a property that it contains subfield
$L/\mathbb{Q}(\sqrt{13},\sqrt{3})$ such that $L/\mathbb{Q}$ is $D_{8}$
extension. Lemma 4.4 implies that there exists a map
$\Gamma:\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})}\rightarrow\mu_{2}$
defined over $L_{H,F_{2}}$ such that $d\Gamma=\chi_{13}\cup\chi_{3}=H^{-1}\cup
F_{2}$. One can check that $L_{H,F_{2}}/\mathbb{Q}$ is unramified outside the
set $\\{2,3,13\\}$ of primes of bad reduction of $E$, hence it follows from
the proof of Theorem 3.2. in [Smi16] that $L_{H,F_{2}}$ is governing field of
$H^{-1}$ and $F_{2}$. The choice of field $L_{H^{-1},F_{2}}$ is particularly
nice since it is easy to check that for prime $p$ the Cassels-Tate pairing
$\langle H^{-p},F_{2}^{p}\rangle_{CT}$ is equal to $0$ if and only if $p$
splits completely in $L_{H^{-1},F_{2}}$ provided that $H^{-p}$ and $F_{2}^{p}$
define an element in $\operatorname{Sel^{(2)}}(E^{-p})$. It follows from
Proposition 2.3 that $H^{-p}$ and $F_{2}^{p}$ are ELS if and only if $p=13$ or
$p$ splits completely in the field of definition
$K_{H^{-1},F_{2}}=\mathbb{Q}(\sqrt{13},\sqrt{-1},\sqrt{-3})$.

###  4.2.
$L_{H_{1},H_{2}}=\mathbb{Q}(\sqrt{3},\sqrt{-1},\sqrt{2})(\sqrt{8(1+\sqrt{3})(4+2\sqrt{3})})$

It follows from Lemma 3.3 that $H_{1}(\sigma)=\chi_{3}(\sigma)P_{1}$ and
$H_{2}(\sigma)=\chi_{-1}(\sigma)P_{1}+\chi_{-2}(\sigma)P_{2}$ for all
$\sigma\in\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})}$, thus
$H_{1}\cup H_{2}=\chi_{3}\cup\chi_{-2}$. Since $L_{H_{1},H_{2}}$ is unramified
outside $\\{2,3,13\\}$ and since $L_{H_{1},H_{2}}$ contains a degree two
extension $L$ of $\mathbb{Q}(\sqrt{3},\sqrt{-2})$ such that $L/\mathbb{Q}$ is
Galois with Galois group $D_{8}$, same as in 4.1, we can conclude that
$L_{H_{1},H_{2}}$ is governing field of $H_{1}$ and $H_{2}$. Moreover, for $p$
prime such that $H_{1}^{p}$ and $H_{2}^{p}$ define an element in
$\operatorname{Sel^{(2)}}(E^{p})$ (or equivalently for prime $p$ which splits
completely in $K_{H_{1},H_{2}}=\mathbb{Q}(\sqrt{3},\sqrt{-1},\sqrt{2})$), we
have that $\langle H_{1}^{p},H_{2}^{p}\rangle_{CT}$ is equal to $0$ if and
only if $p$ splits completely in $L_{H_{1},H_{2}}$.

###  4.3.
$L_{F_{1},F_{2}}=\mathbb{Q}(\sqrt{3},\sqrt{-1},\sqrt{2})(\sqrt{8(1+\sqrt{3})(4+2\sqrt{3})})$

Here conclusion is the same as in 4.2, for $p$ prime such that $F_{1}^{p}$ and
$F_{2}^{p}$ define an element in $\operatorname{Sel^{(2)}}(E^{-p})$ (or
equivalently for prime $p$ which splits completely in
$K_{F_{1},F_{2}}=\mathbb{Q}(\sqrt{3},\sqrt{-1},\sqrt{2})$), we have that
$\langle F_{1}^{p},F_{2}^{p}\rangle_{CT}$ is equal to $0$ if and only if $p$
splits completely in $L_{F_{1},F_{2}}=L_{H_{1},H_{2}}$.

###  4.4.
$L_{H,H_{2}}=\mathbb{Q}(\sqrt{-1},\sqrt{2},\sqrt{13})(\sqrt{4+2\sqrt{13}})$

Lemma 3.3 implies that $H(\sigma)=\chi_{13}(\sigma)P_{2}$ and
$H_{2}(\sigma)=\chi_{-1}(\sigma)P_{1}+\chi_{-2}(\sigma)P_{2}$ for all
$\sigma\in\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})}$, thus $H\cup
H_{2}=\chi_{13}\cup\chi_{-1}$. Since $L_{H,H_{2}}$ is unramified outside
$\\{2,3,13\\}$ and since $L_{H,H_{2}}$ contains a degree two extension $L$ of
$\mathbb{Q}(\sqrt{13},\sqrt{-1})$ such that $L/\mathbb{Q}$ is Galois with
Galois group $D_{8}$, same as in 4.1 we can conclude that $L_{H,H_{2}}$ is
governing field of $H$ and $H_{2}$. Also, for $p$ prime such that $H^{p}$ and
$H_{2}^{p}$ define an element in $\operatorname{Sel^{(2)}}(E^{p})$ (or
equivalently for prime $p$ which splits completely in
$K_{H,H_{2}}=\mathbb{Q}(\sqrt{13},\sqrt{-1},\sqrt{2})$), we have that $\langle
H^{p},H_{2}^{p}\rangle_{CT}$ is equal to $0$ if and only if $p$ splits
completely in $L_{H,H_{2}}$.

###  4.5. $L_{H,H_{1}}=\mathbb{Q}(\sqrt{3},\sqrt{13})(\sqrt{4+\sqrt{13}})$

Lemma 3.3 implies that $H(\sigma)=\chi_{13}(\sigma)P_{2}$ and
$H_{1}(\sigma)=\chi_{3}(\sigma)P_{1}$ for all
$\sigma\in\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})}$, thus $H\cup
H_{1}=\chi_{13}\cup\chi_{3}$. Since $L_{H,H_{1}}$ is unramified outside
$\\{2,3,13\\}$ and since $L_{H,H_{1}}/\mathbb{Q}$ is $D_{8}$ extension same as
in 4.1 we conclude that $L_{H,H_{1}}$ is governing field of $H$ and $H_{1}$.
Also, for $p$ prime such that $H^{p}$ and $H_{1}^{p}$ define an element in
$\operatorname{Sel^{(2)}}(E^{p})$, we have that $\langle
H^{p},H_{1}^{p}\rangle_{CT}$ is equal to $0$ if and only if $p$ splits
completely in $L_{H,H_{1}}$. Note that $H^{p}$ and $H_{1}^{p}$ are ELS if and
only if $p=13$ or $p$ splits completely in
$K_{H,H_{1}}=\mathbb{Q}(\sqrt{13},\sqrt{3})$ and $p\equiv 1\pmod{4}$.

###  4.6. $L_{H^{-1},F_{1}}=\mathbb{Q}(\sqrt{-2},\sqrt{13})(\sqrt{-1})$

It follows from Lemma 3.3 that for all
$\sigma\in\operatorname{Gal(\overline{\mathbb{Q}}/\mathbb{Q})}$ we have that
$H^{-1}(\sigma)=\chi_{13}(\sigma)P_{1}+\chi_{13}(\sigma)P_{2}=\chi_{13}(\sigma)P_{3}$
and
$F_{1}(\sigma)=\chi_{-2}(\sigma)P_{1}+\chi_{-2}(\sigma)P_{2}=\chi_{-2}(\sigma)P_{3}$,
thus $e_{2}(H^{-1}(\sigma),F_{1}(\sigma))=1$. Therefore $H^{-1}\cup F_{1}$ is
alternating (see Lemma 3.1. in [Smi16]) and $\langle
H^{-d_{1}},F_{1}^{d_{1}}\rangle_{CT}=\langle
H^{-d_{2}},F_{1}^{d_{2}}\rangle_{CT}$ for all pairs
$(d_{1},d_{2})\in\mathcal{D}$ from Theorem 4.1. For $p$ prime such that
$H^{-p}$ and $F_{1}^{p}$ define an element in
$\operatorname{Sel^{(2)}}(E^{-p})$, we can check by computing set
$\mathcal{D}$ that $\langle H^{-p},F_{1}^{p}\rangle_{CT}$ is equal to $0$ if
and only if $p$ splits completely in $L_{H^{-1},F_{1}}$. Note that $H^{-p}$
and $F_{1}^{p}$ are ELS if and only if $p$ splits completely in
$K_{H^{-1},F_{1}}=\mathbb{Q}(\sqrt{13},\sqrt{-2})$, thus, as before, the
splitting behaviour of $p$ in $L_{H^{-1},F_{1}}$ determines Cassels-Tate
pairing even though $L_{H^{-1},F_{1}}$ is not a governing field of $H^{-1}$
and $F_{1}$.

##  5\. Proofs of main results

###### Proof of Theorem 1.1.

From Section 4 (see also Table 1), we see that the governing field of the pair
$(H^{-1},F_{1})$ is
$L_{H^{-1},F_{1}}=\mathbb{Q}(\sqrt{-2},\sqrt{13})(\sqrt{-1})$. In particular,

$\langle H^{d},F_{1}^{-d}\rangle_{CT}=\begin{cases}0&\textrm{ if $|d|$ splits
completely in }L_{H^{-1},F_{1}},\\\ 1&\textrm{ otherwise }.\end{cases}$

For $d<0$, it follows from the description of set $T$ that $\langle
H^{d},F_{1}^{-d}\rangle_{CT}=1$ if $d\equiv 1\pmod{4}$ and $\langle
H^{d},F_{1}^{-d}\rangle_{CT}=0$ if $d\equiv 3\pmod{4}$. Hence a) follows. For
b), assume that $d\equiv 3\pmod{4}$ and $\iota(H^{d})\neq 0$. As argued in the
introduction, there is $L\in\operatorname{Sel^{(2)}}(E^{d})$ such that
$\langle H^{d},L\rangle_{CT}=1$. Since $\langle
H^{d},F_{1}^{-d}\rangle_{CT}=0$, from the bilinearity of the Cassels-Tate
pairing it follows that $\langle H^{d},F_{2}^{-d}\rangle_{CT}=1$ (as $F_{2}$
is remaining generator of $\operatorname{Sel^{(2)}}(E^{d})$). The other
implication in b) is obvious. Part c) is proved similarly. The only difference
here is that in $d>0$ case, $\operatorname{Sel^{(2)}}(E^{d})$ is, in addition
to torsion classes, generated by $H^{d},H_{1}^{d}$, and $H_{2}^{d}$. ∎

###### Proof of Corollary 1.6.

First we count the contribution to $S(X)$ of $d=\pm p$ for which $d\notin T$.
It follows from Conjectures 1 and 2, and Propositions 3.4 and 3.1 that the
only significant case is when $w(E^{d})=-1$ (assuming $H^{d}$ is ELS) in which
case $\Sha(E^{d})[2]$ is trivial. It follows from Propositions 2.1, 2.4 and
3.1 that this is equivalent to $\genfrac{(}{)}{}{}{d}{13}=1$,
$\genfrac{(}{)}{}{}{d}{2}\cdot\genfrac{(}{)}{}{}{d}{3}\cdot\genfrac{(}{)}{}{}{d}{13}=\textrm{sgn}(d)$
and $d\not\equiv 1\pmod{8}$ if $d>0$ or $d\not\equiv 5,7\pmod{8}$ if $d<0$.
Thus if

$d\equiv
29,35,53,55,77,79,101,103,107,127,131,155,173,179,199,251,269,295\pmod{8\cdot
3\cdot 13}$

when $d>0$ or if $d<0$ and

$d\equiv 17,43,113,139,185,209,211,233,235,257,259,283\pmod{8\cdot 3\cdot
13},$

then $H^{d}(\mathbb{Q})\neq\emptyset$. There are $18$ residue classes in the
first case, and $12$ in the second, thus by Dirichlet’s theorem on arithmetic
progressions, the contribution to $C_{1}$ is $\frac{30}{2\phi(8\cdot 3\cdot
13)}=\frac{5}{32}$.

Next, consider the case $d>0$, $d\in T$ and $\Sha(E^{d})[2]\neq
2\Sha(E^{d})[4]$. Corollary 1.2 together with Proposition 3.9 implies that in
this case $H^{d}(\mathbb{Q})\neq\emptyset$ if and only if $d$ does not split
completely in $L_{H_{1},H_{2}}$, and splits completely in $L_{H,H_{1}}$ and
$L_{H,H_{2}}$. One can check that the assumption $d>0$ and $d\in T$ is
equivalent to the requirement that $d$ splits completely in $K_{H,H_{1}}$,
$K_{H,H_{2}}$ and $K_{H_{1},H_{2}}$, thus we need to find a density of $d$’s
such that $d$ splits completely in composition
$K=L_{H,H_{1}}L_{H,H_{2}}K_{H_{1},H_{2}}$ but not in its degree two extension
$L=L_{H,H_{1}}L_{H,H_{2}}L_{H_{1},H_{2}}$. By Chebotarev density theorem the
density of such $d$’s is $\frac{1}{\deg{K}}\cdot\frac{1}{2}$. From Table 1 we
see that $K_{H_{1},H_{2}}$ is contained in $L_{H,H_{1}}L_{H,H_{2}}$. Moreover,
one can check that $\deg{L_{H,H_{1}}L_{H,H_{2}}}=64$, thus in this case the
contribution to $C_{1}$ is equal to $\frac{1}{2}\cdot\frac{1}{128}$ (we have
extra $\frac{1}{2}$ since $C_{1}$ is a lower bound for $\frac{S(X)}{2\pi(X)}$
and not $\frac{S(X)}{\pi(X)}$ ).

Finally, consider the case $d<0$, $d\in T$ and $\Sha(E^{d})[2]\neq
2\Sha(E^{d})[4]$. Corollary 1.2 together with Proposition 3.9 implies that
$H^{d}(\mathbb{Q})\neq\emptyset$ if and only if $d=-p$, where $p\equiv
1\pmod{4}$, does not split completely in $L_{F_{1},F_{2}}$ and splits
completely in $L_{H^{-1},F_{2}}$. One can check that assumption $p\equiv
1\pmod{4}$ and $-p\in T$ is equivalent to $p$ splits completely in
$K_{H^{-1},F_{2}}$ (we see in Table 1 that $\mathbb{Q}(\sqrt{-1})\subset
K_{H^{-1},F_{2}})$ and $K_{F_{1},F_{2}}$. As in the previous case, we need to
compute the density of primes which split completely in composition
$L_{H^{-1},F_{2}}K_{F_{1},F_{2}}$, but not in its degree two extension
$L_{H^{-1},F_{2}}L_{F_{1},F_{2}}$. Since
$\deg{L_{H^{-1},F_{2}}K_{F_{1},F_{2}}}=32$, in this case the contribution to
$C_{1}$ is equal to $\frac{1}{2}\cdot\frac{1}{64}$. Hence it follows that
$C_{1}=\frac{5}{32}+\frac{1}{256}+\frac{1}{128}=\frac{43}{256}$.

To compute the upper bound $C_{2}$, we need to find the density of the
remaining case, $d\in T$ and $\Sha(E^{d})[2]=2\Sha(E^{d})[4]$, in which our
method does not provide us an answer. If $d>0$, by Proposition 3.9 it is
enough to compute the density of primes $p$ which splits completely in
$L_{H,H_{1}}$, $L_{H,H_{2}}$ and $L_{H_{1},H_{2}}$. From Table 1, we see that
the composition of these three fields have degree $128$, hence by Chebotarev
density theorem the density of primes with this splitting property in $1/128$,
hence contribution to $C_{2}-C_{1}$ is $1/256$.

If $d<0$ and $d\equiv 7\pmod{8}$, then $p$ must split completely in
$L_{F_{1},F_{2}},L_{H,F_{2}}$ and $K=\mathbb{Q}(\sqrt{-2},\sqrt{13})$ (see
Table 1), and furthermore it must either split completely in

$L=\mathbb{Q}(\sqrt{-2},\sqrt{13})(\sqrt{4+2\sqrt{13}})$

or none of its factors in $K$ splits further in $L$ (note that $L/\mathbb{Q}$
is not Galois extension). One can check that this condition is equivalent for
$p$ to split completely in composition $L_{F_{1},F_{2}}L_{H,F_{2}}$ which is
of degree $64$, hence the density of such primes is $1/64$, and contribution
to $C_{2}-C_{1}$ is equal to $1/128$. Hence $C_{2}=C_{1}+1/256+1/128=46/256$.
∎

##  6\. Future work

This paper left us with some interesting questions which may be addressed in
the future projects:

  1. a)

What information can be obtained about $H^{d}(\mathbb{Q})$ in the case when
$\Sha(E^{d})[2]=2\Sha(E^{d})[4]$?

  2. b)

What can one say about $H^{d}(\mathbb{Q})\neq\emptyset$ for some larger class
of $d$’s? The main reason why we considered only $d$’s for which $|d|$ is
prime is that in this case we can control the $2$-Selmer group of quadratic
twists $E^{d}$ \- we have explicit generators. This might also be the case,
for example, for the set of $d$’s which are the products of two primes.

  3. c)

Can one obtain similar results for the quartics other that $H$? It seems this
could be within the reach of this method provided that, as in b), we have
explicit description of $2$-Selmer groups of quadratic twists.

## Acknowledgments

The author was supported by the Croatian Science Foundation under the project
no. IP-2018-01-1313, and by the QuantiXLie Center of Excellence, a project co-
financed by the Croatian Government and European Union through the European
Regional Development Fund - the Competitiveness and Cohesion Operational
Programme (Grant KK.01.1.1.01.0004).

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