2. A quadratic twist and three higher-degree twists

Let $f(x)\in \mathbb{Q}[x]$ be a square-free polynomial of degree at least $3$ . Fix odd integers $m_1,m_2,m_3\ge 3$ . We are investigating the existence of non-square rational numbers $D$ such that the Jacobians of the (hyperelliptic) curves

\begin{equation*} C\,:\, Dy^2=f(x),\quad C_1\,:\,y^2=Dx^{m_1}+a,\quad C_2\,:\,y^2=Dx^{m_2}+b,\quad C_3\,:\,y^2=Dx^{m_3}+c \end{equation*}

are of positive Mordell–Weil rank over $\mathbb{Q}$ .

We set $M=\textrm {lcm}(m_1,m_2,m_3)$ and $M_{i}=M/m_i$ , $i=1,2,3$ . The rational number $D$ can be found by solving the following system of equations:

(2.1) \begin{equation} \frac {y_1^2-a}{x_1^{m_1}}=\frac {y_2^2-b}{x_2^{m_2}}=\frac {y_3^2-c}{x_3^{m_3}}=\frac {f(x_4)}{y_4^2}. \end{equation}

More precisely, we are looking for solutions $x_i,y_i$ , $i=1,2,3,4$ , for the system (2.1) where

\begin{equation*}x_1x_2x_3\big(y_1^2 -a\big)\big(y_2^2-b\big)\big(y_3^2-c\big)y_4\ f(x_4) \ne 0.\end{equation*}

In what follows, we suggest how one can obtain a family of parametric solutions to (2.1). We set

\begin{equation*} x_1=\frac {1}{v_1^{2}T^{M_1}},\quad x_2=\frac {1}{v_2^2T^{M_2}},\quad x_3=\frac {1}{v_3^2T^{M_3}},\quad x_4 =u,\quad y_1=p,\quad y_2=q,\quad y_3=r,\quad y_4=\frac {1}{T^{(M-1)/2}} \end{equation*}

where $u,v_1,v_2,v_3$ are rational parameters. We observe that

\begin{eqnarray*} T&=&\frac {f(u)}{v_3^{2m_3}(r^2-c)},\\ v_1^{2m_1}T^M(p^2-a)&=&v_2^{2m_2}T^M(q^2-b)=v_3^{2m_3}T^M(r^2-c). \end{eqnarray*}

In other words, the system (2.1) can be simplified to

\begin{equation*} T=\frac {f(u)}{v_3^{2m_3}(r^2-c)},\quad v_1^{2m_1}(p^2-a)=v_2^{2m_2}(q^2-b)=v_3^{2m_3}(r^2-c). \end{equation*}

Geometrically, the second equation defines an intersection $\mathcal{C}_{a,b,c}$ of two quadratic surfaces in $\mathbb{P}^3$ over $\mathbb{Q}(v_1,v_2,v_3)$ . Since $[p\,:\,q\,:\,r\,:\,s]=[\pm v_2^{m_2}v_3^{m_3}\,:\,\pm v_1^{m_1}v_3^{m_3}\,:\,\pm v_1^{m_1}v_2^{m_2}\,:\,0]$ are rational points on $\mathcal{C}_{a,b,c}$ , it follows that $\mathcal{C}_{a,b,c}$ is an elliptic curve over $\mathbb{Q}(v_1,v_2,v_3)$ . One may check that the latter four projective rational points form a subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ .

We now assume, further, that $a$ , $b$ , and $c$ are non-zero rational squares. After a transformation, the curve $\mathcal C_{a,b,c}$ is defined by

\begin{equation*}x^2-T_1^2=y^2-T_2^2=z^2-T_3^2,\end{equation*}

where $T_1=av_1^{m_1}$ , $T_2=bv_2^{m_2}$ , and $T_3=cv_3^{m_3}$ .

Proposition 2.1. Let $C_{T_1,T_2,T_3}$ be the elliptic curve defined by

\begin{equation*}x^2-T_1^2= y^2-T_2^2=z^2-T_3^2\end{equation*}

over $\mathbb{Q}(T_1,T_2,T_3)$ . The curve $C_{T_1,T_2,T_3}$ has positive Mordell–Weil rank over $\mathbb{Q}(T_1,T_2,T_3)$ . In particular, except for a thin set of triples $(t_1,t_2,t_3)\in \mathbb{Q}\times \mathbb{Q}\times \mathbb{Q}$ , the curve $C_{t_1,t_2,t_3}$ has positive Mordell–Weil rank over $\mathbb{Q}$ .

Proof. The point $[T_1\,:\,T_2\,:\,T_3\,:\,1]\in C_{T_1,T_2,T_3}(\mathbb{Q}(T_1,T_2,T_3))$ . In addition, it specializes to a point of infinite order on $C_{1,2,3}$ when $[1\,:\,1\,:\,1\,:\,0]$ is regarded as the identity element, Magma [Reference Bosma, Cannon and Playoust2]. For the convenience of the reader, we include the code as it will be used later.

Now the statement follows from Silverman’s specialization theorem [Reference Silverman7, Section 20, Theorem 20.1].

One remarks that when the point $[w_1\,:\,w_2\,:\,w_3\,:\,1]$ is a point of infinite order in $\mathcal C_{a,b,c}$ , then any of its multiples can be used to generate a solution to the system (2.1).

Doubling the point $P\ {:}{=}\ [a\,:\,b\,:\,c\,:\,1]\in \mathcal{C}_{a^2,b^2,c^2}(\mathbb{Q}(v_1,v_2,v_3))$ yields that $2P=[w_1\,:\,w_2\,:\,w_3\,:\,1]$ is given by

\begin{eqnarray*} \bigg [ \frac {a^2 b^2 v_1^{2m_1} v_2^{2m_2} + a^2 c^2 v_1^{2m_1} v_3^{2m_3} - b^2 c^2 v_2^{2m_2} v_3^{2m_3}}{2a b c v_1^{2m_1} v_2^{m_2} v_3^{m_3}}&:&\frac {a^2 b^2 v_1^{2m_1} v_2^{2m_2} - a^2 c^2 v_1^{2m_1} v_3^{2m_3} + b^2 c^2 v_2^{2m_2} v_3^{2m_3}}{2a b c v_1^{m_1} v_2^{2m_2} v_3^{m_3}}\\ &:& \frac {-a^2 b^2 v_1^{2m_1} v_2^{2m_2} + a^2 c^2 v_1^{2m_1} v_3^{2m_3} + b^2 c^2 v_2^{2m_2} v_3^{2m_3}}{2 a b c v_1^{m_1} v_2^{m_2} v_3^{2m_3}}\,:\,1 \bigg ]. \end{eqnarray*}

We now set

\begin{equation*} T=\frac {f(u)}{v_3^{2m_3}\big(w_3^2-c^2\big)},\quad D=\big(w_1^2-a^2\big)v_1^{2m_1}T^M. \end{equation*}

We consider the following twists of hyperelliptic curves:

\begin{equation*}C\,:\,D y^2=f(x),\quad C_1\,:\,y^2=Dx^{m_1}+a^2,\quad C_2\,:\,y^2=Dx^{m_2}+b^2,\quad C_3\,:\,y^2=Dx^{m_3}+c^2.\end{equation*}

Now, the points $P_i\ {:}{=}\ (x_i,y_i)=\big(\frac {1}{v_i^2T^{M_i}},w_i\big)$ are $\mathbb{Q}(v_1,v_2,v_3)$ -rational points in $C_i$ , $i=1,2,3$ , whereas $P\ {:}{=}\ (x_4,y_4)=\big (u,\frac {1}{T^{(M-1)/2}}\big )$ is a $\mathbb{Q}(u,v_1,v_2,v_3)$ -rational point in $C$ .

In [Reference Ulas8, Reference Ulas9], explicit constructions of families of triples and quadruples of elliptic curves with $j$ -invariants $0$ or $1728$ were given such that there is a rational function $D$ for which the twists of the curves in each tuple by $D$ are of positive Mordell–Weil rank. This was extended to pairs and triples of Jacobians of hyperelliptic curves of the same genus in [Reference Jȩdrzejak, Top and Ulas5, Reference Jȩdrzejak and Ulas6]. Following the argument above, we obtain the following theorem which can be considered as an extension of the aforementioned results to quadruples of Jacobians of hyperelliptic curves that are not necessarily of the same genus.

Theorem 2.2. Let $f \in \mathbb{Q}[x]$ be a square-free polynomial of degree at least $3$ . Let $m_1,m_2,m_3\ge 3$ be odd integers. Consider the hyperelliptic curves

\begin{equation*}E\,:\,y^2=f(x),\quad E_1\,:\,y^2=x^{m_1}+a^2,\quad E_2\,:\,y^2=x^{m_2}+b^2,\quad E_3\,:\,y^2=x^{m_3}+c^2,\end{equation*}

where $a,b,c$ are non-zero rational numbers. Then there exists a rational function $D_{2,m_1,m_2,m_3} \in \mathbb{Q}(u, v_1,v_2,v_3)$ such that the Jacobian of the quadratic twist of the curve $E$ by $D_{2,m_1,m_2,m_3}$ and the Jacobians of the $m_i$ -twists of the curves $E_i$ , $i=1,2,3$ , by $D_{2,m_1,m_2,m_3}$ have positive Mordell–Weil rank over $\mathbb{Q}(u, v_1, v_2,v_3)$ .

Example 2.3. The curves $E\,{:}\, D y^2=x^5+x+1$ and $E_i\,:\,y^2=D x^{4i+1}+i^2$ , $i=1,2,3$ , are of positive Mordell–Weil rank over $\mathbb{Q}$ , where

\begin{eqnarray*} D\,:&=&D(u,v_1,v_2,v_3)=\frac {\big(u^5+u+1\big)^{585}\big(w_1^2-1\big)v_1^{10}}{v_3^{15210}(w_3^2-9)^{585}}\textrm { where} \\ w_1&=&\frac {3 v_3^{13}}{4 v_2^9} + v_2^9\left ( \frac {1}{3 v_3^{13}} - \frac {3 v_3^{13}}{v_1^{10}}\right)\!,\\ w_2&=&\frac {3 v_3^{13}}{v_1^5} + v_1^5\left ( \frac {1}{3 v_3^{13}} - \frac {3 v_3^{13}}{4v_2^{18}}\right)\!,\\ w_3&=&\frac {3 v_2^{9}}{v_1^5} + v_1^5\left ( \frac {3}{4 v_2^{9}} - \frac {v_2^{9}}{3v_3^{26}}\right)\!,\quad u,v_1,v_2,v_3\in \mathbb{Q}\setminus \{0\}. \end{eqnarray*}

Due to Silverman’s specialization theorem of abelian varieties, one knows that for all but a thin set of values of $u,v_1,v_2,v_3$ in $\mathbb{Q}$ , the points $\left (u,\frac {v_3^{7592}(w_3^2-9)^{292}}{(u^5+u+1)^{292}}\right )-(\infty )$ on the Jacobian of $E$ and the point $(P_i)-(\infty )$ , $i=1,2,3$ , on the Jacobian of $E_i$ , where

\begin{equation*}P_i=\left (\frac {v_3^{26M_i}(w_3^2-9)^{M_i}}{v_i^2 (u^5+u+1)^{M_i}},w_i\right)\!, \,M_1=117,\,M_2=65,\,M_3=45,\end{equation*}

are points of infinite order.

In Corollaries 3.2 and 4.2 of [Reference Jȩdrzejak and Ulas6], the authors established the existence of pairs of hyperelliptic curves $C_1$ and $C_2$ together with a rational function $D$ such that the Mordell–Weil rank of the Jacobian of a twist of $C_1$ by $D$ is positive, whereas the Mordell–Weil rank of the Jacobian of a twist of $C_2$ by $D$ is at least $2$ . In the following theorem, we present such examples of pairs of hyperelliptic curves where the Mordell–Weil rank of the Jacobian of a twist of $C_2$ by $D$ is at least $3$ .

Theorem 2.4. Let $f \in \mathbb{Q}[x]$ be a square-free polynomial of degree at least $3$ . Let $m\ge 3$ be an odd integer. Consider the hyperelliptic curves

\begin{equation*}E\,:\,y^2=f(x),\quad E'\,:\,y^2=x^{m}+a^2,\quad a\in \mathbb{Q}\setminus \{0\}.\end{equation*}

Then there exists a rational function $D_{2,m} \in \mathbb{Q}(u, v_1,v_2,v_3)$ such that the Jacobian of the quadratic twist of the curve $E$ by $D_{2,m}$ is of positive Mordell–Weil rank and the Mordell–Weil rank of the Jacobian of the $m$ -twist of the curve $E'$ by $D_{2,m}$ over $\mathbb{Q}(u, v_1, v_2,v_3)$ is at least $3$ .

Proof. In Theorem 2.2, we set $a=b=c$ and $m\ {:}{=}\ m_1=m_2=m_3$ . Choosing $(p,q,r)=(w_1,w_2,w_3)$ as defined above, we obtain the three rational points $P_i=\left (\frac {1}{v_i^2},w_i\right )$ , $i=1,2,3$ , in $E'(\mathbb{Q})$ where $E'$ is defined by $y^2=D x^m+a$ .

We consider the automorphisms $\phi _{i}$ , $1\le i\le 3$ , of $K=\mathbb{Q}(u,v_1,v_2,v_3)$ defined by

\begin{eqnarray*} \phi _{i}(u)=u, \quad \phi _i(v_i)=-v_i,\quad \phi _i(v_j)=v_j\textrm { for }i\ne j. \end{eqnarray*}

Since $\phi _i(T)=T$ and $\phi _i(D)=D$ , the automorphism $\phi _i$ induces a map $\Phi _i$ on $E'$ and its Jacobian $J$ . In particular,

\begin{equation*}\Phi _i(P_i)=P_i, \,i=1,2,3,\quad \Phi _i(P_j)=\left (\frac {1}{v_j^2},-w_j\right)\!,\,i\ne j. \end{equation*}

We define the $K$ -rational divisors $D_i\ {:}{=}\ (P_i)-(\infty )$ , $i=1,2,3$ , on $E'$ . In accordance with the proof of Theorem 2.2, the divisors $D_i$ define rational points of infinite order in $J$ . In addition, one has

\begin{equation*}\Phi _i(D_i)=D_i,\,i=1,2,3,\qquad \Phi _i(D_j)=\left (\left (\frac {1}{v_j^2},-w_j\right )\right )-(\infty )\sim -D_j,\,i\ne j.\end{equation*}

We now assume that there are integers $\alpha ,\beta ,\gamma$ such that $\alpha D_1 +\beta D_2+\gamma D_3\sim 0$ . Applying the automorphism $\Phi _1$ , one gets $\alpha D_1-\beta D_2-\gamma D_3\sim 0$ . Adding the two linear equivalence relations, one obtains $2\alpha D_1\sim 0$ , hence $\alpha =0$ . Similarly, one applies the automorphism $\Phi _2$ and obtains $\beta =0$ . It follows that $D_1$ , $D_2$ , and $D_3$ are linearly independent.

3. A quadratic twist and three twists of even degrees

Let $f(x)\in \mathbb{Q}[x]$ be a square-free polynomial of degree at least $3$ . Fix odd integers $m_1,m_2,m_3\ge 3$ . We are investigating the existence of non-square rational numbers $D$ such that the Jacobians of the (hyperelliptic) curves

\begin{equation*} H\,:\,Dy^2=f(x),\quad H_1\,:\,y^2=x^{m_1}+a D,\quad H_2\,:\,y^2=x^{m_2}+b D,\quad H_3\,:\,y^2=x^{m_3}+c D \end{equation*}

are of positive Mordell–Weil rank over $\mathbb{Q}$ .

Setting $M=\textrm {lcm}(m_1,m_2,m_3)$ and $M_{i}=M/m_i$ , $i=1,2,3$ . We find the rational number $D$ by solving the following system of equations

(3.1) \begin{equation} \frac {y_1^2-x_1^{m_1}}{a}=\frac {y_2^2-x_2^{m_2}}{b}=\frac {y_3^2-x_3^{m_3}}{c}=\frac {f(x_4)}{y_4^2}, \end{equation}

where the solutions $x_i,y_i$ , $i=1,2,3,4$ , satisfy $x_i y_i\ f(x_4)(y_1^2-x_1^{m_1})(y_2^2-x_2^{m_2})(y_3^2-x_3^{m_3})\ne 0$ . To produce a family of parametric solutions to the system (3.1), we set

\begin{eqnarray*} x_1&=&v_1^{2}T^{M_1},\, x_2=v_2^2T^{M_2},\, x_3=v_3^2T^{M_3},\, x_4 =u,\\ y_1&=&pT^{(M-1)/2},\, y_2=qT^{(M-1)/2},\,y_3=rT^{(M-1)/2},\,y_4=\frac {1}{T^{(M-1)/2}}, \end{eqnarray*}

where $u,v_1,v_2,v_3$ are rational parameters. After simplification, one obtains

\begin{equation*}T=\frac {p^2-af(u)}{v_1^{2m_1}}=\frac {q^2-bf(u)}{v_2^{2m_2}}=\frac {r^2-cf(u)}{v_3^{2m_3}}.\end{equation*}

It follows that solving the system (3.1) is equivalent to finding rational points on the curve

\begin{equation*}v_2^{2m_2}v_3^{2m_3}(p^2-af(u))=v_1^{2m_1}v_3^{2m_3}(q^2-bf(u))=v_1^{2m_1}v_2^{2m_2}(r^2-cf(u))\end{equation*}

which is an intersection $\mathcal H_{a,b,c}$ of two quadratic surfaces in $\mathbb{P}^3$ over $\mathbb{Q}(v_1,v_2,v_3)$ . In addition, the curve $\mathcal H_{a,b,c}$ possesses the four projective rational points $[p\,:\,q\,:\,r\,:\,s]=[\!\pm v_1^{m_1}\,:\,\pm v_2^{m_2}\,:\,\pm v_3^{m_3}\,:\,0]$ . Thus, $\mathcal H_{a,b,c}$ is an elliptic curve.

Theorem 3.1. Let $E$ be an elliptic curve defined by $y^2=f(x)$ where $f \in \mathbb{Q}[x]$ is a polynomial of degree $3$ or $4$ . Assume, moreover, that the Mordell–Weil rank of $E$ is positive. Let $m_1,m_2,m_3\ge 3$ be odd integers. Consider the hyperelliptic curves

\begin{equation*}E_1\,:\,y^2=x^{m_1}+a^2,\qquad E_2\,:\, y^2=x^{m_2}+b^2,\qquad E_3\,:\,y^2=x^{m_3}+c^2,\end{equation*}

where $a,b,c$ are non-zero rational numbers. Then there exists a rational function $D_{2,2m_1,2m_2,2m_3} \in \mathbb{Q}( v_1,v_2,v_3)$ such that the quadratic twist of the curve $E$ and the Jacobians of the $2m_i$ -twists of the curves $E_i$ , $i=1,2,3$ , by $D_{2,2m_1,2m_2,2m_3}$ have positive Mordell–Weil rank over $\mathbb{Q}(v_1, v_2,v_3)$ .

Proof. Following the discussion above, one needs to find a rational point on the elliptic curve

\begin{equation*}\mathcal H_{a^2,b^2,c^2}\,:\, v_2^{2m_2}v_3^{2m_3}(p^2-a^2f(u))=v_1^{2m_1}v_3^{2m_3}(q^2-b^2f(u))=v_1^{2m_1}v_2^{2m_2}(r^2-c^2f(u)).\end{equation*}

Since $E$ is of positive Mordell–Weil rank, we choose $u$ such that $(u,y_u)\in E(\mathbb{Q})$ is of infinite order. Now, the elliptic curve $\mathcal H_{a^2,b^2,c^2}$ is birational to

\begin{equation*}x^2-T_{1}^2=y^2-T_{2}^2=z^2-T_{3}^2\end{equation*}

where $T_{1}=av_2^{m_2}v_3^{m_3}y_u$ , $T_{2}=bv_1^{m_1}v_3^{m_3}y_u$ , and $T_{3}=cv_1^{m_1}v_2^{m_2}y_u$ . According to Proposition2.1, the latter curve is of positive Mordell–Weil rank over $\mathbb{Q}(T_1,T_2,T_3)$ . Now, the proof is similar to the proof of Theorem 2.2.

We remark that doubling the point of infinite order $[a y_u\,:\,by_u\,:\,cy_u\,:\,1]\in \mathcal H_{a^2,b^2,c^2}(\mathbb{Q}(v_1,v_2,v_3))$ yields the point $[w_1\,:\,w_2\,:\,w_3\,:\,1]$ given by

\begin{eqnarray*} \bigg [\frac {a^2 b^2 v_3^{2m_3} y_u + a^2 c^2 v_2^{2m_2} y_u - b^2 c^2 v_1^{2m_1} y_u}{2 a b c v_2^{m_2} v_3^{m_3}} &:& \frac {a^2 b^2 v_3^{2m_3} y_u - a^2 c^2 v_2^{2m_2}y_u + b^2 c^2 v_1^{2m_1} y_u}{2a b c v_1^{m_1}v_3^{m_3}} \\ &:& \frac {-a^2 b^2 v_3^{2m_3} y_u + a^2 c^2 v_2^{2m_2}y_u + b^2 c^2 v_1^{2m_1}y_u}{2a b c v_1^{m_1}v_2^{m_2}} \,:\, 1 \bigg ]\in \mathcal H_{a^2,b^2,c^2}(\mathbb{Q}(v_1,v_2,v_3)). \end{eqnarray*}

One obtains the following result.

Proof. We set $m\ {:}{=}\ m_1=m_2=m_3$ and $a=b=c$ in Theorem 3.1. The three $\mathbb{Q}(v_1,v_2,v_3)$ -rational points $P_i\ {:}{=}\ \left (v_i^2, w_i T^{(m-1)/2}\right )$ , $i=1,2,3$ , on the curve $y^2=x^m+a D$ are of infinite order.

We consider the automorphisms $\phi _{i}$ , $ i=1,2,3$ , of $\mathbb{Q}(v_1,v_2,v_3)$ defined by

\begin{equation*} \phi _i(v_i)=-v_i,\quad \phi _i(v_j)=v_j\textrm { for }i\ne j. \end{equation*}

Now the proof follows as in the proof of Theorem 2.4.

4. One even-degree twist and three odd-degree twists

Given the curves

\begin{equation*}y^2=x^{m_1}+a,\quad y^2=x^{m_2}+b,\quad y^2=x^{m_3}+c,\quad y^2=x^{m_4}+d,\end{equation*}

where $a,b,c,d$ are non-zero rationals and $m_i$ , $i=1,2,3,4$ , are odd integers, we investigate the existence of a rational function $D$ such that the Jacobians of the following twists are of positive Mordell–Weil rank:

\begin{equation*}y^2=Dx^{m_1}+a,\quad y^2=Dx^{m_2}+b,\quad y^2=Dx^{m_3}+c,\quad y^2=x^{m_4}+dD.\end{equation*}

The rational $D$ can be obtained by solving the system:

\begin{equation*}\frac {y_1^2-a}{x_1^{m_1}}=\frac {y_2^2-b}{x_2^{m_2}}=\frac {y_3^2-c}{x_3^{m_3}}=\frac {y_4^2-x_4^{m_4}}{d}\end{equation*}

where $x_iy_i(y_1^2-a)(y_2^2-b)(y_3^2-c)(y_4^2-x_4^{m_4})\ne 0$ . Let $M=\textrm {lcm}(m_1,m_2,m_3,m_4)$ with $M_i=M/m_i$ , $i=1,2,3,4$ . We set

\begin{eqnarray*} x_1&=&\frac {1}{v_1^2T^{M_1}},\quad x_2=\frac {1}{v_2^2T^{M_2}},\quad x_3=\frac {1}{v_3^2T^{M_3}},\quad x_4=v_4^2 T^{m_4}, \\ y_1&=&p,\quad y_2=q,\quad y_3=r,\quad y_4=uT^{(M-1)/2} \end{eqnarray*}

where $u,v_1,v_2,v_3,v_4$ are rational parameters. One now obtains that

\begin{equation*}(p^2-a)v_1^{2m_1}T=(q^2-b)v_2^{2m_2}T=(r^2-c)v_3^{2m_3}T=\frac {u^2-v_4^{2m_4}T}{d}.\end{equation*}

In other words, one needs to solve the system:

\begin{equation*}T=\frac {u^2}{v_4^{2m_4}+d(r^2-c)v_3^{2m_3}},\quad v_1^{2m_1}(p^2-a)=v_2^{2m_2}(q^2-b)=v_3^{2m_3}(r^2-c).\end{equation*}

We now obtain the following result.

Theorem 4.1. Let $m_1,m_2,m_3,m_4\ge 3$ be odd integers. Consider the hyperelliptic curves

\begin{equation*} E_1\,:\,y^2=x^{m_1}+a^2,\quad E_2\,:\, y^2=x^{m_2}+b^2,\quad E_3\,:\,y^2=x^{m_3}+c^2,\quad E_4\,:\,y^2=x^{m_4}+d\end{equation*}

where $a,b,c,d$ are non-zero rational numbers. Then there exists a rational function $D_{m_1,m_2,m_3,2m_4} \in \mathbb{Q}(u, v_1,v_2,v_3)$ such that the Jacobian of the $m_i$ -twists of the curves $E_i$ , $i=1,2,3$ by $D_{m_1,m_2,m_3,2m_4}$ and the Jacobian of the $2m_4$ -twist of $E_4$ by $D_{m_1,m_2,m_3,2m_4}$ have positive Mordell–Weil rank over $\mathbb{Q}(u, v_1, v_2,v_3)$ .