Let $f(t)$ be a rational normal cubic curve in $\mathbb{P}^3$ (it is not contained in any plane) and also we assume that this cubic curve passes through two points $(0,0,0)$ and $(1,0,0)$. By an easy calculation one can see that $f(t)=(f_1(t),f_2(t),f_3(t))$ is of the following form:

\begin{align*} f_1(t) &= t^3 + at^2 -at\\ f_2(t) &= b t^3 + c t^2 - (b+c)t\\ f_3(t)&= d t^3 + e t^2 - (d+e)t. \end{align*} and we assume that all the coefficients in the above curve is in the number field $K=\mathbb{Q}(\sqrt{k}, \sqrt{k'})$ for two square free integers $k$ and $k'$. Let $t_j \in K$ and Consider the distance function $$D(t)= \big(f_1(t) - f_1(t_j)\big)^2 + \big(f_2(t) - f_2(t_j)\big)^2 + \big(f_3(t) - f_3(t_j)\big)^2, $$ In fact $D(t)$ is the square of the distance to a fixed point on the cubic curve corresponding to the parameter $t_j$. The polynomial $D(t)$ is a degree $6$ polynomial which has a root $t=t_j$ of multiplicity $2$, so it can be written in the form \begin{align*} D(t)= (t-t_j)^2 Q_j(t) \end{align*}
where $Q_j(t)= c_4(t_j)t^4 + c_3(t_j)t^3 + c_2(t_j)t^2 + c_1(t_j)t + c_0(t_j)$ and by an easy calculation we see that the coefficient $c_i(t_j)$ is a polynomial of degree $4 - i$ in terms of $t_j$ as the following.

\begin{align*} c_0(t_j)&=(1+b^2 +d^2)t_j^4 + 2 (a +bc + ed)t_j^3 + (a^2 + c^2 +e^2 -2a -2bc -2de -2d^2 )t_j^2 \\ &+ (a^2 + c^2 + e^2 -2bc -2de)t_j + a^2 + (b+c)^2 + (d+e)^2\\ c_1(t_j) &= 2(1+b^2 + d^2)t_j^3 + 4(a + bc + de )t_j^2 + 2(a^2 + c^2 + e^2 - b^2 - d^2 - a- bc - de )t_j \\ &-2 (a^2 + c^2 + e^2 + bc + de)\\ c_2(t_j)&= 3(1+b^2 + d^2)t_j^2 + 2(bc + de + a )t_j -2 (a^2 + b^2 + c^2 + d^2 + e^2 + a -bc -de)\\ c_3(t_j)&= 2(1+b^2 + d^2)t_j + 2(a +bc + de)\\ c_4(t_j)&=1+b^2 + d^2, \end{align*} Since $D(t)$ is the distance polynomial it has no other real roots. Thus the roots of the polynomials $Q_j(t)$ are all complex roots. I want to prove that there exists two parameters $t_1, t_2 \in K$ for which the corresponding polynomials $Q_1$ and $Q_2$ has no common root. A natural way to think about this problem is to consider the resultant of two polynomials $Q_1$ and $Q_2$. One can see that the Resultant Res$(Q_1,Q_2)$ is a degree $16$ polynomial in therms of two variable $t_1$ and $t_2$, which always has the degree $4$ factor $(t_1-t_2)^4$. So if we mode out the factor $(t_1-t_2)^4$ we arrive at a degree $12$ plane algebraic curve $C: P(t_1,t_2)=0$ over the number field $K$. I need to prove that the genus of the degree $12$ curve $C$ is always greater than $1$  ,i.e.$g\ge2$, so by Fallings theorem there is only finitely many $K$-rational points on the curve $C$. Hence we can find two parameters $t_1, t_2 \in K$ for which we have Res$(Q_1,Q_2)(t_1,t_2)\neq 0$ and hence we find two polynomials $Q_1$ and $Q_2$ with no common root.\ I use the genus command in Maple and compute the genus of the above resultant for some choice of parameters $a,b,c,d,e$. For example for the choice $a=0, b=e=1, c=d=0$ which corresponds to the cubic curve $E: f(t)=(t^3,t^3-t, t^2-t)$ and computing the corresponding resultant $E'$ with Maple and the the genus, we come up with $g(E')=9$! with the choice $a=b=c=d=1 , e=0$ the genus of the corresponding resultant becomes $19$. by some other computation with Maple, It seems that the minimum genus for choosing the parameters $a,b,c,d,e$ is $9$. Note that I only need to prove that the above genus is greater than $1$. Experimentally this true but I want to prove it theoretically.\

One raw idea: If somehow we could construct a morphism from the curve $C$ to the curve $E'$, $ \phi : C \rightarrow E'$ then since $g(E')=9$ by Riemann-Hurwits formula one can easily show that $g(C) > 1$ in order to apply the Faltings theorem.

I appreciate it if any one could have any idea/comment on this problem.

Let $f(t)$ be a rational normal cubic curve in $\mathbb{P}^3$ (it is not contained in any plane) and also we assume that this cubic curve passes through two points $(0,0,0)$ and $(1,0,0)$. By an easy calculation one can see that $f(t)=(f_1(t),f_2(t),f_3(t))$ is of the following form:

\begin{align*} f_1(t) &= t^3 + at^2 -at\\ f_2(t) &= b t^3 + c t^2 - (b+c)t\\ f_3(t)&= d t^3 + e t^2 - (d+e)t. \end{align*} and we assume that all the coefficients in the above curve is in the number field $K=\mathbb{Q}(\sqrt{k}, \sqrt{k'})$ for two square free integers $k$ and $k'$. Let $t_j \in K$ and Consider the distance function $$D(t)= \big(f_1(t) - f_1(t_j)\big)^2 + \big(f_2(t) - f_2(t_j)\big)^2 + \big(f_3(t) - f_3(t_j)\big)^2, $$ In fact $D(t)$ is the square of the distance to a fixed point on the cubic curve corresponding to the parameter $t_j$. The polynomial $D(t)$ is a degree $6$ polynomial which has a root $t=t_j$ of multiplicity $2$, so it can be written in the form \begin{align*} D(t)= (t-t_j)^2 Q_j(t) \end{align*}
where $Q_j(t)= c_4(t_j)t^4 + c_3(t_j)t^3 + c_2(t_j)t^2 + c_1(t_j)t + c_0(t_j)$ and by an easy calculation we see that the coefficient $c_i(t_j)$ is a polynomial of degree $4 - i$ in terms of $t_j$ as the following.

\begin{align*} c_0(t_j)&=(1+b^2 +d^2)t_j^4 + 2 (a +bc + ed)t_j^3 + (a^2 + c^2 +e^2 -2a -2bc -2de -2d^2 )t_j^2 \\ &+ (a^2 + c^2 + e^2 -2bc -2de)t_j + a^2 + (b+c)^2 + (d+e)^2\\ c_1(t_j) &= 2(1+b^2 + d^2)t_j^3 + 4(a + bc + de )t_j^2 + 2(a^2 + c^2 + e^2 - b^2 - d^2 - a- bc - de )t_j \\ &-2 (a^2 + c^2 + e^2 + bc + de)\\ c_2(t_j)&= 3(1+b^2 + d^2)t_j^2 + 2(bc + de + a )t_j -2 (a^2 + b^2 + c^2 + d^2 + e^2 + a -bc -de)\\ c_3(t_j)&= 2(1+b^2 + d^2)t_j + 2(a +bc + de)\\ c_4(t_j)&=1+b^2 + d^2, \end{align*} Since $D(t)$ is the distance polynomial it has no other real roots. Thus the roots of the polynomials $Q_j(t)$ are all complex roots. I want to prove that there exists two parameters $t_1, t_2 \in K$ for which the corresponding polynomials $Q_1$ and $Q_2$ has no common root. A natural way to think about this problem is to consider the resultant of two polynomials $Q_1$ and $Q_2$. One can see that the Resultant Res$(Q_1,Q_2)$ is a degree $16$ polynomial in therms of two variable $t_1$ and $t_2$, which always has the degree $4$ factor $(t_1-t_2)^4$. So if we mode out the factor $(t_1-t_2)^4$ we arrive at a degree $12$ plane algebraic curve $C: P(t_1,t_2)=0$ over the number field $K$. I need to prove that the genus of the degree $12$ curve $C$ is always greater than $1$, i.e.  $g\ge2$, so by Fallings theorem there is only finitely many $K$-rational points on the curve $C$. Hence we can find two parameters $t_1, t_2 \in K$ for which we have Res$(Q_1,Q_2)(t_1,t_2)\neq 0$ and hence we find two polynomials $Q_1$ and $Q_2$ with no common root.\ I use the genus command in Maple and compute the genus of the above resultant for some choice of parameters $a,b,c,d,e$. For example for the choice $a=0, b=e=1, c=d=0$ which corresponds to the cubic curve $E: f(t)=(t^3,t^3-t, t^2-t)$ and computing the corresponding resultant $E'$ with Maple and the the genus, we come up with $g(E')=9$! with the choice $a=b=c=d=1 , e=0$ the genus of the corresponding resultant becomes $19$. by some other computation with Maple, It seems that the minimum genus for choosing the parameters $a,b,c,d,e$ is $9$. Note that I only need to prove that the above genus is greater than $1$. Experimentally this true but I want to prove it theoretically.\

One raw idea: If somehow we could construct a morphism from the curve $C$ to the curve $E'$, $ \phi : C \rightarrow E'$ then since $g(E')=9$ by Riemann-Hurwits formula one can easily show that $g(C) > 1$ in order to apply the Faltings theorem.

I appreciate it if any one could have any idea/comment on this problem.

Let $f(t)$ be a rational normal cubic curve in $\mathbb{P}^3$ (it is not contained in any plane) and also we assume that this cubic curve passes through two points $(0,0,0)$ and $(1,0,0)$. By an easy calculation one can see that $f(t)=(f_1(t),f_2(t),f_3(t))$ is of the following form:

\begin{align*} f_1(t) &= t^3 + at^2 -at\\ f_2(t) &= b t^3 + c t^2 - (b+c)t\\ f_3(t)&= d t^3 + e t^2 - (d+e)t. \end{align*} and we assume that all the coefficients in the above curve is in the number field $K=\mathbb{Q}(\sqrt{k}, \sqrt{k'})$ for two square free integers $k$ and $k'$. Let $t_j \in K$ and Consider the distance function $$D(t)= \big(f_1(t) - f_1(t_j)\big)^2 + \big(f_2(t) - f_2(t_j)\big)^2 + \big(f_3(t) - f_3(t_j)\big)^2, $$ In fact $D(t)$ is the square of the distance to a fixed point on the cubic curve corresponding to the parameter $t_j$. The polynomial $D(t)$ is a degree $6$ polynomial which has a root $t=t_j$ of multiplicity $2$, so it can be written in the form \begin{align*} D(t)= (t-t_j)^2 Q_j(t) \end{align*}
where $Q_j(t)= c_4(t_j)t^4 + c_3(t_j)t^3 + c_2(t_j)t^2 + c_1(t_j)t + c_0(t_j)$ and by an easy calculation we see that the coefficient $c_i(t_j)$ is a polynomial of degree $4 - i$ in terms of $t_j$ as the following.

\begin{align*} c_0(t_j)&=(1+b^2 +d^2)t_j^4 + 2 (a +bc + ed)t_j^3 + (a^2 + c^2 +e^2 -2a -2bc -2de -2d^2 )t_j^2 \\ &+ (a^2 + c^2 + e^2 -2bc -2de)t_j + a^2 + (b+c)^2 + (d+e)^2\\ c_1(t_j) &= 2(1+b^2 + d^2)t_j^3 + 4(a + bc + de )t_j^2 + 2(a^2 + c^2 + e^2 - b^2 - d^2 - a- bc - de )t_j \\ &-2 (a^2 + c^2 + e^2 + bc + de)\\ c_2(t_j)&= 3(1+b^2 + d^2)t_j^2 + 2(bc + de + a )t_j -2 (a^2 + b^2 + c^2 + d^2 + e^2 + a -bc -de)\\ c_3(t_j)&= 2(1+b^2 + d^2)t_j + 2(a +bc + de)\\ c_4(t_j)&=1+b^2 + d^2, \end{align*} Since $D(t)$ is the distance polynomial it has no other real roots. Thus the roots of the polynomials $Q_j(t)$ are all complex roots. I want to prove that there exists two parameters $t_1, t_2 \in K$ for which the corresponding polynomials $Q_1$ and $Q_2$ has no common root. A natural way to think about this problem is to consider the resultant of two polynomials $Q_1$ and $Q_2$. One can see that the Resultant Res$(Q_1,Q_2)$ is a degree $16$ polynomial in therms of two variable $t_1$ and $t_2$, which always has the degree $4$ factor $(t_1-t_2)^4$. So if we mode out the factor $(t_1-t_2)^4$ we arrive at a degree $12$ plane algebraic curve $C: P(t_1,t_2)=0$ over the number field $K$. I need to prove that the genus of the degree $12$ curve $C$ is always greater than $1$ ,i.e.$g\ge2$, so by Fallings theorem there is only finitely many $K$-rational points on the curve $C$. Hence we can find two parameters $t_1, t_2 \in K$ for which we have Res$(Q_1,Q_2)(t_1,t_2)\neq 0$ and hence we find two polynomials $Q_1$ and $Q_2$ with no common root.\ I use the genus command in Maple and compute the genus of the above resultant for some choice of parameters $a,b,c,d,e$. For example for the choice $a=0, b=e=1, c=d=0$ which corresponds to the cubic curve $E: f(t)=(t^3,t^3-t, t^2-t)$ and computing the corresponding resultant $E'$ with Maple and the the genus, we come up with $g(E')=9$! with the choice $a=b=c=d=1 , e=0$ the genus of the corresponding resultant becomes $19$. by some other computation with Maple, It seems that the minimum genus for choosing the parameters $a,b,c,d,e$ is $9$. Note that I only need to prove that the above genus is greater than $1$. Experimentally this true but I want to prove it theoretically.\

One idea: If somehow we could construct a morphism from any the curve $ C \rightarrow E'$ then since $g(E')=9$ by Riemann-Hurwits formula one can easily show that $g(C) > 1$ in order to apply the Faltings theorem.

I appreciate it if any one could have any idea/comment on this problem.

Let $f(t)$ be a rational normal cubic curve in $\mathbb{P}^3$ (it is not contained in any plane) and also we assume that this cubic curve passes through two points $(0,0,0)$ and $(1,0,0)$. By an easy calculation one can see that $f(t)=(f_1(t),f_2(t),f_3(t))$ is of the following form:

\begin{align*} f_1(t) &= t^3 + at^2 -at\\ f_2(t) &= b t^3 + c t^2 - (b+c)t\\ f_3(t)&= d t^3 + e t^2 - (d+e)t. \end{align*} and we assume that all the coefficients in the above curve is in the number field $K=\mathbb{Q}(\sqrt{k}, \sqrt{k'})$ for two square free integers $k$ and $k'$. Let $t_j \in K$ and Consider the distance function $$D(t)= \big(f_1(t) - f_1(t_j)\big)^2 + \big(f_2(t) - f_2(t_j)\big)^2 + \big(f_3(t) - f_3(t_j)\big)^2, $$ In fact $D(t)$ is the square of the distance to a fixed point on the cubic curve corresponding to the parameter $t_j$. The polynomial $D(t)$ is a degree $6$ polynomial which has a root $t=t_j$ of multiplicity $2$, so it can be written in the form \begin{align*} D(t)= (t-t_j)^2 Q_j(t) \end{align*}
where $Q_j(t)= c_4(t_j)t^4 + c_3(t_j)t^3 + c_2(t_j)t^2 + c_1(t_j)t + c_0(t_j)$ and by an easy calculation we see that the coefficient $c_i(t_j)$ is a polynomial of degree $4 - i$ in terms of $t_j$ as the following.

\begin{align*} c_0(t_j)&=(1+b^2 +d^2)t_j^4 + 2 (a +bc + ed)t_j^3 + (a^2 + c^2 +e^2 -2a -2bc -2de -2d^2 )t_j^2 \\ &+ (a^2 + c^2 + e^2 -2bc -2de)t_j + a^2 + (b+c)^2 + (d+e)^2\\ c_1(t_j) &= 2(1+b^2 + d^2)t_j^3 + 4(a + bc + de )t_j^2 + 2(a^2 + c^2 + e^2 - b^2 - d^2 - a- bc - de )t_j \\ &-2 (a^2 + c^2 + e^2 + bc + de)\\ c_2(t_j)&= 3(1+b^2 + d^2)t_j^2 + 2(bc + de + a )t_j -2 (a^2 + b^2 + c^2 + d^2 + e^2 + a -bc -de)\\ c_3(t_j)&= 2(1+b^2 + d^2)t_j + 2(a +bc + de)\\ c_4(t_j)&=1+b^2 + d^2, \end{align*} Since $D(t)$ is the distance polynomial it has no other real roots. Thus the roots of the polynomials $Q_j(t)$ are all complex roots. I want to prove that there exists two parameters $t_1, t_2 \in K$ for which the corresponding polynomials $Q_1$ and $Q_2$ has no common root. A natural way to think about this problem is to consider the resultant of two polynomials $Q_1$ and $Q_2$. One can see that the Resultant Res$(Q_1,Q_2)$ is a degree $16$ polynomial in therms of two variable $t_1$ and $t_2$, which always has the degree $4$ factor $(t_1-t_2)^4$. So if we mode out the factor $(t_1-t_2)^4$ we arrive at a degree $12$ plane algebraic curve $C: P(t_1,t_2)=0$ over the number field $K$. I need to prove that the genus of the degree $12$ curve $C$ is always greater than $1$ ,i.e.$g\ge2$, so by Fallings theorem there is only finitely many $K$-rational points on the curve $C$. Hence we can find two parameters $t_1, t_2 \in K$ for which we have Res$(Q_1,Q_2)(t_1,t_2)\neq 0$ and hence we find two polynomials $Q_1$ and $Q_2$ with no common root.\ I use the genus command in Maple and compute the genus of the above resultant for some choice of parameters $a,b,c,d,e$. For example for the choice $a=0, b=e=1, c=d=0$ which corresponds to the cubic curve $E: f(t)=(t^3,t^3-t, t^2-t)$ and computing the corresponding resultant $E'$ with Maple and the the genus, we come up with $g(E')=9$! with the choice $a=b=c=d=1 , e=0$ the genus of the corresponding resultant becomes $19$. by some other computation with Maple, It seems that the minimum genus for choosing the parameters $a,b,c,d,e$ is $9$. Note that I only need to prove that the above genus is greater than $1$. Experimentally this true but I want to prove it theoretically.\

One raw idea: If somehow we could construct a morphism from the curve $C$ to the curve $E'$, $ \phi : C \rightarrow E'$ then since $g(E')=9$ by Riemann-Hurwits formula one can easily show that $g(C) > 1$ in order to apply the Faltings theorem.

I appreciate it if any one could have any idea/comment on this problem.

Let $f(t)$ be a rational normal cubic curve in $\mathbb{P}^3$ (it is not contained in any plane) and also we assume that this cubic curve passes through two points $(0,0,0)$ and $(1,0,0)$. By an easy calculation one can see that $f(t)=(f_1(t),f_2(t),f_3(t))$ is of the following form:

\begin{align*} f_1(t) &= t^3 + at^2 -at\\ f_2(t) &= b t^3 + c t^2 - (b+c)t\\ f_3(t)&= d t^3 + e t^2 - (d+e)t. \end{align*} and we assume that all the coefficients in the above curve is in the number field $K=\mathbb{Q}(k,k')$. Let $t_j \in K$ and Consider the distance function $$D(t)= \big(f_1(t) - f_1(t_j)\big)^2 + \big(f_2(t) - f_2(t_j)\big)^2 + \big(f_3(t) - f_3(t_j)\big)^2, $$ In fact $D(t)$ is the square of the distance to a fixed point on the cubic curve corresponding to the parameter $t_j$. The polynomial $D(t)$ is a degree $6$ polynomial which has a root $t=t_j$ of multiplicity $2$, so it can be written in the form \begin{align*} D(t)= (t-t_j)^2 Q_j(t) \end{align*}
where $Q_j(t)= c_4(t_j)t^4 + c_3(t_j)t^3 + c_2(t_j)t^2 + c_1(t_j)t + c_0(t_j)$ and by an easy calculation we see that the coefficient $c_i(t_j)$ is a polynomial of degree $4 - i$ in terms of $t_j$ as the following.

\begin{align*} c_0(t_j)&=(1+b^2 +d^2)t_j^4 + 2 (a +bc + ed)t_j^3 + (a^2 + c^2 +e^2 -2a -2bc -2de -2d^2 )t_j^2 \\ &+ (a^2 + c^2 + e^2 -2bc -2de)t_j + a^2 + (b+c)^2 + (d+e)^2\\ c_1(t_j) &= 2(1+b^2 + d^2)t_j^3 + 4(a + bc + de )t_j^2 + 2(a^2 + c^2 + e^2 - b^2 - d^2 - a- bc - de )t_j \\ &-2 (a^2 + c^2 + e^2 + bc + de)\\ c_2(t_j)&= 3(1+b^2 + d^2)t_j^2 + 2(bc + de + a )t_j -2 (a^2 + b^2 + c^2 + d^2 + e^2 + a -bc -de)\\ c_3(t_j)&= 2(1+b^2 + d^2)t_j + 2(a +bc + de)\\ c_4(t_j)&=1+b^2 + d^2, \end{align*} Since $D(t)$ is the distance polynomial it has no other real roots. Thus the roots of the polynomials $Q_j(t)$ are all complex roots. I want to prove that there exists two parameters $t_1, t_2 \in K$ for which the corresponding polynomials $Q_1$ and $Q_2$ has no common root. A natural way to think about this problem is to consider the resultant of two polynomials $Q_1$ and $Q_2$. One can see that the Resultant Res$(Q_1,Q_2)$ is a degree $16$ polynomial in therms of two variable $t_1$ and $t_2$, which always has the degree $4$ factor $(t_1-t_2)^4$. So if we mode out the factor $(t_1-t_2)^4$ we arrive at a degree $12$ plane algebraic curve $C: P(t_1,t_2)=0$ over the number field $K$. I need to prove that the genus of the degree $12$ curve $C$ is always greater than $1$ ,i.e.$g\ge2$, so by Fallings theorem there is only finitely many $K$-rational points on the curve $C$. Hence we can find two parameters $t_1, t_2 \in K$ for which we have Res$(Q_1,Q_2)(t_1,t_2)\neq 0$ and hence we find two polynomials $Q_1$ and $Q_2$ with no common root.\ I use the genus command in Maple and compute the genus of the above resultant for some choice of parameters $a,b,c,d,e$. For example for the choice $a=0, b=e=1, c=d=0$ which corresponds to the cubic curve $E: f(t)=(t^3,t^3-t, t^2-t)$ and computing the corresponding resultant $E'$ with Maple and the the genus, we come up with $g(E')=9$! with the choice $a=b=c=d=1 , e=0$ the genus of the corresponding resultant becomes $19$. by some other computation with Maple, It seems that the minimum genus for choosing the parameters $a,b,c,d,e$ is $9$. Note that I only need to prove that the above genus is greater than $1$. Experimentally this true but I want to prove it theoretically.\

One idea: If somehow we could construct a morphism from any the curve $ C \rightarrow E'$ then since $g(E')=9$ by Riemann-Hurwits formula one can easily show that $g(C) > 1$ in order to apply the Faltings theorem.

I appreciate it if any one could have any idea/comment on this problem.

Let $f(t)$ be a rational normal cubic curve in $\mathbb{P}^3$ (it is not contained in any plane) and also we assume that this cubic curve passes through two points $(0,0,0)$ and $(1,0,0)$. By an easy calculation one can see that $f(t)=(f_1(t),f_2(t),f_3(t))$ is of the following form:

\begin{align*} f_1(t) &= t^3 + at^2 -at\\ f_2(t) &= b t^3 + c t^2 - (b+c)t\\ f_3(t)&= d t^3 + e t^2 - (d+e)t. \end{align*} and we assume that all the coefficients in the above curve is in the number field $K=\mathbb{Q}(\sqrt{k}, \sqrt{k'})$ for two square free integers $k$ and $k'$. Let $t_j \in K$ and Consider the distance function $$D(t)= \big(f_1(t) - f_1(t_j)\big)^2 + \big(f_2(t) - f_2(t_j)\big)^2 + \big(f_3(t) - f_3(t_j)\big)^2, $$ In fact $D(t)$ is the square of the distance to a fixed point on the cubic curve corresponding to the parameter $t_j$. The polynomial $D(t)$ is a degree $6$ polynomial which has a root $t=t_j$ of multiplicity $2$, so it can be written in the form \begin{align*} D(t)= (t-t_j)^2 Q_j(t) \end{align*}
where $Q_j(t)= c_4(t_j)t^4 + c_3(t_j)t^3 + c_2(t_j)t^2 + c_1(t_j)t + c_0(t_j)$ and by an easy calculation we see that the coefficient $c_i(t_j)$ is a polynomial of degree $4 - i$ in terms of $t_j$ as the following.

\begin{align*} c_0(t_j)&=(1+b^2 +d^2)t_j^4 + 2 (a +bc + ed)t_j^3 + (a^2 + c^2 +e^2 -2a -2bc -2de -2d^2 )t_j^2 \\ &+ (a^2 + c^2 + e^2 -2bc -2de)t_j + a^2 + (b+c)^2 + (d+e)^2\\ c_1(t_j) &= 2(1+b^2 + d^2)t_j^3 + 4(a + bc + de )t_j^2 + 2(a^2 + c^2 + e^2 - b^2 - d^2 - a- bc - de )t_j \\ &-2 (a^2 + c^2 + e^2 + bc + de)\\ c_2(t_j)&= 3(1+b^2 + d^2)t_j^2 + 2(bc + de + a )t_j -2 (a^2 + b^2 + c^2 + d^2 + e^2 + a -bc -de)\\ c_3(t_j)&= 2(1+b^2 + d^2)t_j + 2(a +bc + de)\\ c_4(t_j)&=1+b^2 + d^2, \end{align*} Since $D(t)$ is the distance polynomial it has no other real roots. Thus the roots of the polynomials $Q_j(t)$ are all complex roots. I want to prove that there exists two parameters $t_1, t_2 \in K$ for which the corresponding polynomials $Q_1$ and $Q_2$ has no common root. A natural way to think about this problem is to consider the resultant of two polynomials $Q_1$ and $Q_2$. One can see that the Resultant Res$(Q_1,Q_2)$ is a degree $16$ polynomial in therms of two variable $t_1$ and $t_2$, which always has the degree $4$ factor $(t_1-t_2)^4$. So if we mode out the factor $(t_1-t_2)^4$ we arrive at a degree $12$ plane algebraic curve $C: P(t_1,t_2)=0$ over the number field $K$. I need to prove that the genus of the degree $12$ curve $C$ is always greater than $1$ ,i.e.$g\ge2$, so by Fallings theorem there is only finitely many $K$-rational points on the curve $C$. Hence we can find two parameters $t_1, t_2 \in K$ for which we have Res$(Q_1,Q_2)(t_1,t_2)\neq 0$ and hence we find two polynomials $Q_1$ and $Q_2$ with no common root.\ I use the genus command in Maple and compute the genus of the above resultant for some choice of parameters $a,b,c,d,e$. For example for the choice $a=0, b=e=1, c=d=0$ which corresponds to the cubic curve $E: f(t)=(t^3,t^3-t, t^2-t)$ and computing the corresponding resultant $E'$ with Maple and the the genus, we come up with $g(E')=9$! with the choice $a=b=c=d=1 , e=0$ the genus of the corresponding resultant becomes $19$. by some other computation with Maple, It seems that the minimum genus for choosing the parameters $a,b,c,d,e$ is $9$. Note that I only need to prove that the above genus is greater than $1$. Experimentally this true but I want to prove it theoretically.\

One idea: If somehow we could construct a morphism from any the curve $ C \rightarrow E'$ then since $g(E')=9$ by Riemann-Hurwits formula one can easily show that $g(C) > 1$ in order to apply the Faltings theorem.

I appreciate it if any one could have any idea/comment on this problem.