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%\title[On elliptic curves arising from Heron triangles]{On elliptic curves arising from Heron triangles}
\title[On elliptic curves via Heron triangles and Diophantine triples]{On elliptic curves via Heron triangles and Diophantine triples}

\author[F. Izadi]{F. Izadi}
\address{Department of pure Mathematics, Azarbaijan  Shahid Madani University, Tabriz 53751-71379, Iran}
\email{izadi@azaruniv.edu}

\author[F. Khoshnam]{F. Khoshnam}
%\address{MDepartment of pure Mathematics, Azarbaijan  Shahid Madani University, Tabriz 53751-71379, Iran}
\email{khoshnam@azaruniv.edu}


\begin{abstract}
In this article, we construct families of elliptic curves arising from the Heron triangles and Diophantine triples with the Mordell-Weil torsion subgroup of $\Bbb{Z}/2\Bbb{Z}\times \Bbb{Z}/2\Bbb{Z}$. These families  have ranks at least 2 and 3, respectively, and contain particular examples with rank equal to $7$.
\end{abstract}

\subjclass[2010]{Primary 14H52; Secondary 11G05, 14G05}

\keywords{Diophantine triple, elliptic curve, family of elliptic curves, the Heron triangle, specialization, rank, torsion group}

\maketitle

\section{Introduction}
Triangles with integral sides and area have been considered by Indian  mathematician  Brahmagupta (598-668 A.D.).
In general, the  sides  and  area  are  related  by  a  formula  first  proved  by  Greek
mathematician Heron of Alexandria (c. 10 A.D -  c. 75 A.D.) as
$$S=\sqrt{P(P-a)(P-b)(P-c)},$$
where $P=(a+b+c)/2$ is the semi perimeter.

Triangles with rational sides and area are known as the Heron triangles
(for more information and fundamental results on Heron triangles, see  \cite{Fin,GM,Rus}).

Goins and Maddox have studied Heron triangles by considering the elliptic curve
$$E_{\tau}^{(n)}: y^2=x(x -n\tau)(x+n\tau^{-1})$$
as a generalization of the congruent number problem (see \cite{GM}).
In the same paper, they also have found 4 curves  of rank 3 with  torsion subgroup
$\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$.
%D. J. Rusin in his paper \cite{Rus} has considered a set of triangles in the plane with rational sides
%and a given area $S$, and shown there are infinitely many such triangles for each
%possible area $S$.  He also has shown that infinitely many such triangles may be
%constructed from a given one, all sharing a side of the original triangle, unless
%the  original  is  equilateral.
Campbell and Goins \cite{CG} by analyzing the elliptic curve
$$E_t: y^2= x^3+(t^2+2)x^2+x$$
defined over the rational function field $\Bbb Q(t)$ described connections between the problem of finding Heron triangles with
a given area possessing at least one side of a particular length and rational Diophantine quadruples
and quintuples. They also have studied the relation
between these problems and elliptic curves with torsion subgroup $\Bbb Z/2\Bbb Z\times\Bbb Z/8\Bbb Z$,
and found a new elliptic curve with this torsion having rank 3 and  an infinite family of elliptic curves
with torsion subgroup  $\Bbb Z/2\Bbb Z\times \Bbb Z/8\Bbb Z$ and rank
at least 1.
Having constructed a family of Diophantine triples  such that the correspondent elliptic curve over $\Bbb Q$ has torsion subgroup $\Bbb{Z}/2\Bbb{Z}\times \Bbb{Z}/2\Bbb{Z}$ and rank 5, Aguirre et al. \cite{ADP} have obtained two examples of elliptic curves over $\Bbb Q$ with
torsion subgroup $\Bbb{Z}/2\Bbb{Z}\times \Bbb{Z}/2\Bbb{Z}$ and rank equal to 11.
Dujella and Peral in a joint work \cite{DP} have created subfamilies of elliptic curves coming from the Heron triangles of ranks at least 3, 4,  and 5 - as a matter of fact, the rank 3 family is exactly the same as this paper (but with different notation). We are not sure that this is an accident or restatement of our rank 3 family via some transformations of parameter as they cited a preprint of this paper as a main reference.
 We will discuss this connection in the desired section. They also have given examples of elliptic curves over $\Bbb Q$ with rank equal to 9 and 10. %In the sequel, we see that our approach differs from others.

This paper is organized as follows.
In Section 2,  a family of elliptic curves arising from Heron triangles
introduced by Fine \cite{Fin} is considered and shown that the family has torsion subgroup
$\Bbb{Z}/2\Bbb{Z}\times \Bbb{Z}/2\Bbb{Z}$, and rank at least 2, and a subfamily of rank~$\geq3$.
In Theorem 2.6, a subfamily of $Y^2=(aX+1)(bX+1)(cX+1)$ of rank~$\geq2$ is given.
%It will be seen although the subfamily has a feauture like Fine's curve, its structure does not come from Heron triangles.
This is a generalization of Dujella's work done in \cite{Duj1}. Therein, Dujella extended the Diophantine triple $(a,b,c)=(k-1,k+1,4k)$ to a quadruple  by studying $Y^2=(aX+1)(bX+1)(cX+1),$
and proved that this elliptic curve has generic rank 1 over $\Bbb Q$. In Section 3, some examples of elliptic curves with rank 7 are given.
%All the calculations have been carried out with \verb"SAGE"~\cite{sage}.
\section{Main results}
Let $S$ be area of the triangle $(a,b,c)$, i.e., $S=\sqrt{P(P-a)(P-b)(P-c)},$ where $P=(a+b+c)/2$.
This formula, due to Heron, ensures us to have an elliptic curve
$v^2=u(u-a)(u-b)(u-c)$ with non torsion point $(u,v)=(P,S)$. %This curve is (birationally) equivalent to
%$${\eta}^2=(1-a\zeta)(1-b\zeta)(1-c\zeta),$$
%with the corresponding point $(\zeta,\eta)=\left(P^{-1},SP^{-2}\right)$.
%with the change of coordinates, $u=\zeta^{-1},$ $v=\zeta^{-2}\eta$.
The curve therefore is birationally equivalent to $y^2=(x+ab)(x+bc)(x+ac),$ with corresponding (non torsion) point $(x,y)=\left(-abcP^{-1},abcSP^{-2}\right)$, and is equivalent to $Y^2=(aX+1)(bX+1)(cX+1),$
with corresponding point $(X,Y)=\left(-P^{-1},SP^{-2}\right)$. In the sequel, we are going to treat with special families coming from these two kinds of elliptic curves.


Consider the elliptic curve $E_k: y^2=(x+a(k)b(k))(x+b(k)c(k))(x+a(k)c(k))$
associated to the Fine triple:
\begin{equation}\label{2.1}
\left\{
\begin{split}
  & a(k)=10k^2-8k+8,\\
  & b(k)=k(k^2-4k+20),\\
  & c(k)=(k+2)(k^2-4).
\end{split}
\right.
\end{equation}
arising from a Heron triangle which has rational area $4k(k^2-4)^2$ (see \cite{Fin}).
(Note that multiplication of sides in \eqref{2.1} by $(2(k^2-4))^{-1}$ implies that the resulting triangle to have area $k$.)
One can easily check that $E_k$ has three rational points of order two:
$$\left\{
  \begin{array}{l}
    T_1=(-k(10k^2-8k+8)(k^2-4k+20),0), \vspace{.2cm} \\
    T_2=(-k(k+2)(k^2-4k+20)(k^2-4),0), \vspace{.2cm} \\
    T_3=(-(k+2)(10k^2-8k+8)(k^2-4),0).
     \end{array}
\right.
$$
As the change of coordinates $(x,y)\rightarrow(x-a(k)b(k),y)$
does not affect the group structure of $E_k(\mathbb{Q})$, we may consider $E_k$ in the form $y^2=x^3+Ax^2+Bx$,
in which
\begin{equation}\label{2.2}
\begin{array}{l}
A=k^6-12k^5+116k^4-480k^3+304k^2-448k-64,\vspace{.2cm}\\
B=4k(5k^2-4k+4)(k^2-4k+20)(3k^2-12k-4)\\ \qquad \times (k^3-8k^2+4k-16).
\end{array}
\end{equation}


The next proposition \cite[p. 38]{hus} is given to characterize the Mordell-Weil torsion subgroup of $E$  over $Q(k)$ in Theorem 2.2.
\begin{pro}
Let $E$ be an elliptic curve defined over a field $\Bbb K$ of characteristic different from zero by
$$y^2=(x-\alpha)(x-\beta)(x-\gamma)=x^3+ax^2+bx+c,$$
where $\alpha,\beta,\gamma\in\Bbb K$. For $(x',y')\in E(\Bbb K)$ there exists $(x,y)\in E(\Bbb K)$ with $2(x,y)=(x',y')$ if and only if $x'-\alpha$, $x'-\beta$, and $x'-\gamma$ are squares.
\end{pro}
\begin{theorem}
Let $a(k)$, $b(k)$ and $c(k)$ be defined as \eqref{2.1}, where $k$ is an arbitrary rational number
 different from 0, -2, and 2.
 Then the elliptic curve
$$E:y^2=\left(x+a(k)b(k)\right)\left(x+b(k)c(k)\right)\left(x+a(k)c(k)\right)$$
defined over $Q(k)$ has torsion subgroup $\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}$.
\end{theorem}
\begin{proof}
The points $\mathcal{O}$ (the point at infinity), $T_1=(-a(k)b(k),0), T_2=(-b(k)c(k),0)$, and $T_3=(-a(k)c(k),0)$ form a subgroup of the torsion group $E(\Bbb{Q}(k))_{tors}$ isomorphic to
$\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}$.
By Mazur's theorem \cite{Maz} it
suffices to check that there exists no point $E(\Bbb{Q}(k))$ of order four, six or eight.
If there exists a point $T$ on $E(\Bbb{Q}(k))$ such that $2T\in \{T_1,T_2,T_3\}$, then
Proposition 2.1 implies that all of the expressions

\begin{align*}
 -a(k)b(k)+a(k)b(k)&=0,\\
 -a(k)b(k)+b(k)c(k)&= k(k^2-4k+20)(k^3-8k^2+4k-16),\\
 -a(k)b(k)+a(k)c(k)&=4(5k^2-4k+4)(3k^2-12k-4),
 \end{align*}
must be perfect squares. But, it is easily seen that for $k=1$ none of the above expressions are perfect squares. Similarly, if $2T=T_2$ and $2T=T_3,$ then
all of the expressions
\begin{align*}
 -b(k)c(k)+a(k)b(k)&=-k(k^2-4k+20)(k^3-8k^2+4k-16),\\
 -b(k)c(k)+b(k)c(k)&= 0,\\
 -b(k)c(k)+a(k)c(k)&=-(k^2-12k+4)(k-2)^2(k+2)^2,
\end{align*}
as well as
\begin{align*}
 -a(k)c(k)+a(k)b(k)&=-4(5k^2-4k+4)(3k^2-12k-4),\\
 -a(k)c(k)+b(k)c(k)&= (k^2-12k+4)(k-2)^2(k+2)^2,\\
 -a(k)c(k)+a(k)c(k)&=0,
\end{align*}
must be perfect squares.
But, it is easily seen that for $k=1$ none of the above expressions are perfect squares. This contradiction shows that $T\notin \{T_1,T_2,T_3\}.$
Thus, by \cite{Maz} it is to prove that there exists no point $T$ such that $3T\in \{T_1,T_2,T_3\}.$
If there exists a point $T=(x,y)$ on $E(\Bbb{Q}(k))$ such that $3T=T_1,T\neq T_1$,
then from $2T=-T+T_1$, the equation
\begin{equation}\label{2.3}
x^4-6h_1(k)x^2-4h_1(k)h_2(k)x-3h_2(k)^2=0,
\end{equation}
is obtained in which
\begin{align*}
 %h_1(k) &= a(k)c(k)+b(k)c(k)-2a(k)b(k)\\
  h_1(k)&=-12k^5-480k^3+116k^4+304k^2-448k-64+k^6,\\
  h_2(k)&=4k(5k^2-4k+4)(k^2-4k+20)(3k^2-12k-4)\\
  &\qquad\times(k^3-8k^2+4k-16).
\end{align*}

It can be easily seen that for $k=1$, the equation \eqref{2.3}, namely
$$x^4+3498x^2+195841360x-21157921200=0$$
has no rational solution. Similarly it can be checked that there does not exist any point $T$ on $E(\mathbb{Q}(k))$ such that $3T=T_2$, $T\neq T_2$, and $3T=T_3$, $T\neq T_3$. Therefore, $ E(\mathbb{Q}(k))_{tors}=\Bbb{Z}/2\Bbb{Z}\times\Bbb{Z}/2\Bbb{Z}$.
\end{proof}
\begin{theorem}
With the terminology in Theorem 2.2, rank~$E(\mathbb{Q}(k))\geq2$.
\end{theorem}
\begin{proof}
Evidently the non torsion points
$$
\begin{array}{l}
{\mathcal P}_1=(-a(k)b(k)c(k)P^{-1}(k),a(k)b(k)c(k)S(k)P(k)^{-2}),\vspace{.2cm}\\
{\mathcal P}_2=(0,a(k)b(k)c(k)),
\end{array}
$$
lie on  $E(\mathbb{Q}(k))$  , where $P(k)$ and $S(k)$ are respectively the associated semi perimeter and area to
($a(k), b(k), c(k)$).


For $k=1$, the elliptic curve $E(\mathbb{Q}(k))$ turns into
$$E_1:y^2=x^3-\frac{73}{36}x^2-\frac{85}{4}x+\frac{7225}{144},$$
with
$${\mathcal P}_1=\left(\frac{85}{18},\frac{85}{27}\right),\quad {\mathcal P}_2=\left(0,\frac{85}{12}\right).$$
The N\'{e}ron-Tate  height matrix  \cite[p. 230]{washh} associated to these points is of non vanishing determinant $\approx2.30842249514247$  (carried out with \verb"SAGE" \cite{sage}) showing that the points are linearly independent.Therefore the rank of $E$ over $\Bbb Q(k)$ is
 $\geq2$ , and hence, by the specialization theorem of Silverman \cite{sil}, the rank~$E_k(\Bbb Q)\geq2$ "for all but finitely many rational numbers $k.$ "
\end{proof}
\begin{pro}
For each $2\leq r \leq 7,$ there exists some $k$ such that $E_k$ defined in Theorem 2.2
has torsion subgroup ${\Bbb{Z}}/{2\Bbb{Z}}\times{\Bbb{Z}}/{2\Bbb{Z}}$ and rank $r$.
\end{pro}
\begin{proof}
The first part is readily obtained from Theorem 2.2. For the second part, it suffices to note that for the values of $k=$ 4, 7, 19, 11, $98/625$, and $88/31$, the corresponding ranks with using of Mwrank \cite{CER}, are 2, 3, 4, 5, 6, and 7, respectively.
\end{proof}
Now, we are ready to show our main result:
\begin{theorem}
There exists a subfamily of $E_k$ of rank $\geq3$ over $\mathbb{Q}(m).$
\end{theorem}
\begin{proof}
Obviously the non torsion points
$$\begin{array}{l}
{\mathcal P}_1=(4(5k^2-4k+4)(k^2-4k+20), \vspace{.2cm}\\
\qquad\qquad\qquad 8(5k^2-4k+4)(k^2-4k+20)(k-2)^3),\vspace{.2cm}\\
{\mathcal P}_2=(2(5k^2-4k+4)k(k^2-4k+20),\vspace{.2cm}\\
\qquad\qquad\qquad 2k(k-2)(5k^2-4k+4)(k^2-4k+20)(k+2)^2),
\end{array}
$$
lie on the curve $E_k:y^2=x^3+Ax^2+Bx$.
In order to find a subfamily of rank~$\geq3$, we proceed as following.
Let $B_1=2k(3k^2-12k-4)(k^2-4k+20)$, and for some rational numbers $M,N,e$, ${\mathcal P}_3=(B_1M^2/e^2,B_1MN/e^3)$
be on $E_k$.  This implies the quartic equation $B_1M^4+AM^2e^2+B_2e^4=N^2.$
Taking  $M=e=1$, we get $(k-6)(k+2)^5=N^2,$ hence, $(k-6)(k+2)=z^2,$
where $z=N/(k+2)^2$. Using the rational solution $(k,z)=(6,0)$, the parametric solution is then
$(k,z)=(2(3m^2+1)/(m^2-1), 8m/(m^2-1)),$ where $m\in \mathbb{Q}\setminus\{\pm 1\}$.
Therefore, $N=2^9m^5/(m^2-1)^3$ and ${\mathcal P}_3$ turns into
\begin{align*}
{\mathcal P}_3&=(B_1,B_1N)\vspace{.2cm}\\
&=\bigg(\frac{2^{12}(3m^2+1)(m^4+4m^2+1)(m^4+1)}{(m^2-1)^5}, \vspace{.2cm}\\ &\qquad\qquad\qquad\frac{2^{21}(3m^2+1)(m^4+4m^2+1)(m^4+1)m^5}{(m^2-1)^8}\bigg).
\end{align*}
Thus $E_k$ turns into $E_m:y^2=x^3+Ax^2+Bx$ with
$$
\begin{array}{l}
\displaystyle A=\frac{2^{13}(m^{12}+14m^{10}-5m^8+4m^6+11m^4+6m^2+1)}{(m^2-1)^6},\vspace{.2cm}\\
\displaystyle B=-\frac{2^{24}(m^6-5m^4-3m^2-1)M_1M_3}{(-1+m^2)^{10}},
\end{array}
$$
and three non torsion points
$$
\begin{array}{l}
\displaystyle{\mathcal P}_1=\left(\frac{2^{12}M_1}{(m^2-1)^4},\frac{2^{19}M_1(1+m^2)^3}{(m^2-1)^7}\right),\vspace{.2cm}\\
\displaystyle{\mathcal P}_2=\left(\frac{2^{12}(m^4+1)M_2}{(-1+m^2)^5},\frac{2^{20}(m^4+1)m^4(m^2+1)M_2}{(m^2-1)^8}\right),\vspace{.2cm}\\
\displaystyle{\mathcal P}_3=\left(\frac{2^{12}(m^4+1)M_3}{(m^2-1)^5},\frac{2^{21}(m^4+1)m^5M_3}{(m^2-1)^8}\right),
\end{array}
$$
where
$$
\begin{array}{l}
M_1=(m^4+1)(5m^4+4m^2+1),\vspace{.2cm}\\
M_2=(3m^2+1)(5m^4+4m^2+1),\vspace{.2cm}\\
M_3=(3m^2+1)(m^4+4m^2+1).
\end{array}
$$
Regarding the specialization theorem, since for $m=1/2$, the N\'{e}ron-Tate height matrix associated to these points has
non vanishing determinant $\approx11.9727247292862$, then $E_m$ as a subfamily of $E_k$ is of rank~$\geq3$ over $\mathbb{Q}(m).$
\end{proof}
{\bf Remark 1}. We are going to explain how the subfamily in [DP] comes from our subfamily as we mentioned in the introduction.
To this end, first we divide $a(k)$, $b(k)$, and $c(k)$ by $8(k-2)$ and then compute $a(k)+b(k)-c(k)=(k+2)^2/(2k-4)$ and $c(k)/(a(k)+b(k)+c(k))=(k-2)/4.$ We do the same computations for Brahmagupta's triples $a=n(m^2+1$, $b=m(n^2+1$, and $(m+n)(mn-1)$ by letting $k=1$ to get
$a+b-c=2(m+n)$, and $c/(a+b-c)=(mn-1)/2$. By equating the corresponding expressions one can get
$m+n=(k+2)^2/(4k-8)$, and $mn=k/2$. Now the corresponding quadratic equation gives rise to $m=2k/(k-2)$, and $n=(k-2)/4$. By introducing a new variable $w$ satisfying the relation $mn=(w^2+3)/(w^2-1)$ as in [DP, section 2.3], one obtains by substituting
the above values of $m$ and $n,$
$$k=2\times (3+w^2)/(1-w^2)=2 \times(3s^2+1)/(s^2-1),$$ where $s=1/w,$ which is exactly the parametric solution already given in the preprint as well as in the proof of Theorem 2.5 of the present paper.


We say that (\cite{Duj2}) the Diophantine triple $(a,b,c)$ has the property $D(n)$, for any non zero integer $n$, whenever
there exist rational $r,s,$ and $t$ such that
$$ab+n=r^2,\qquad ac+n=s^2,\qquad bc+n=t^2.$$

\begin{theorem}
There exists a subfamily of $C: Y^2=(aX+1)(bX+1)(cX+1)$ over $\mathbb{Q}$ with rank~$\geq2$.
% which essentially does not come from Heron triangles.
\end{theorem}
\begin{proof}
Consider the triple $(a,b,c)=(k-1,k+1,4k)$ with property $D(1)$. The curve $C_k:Y^2=((k-1)X+1)((k+1)X+1)(4kX+1), k\in\Bbb Q,$ has non torsion point ${\mathcal P}_1=(0,1).$
[The triple $(k-1,k+1,4k)$ has the property $D(1)$, but does not form any triangle (note $a(k)+b(k)<c(k)$).]
In order to find a subfamily of $C_k$ of rank~$\geq2,$ let
${\mathcal P}_2=(-P^{-1},P^{-2}S),$ be on the curve, where $P=3k$ and $S=k\sqrt{-3(4k^2-1)}$. This implies to have some rational $u$ such that
$-3(4k^2-1)=u^2.$ Using the rational solution $(k,u)=(1/2,0)$, the parametric solution for $k$ is then
$k=\frac{m^2-12}{2(m^2+12)},$
where $m\in \mathbb{Q}$.
Henceforth, $C_k$ turns into
$$C_m:Y^2=\left(\frac{-m^2-36}{2(m^2+12)}X+1\right)\left(\frac{3(m^2+4)}{2(m^2+12)}X+1\right)  \left(\frac{2(m^2-12)}{m^2+12}X+1\right),
$$
with two non torsion points
$$
\begin{array}{l}
{\mathcal P}_1=(0,1),\vspace{.2cm}\\
\displaystyle{\mathcal P}_2=\left(\frac{-2(m^2+12)}{3(m^2-12)},\frac{8m}{3(m^2-12)}\right).
\end{array}
$$
The associated height matrix to these points at $m=1$ has non vanishing determinant $\approx2.87442404831027$ showing that these points are linearly independent, hence rank~$C_m(\Bbb Q)\geq2$ for all but finitely many $m$'s.
The elliptic curve $y^2=(ax+1)(bx+1)(cx+1)$ with the point $(x,y)$ is isomorphic to $y^2=x^3+(ab+ac+bc)x^2+abc(a+b+c)x+a^2b^2c^2$, with the corresponding point $(abcx,abcy)$.
\end{proof}

{\bf Remark 2.}  We should mention that in \cite{Duj1}, two subfamilies of $C_k$ from Theorem 2.6 with rank  $\geq2$ and one subfamily with rank $\geq3$ were constructed. However they considered the problem for the integer values of $k$ between 1 and 1000, while in our case the values of $k=\frac{m^2-12}{2(m^2+12)},$ are rational numbers less than 1.
\section{Specialization of high rank}
In this stage we want to find curves having large ranks possible. The main idea here is that a curve is more
likely to have large rank if $| E(\mathbb{F}_p)|$ is relatively
large for many primes $p$. We will use the following realization
of this idea. For a prime $p$ we put $a_p =a_p(E) = p + 1 - |
E(\mathbb{F}_p)|$ and
$$SN(E,N) =\sum_{p\leq N, p     \medspace  prime}(1-\frac{p-1}{ |
E(\mathbb{F}_p)|})\log (p)=\sum_{p\leq N, p\medspace
prime}(\frac{-a_p+2}{p+1-a_p})\log (p).$$ This summation is defined as Mestre-Nagao sum.
In order to give examples of high rank for $E_k: y^2= x^3+Ax^2+Bx$ with $A$ and $B$ in the equation (2.2), we observe $k=p/q$, with $\gcd(p,q)=1$, $|p|,|q|<1000$, and Mestre-Nagao sums $SN(1000,E_k)>20$, $SN(10000,E_k)>30$, and $SN(100000,E_k)>40$. Among
these sieved $k$'s, it is considered the ones with high Selmer-rank. Then, rank computations are carried out with \verb"MWrank". This process shows that for $k=\frac{30}{259}$, $\frac{67}{93}$, $\frac{88}{31}$, $\frac{98}{337}$, $\frac{263}{666}$, $\frac{280}{919}$, $\frac{593}{150}$, $\frac{596}{19}$, $\frac{609}{76}$, $\frac{845}{33}$, rank~$E_k(\Bbb Q)=7$.



\begin{thebibliography}{HD}
\normalsize
\baselineskip=17pt

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\bibitem[Duj2]{Duj2} A. Dujella,
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