Ganesh Ujwal
New member
- Joined
- Aug 10, 2014
- Messages
- 32
If \(\displaystyle E/\mathbb{Q}\) the elliptic curve \(\displaystyle y^2=x^3+x^2-25x+29\) and
\(\displaystyle P_1=\left (\frac{61}{4}, \frac{-469}{8}\right ), P_2=\left ( \frac{-335}{81}, \frac{-6868}{729}\right ) , P_3=\left ( 21, 96\right )\) I have to show that these points are \(\displaystyle \mathbb{Z}-\) linearly dependent and indeed that
\(\displaystyle -3P_1-2P_2+6P_3=0\)
To calculate the point \(\displaystyle 3P_1\) , I tried to find firstly \(\displaystyle 2P_1\) :
\(\displaystyle \lambda=\frac{3x_1^2+2x_1-25}{2y_1}, v=\frac{-x_1^3-25x_1+2\cdot 29}{2y_1}\)
\(\displaystyle 2P=(\lambda^2-1-x_1-x_2, -\lambda \cdot x_3-v)\(\displaystyle
\(\displaystyle P_1=\left ( \frac{61}{4}, \frac{-469}{8} \right ) :\)
\(\displaystyle \lambda=\frac{3(\frac{61}{4})^2+2\frac{61}{4}-25}{-2\frac{469}{8}}=\frac{3\frac{61^2}{16}+\frac{61}{2}-25}{-\frac{469}{4}}=\frac{3 \cdot 61^2+ 8 \cdot 61-16 \cdot 25}{- 4 \cdot 469}=-\frac{11251}{1876}, \\ v=\frac{-(\frac{61}{4})^3-25\frac{61}{4}+2\cdot 29}{-2\frac{469}{8}}=\frac{-\frac{61^3}{64}-25\frac{61}{4}+58}{-\frac{469}{4}}=\frac{-61^3-25 \cdot 16 \cdot 61+ 64 \cdot 58}{- 16 \cdot 469}=\frac{-247669}{-7504}=\frac{247669}{7504}\(\displaystyle
\(\displaystyle 2P_1=(x_3, y_3) \\ x_3=\lambda^2-1-2 \frac{61}{4}=\frac{11251^2}{1876^2}-1-\frac{61}{2}=\frac{15724657}{3519376}, \\ y_3= -\lambda \cdot x_3-v=\frac{11251}{1876} \cdot \frac{15724657}{3519376}-\frac{247669}{7504}=-\frac{40991967729}{6602349376}\)
Is it right? Or have I done something wrong?\)\)\)\)
\(\displaystyle P_1=\left (\frac{61}{4}, \frac{-469}{8}\right ), P_2=\left ( \frac{-335}{81}, \frac{-6868}{729}\right ) , P_3=\left ( 21, 96\right )\) I have to show that these points are \(\displaystyle \mathbb{Z}-\) linearly dependent and indeed that
\(\displaystyle -3P_1-2P_2+6P_3=0\)
To calculate the point \(\displaystyle 3P_1\) , I tried to find firstly \(\displaystyle 2P_1\) :
\(\displaystyle \lambda=\frac{3x_1^2+2x_1-25}{2y_1}, v=\frac{-x_1^3-25x_1+2\cdot 29}{2y_1}\)
\(\displaystyle 2P=(\lambda^2-1-x_1-x_2, -\lambda \cdot x_3-v)\(\displaystyle
\(\displaystyle P_1=\left ( \frac{61}{4}, \frac{-469}{8} \right ) :\)
\(\displaystyle \lambda=\frac{3(\frac{61}{4})^2+2\frac{61}{4}-25}{-2\frac{469}{8}}=\frac{3\frac{61^2}{16}+\frac{61}{2}-25}{-\frac{469}{4}}=\frac{3 \cdot 61^2+ 8 \cdot 61-16 \cdot 25}{- 4 \cdot 469}=-\frac{11251}{1876}, \\ v=\frac{-(\frac{61}{4})^3-25\frac{61}{4}+2\cdot 29}{-2\frac{469}{8}}=\frac{-\frac{61^3}{64}-25\frac{61}{4}+58}{-\frac{469}{4}}=\frac{-61^3-25 \cdot 16 \cdot 61+ 64 \cdot 58}{- 16 \cdot 469}=\frac{-247669}{-7504}=\frac{247669}{7504}\(\displaystyle
\(\displaystyle 2P_1=(x_3, y_3) \\ x_3=\lambda^2-1-2 \frac{61}{4}=\frac{11251^2}{1876^2}-1-\frac{61}{2}=\frac{15724657}{3519376}, \\ y_3= -\lambda \cdot x_3-v=\frac{11251}{1876} \cdot \frac{15724657}{3519376}-\frac{247669}{7504}=-\frac{40991967729}{6602349376}\)
Is it right? Or have I done something wrong?\)\)\)\)