Showing $-3P_1-2P_2+6P_3=0$ on an Elliptic Curve

[evinda]

Hello! Merry Christmas! (Beer) (Party) (Bigsmile)If $E/\mathbb{Q}$ the elliptic curve $y^2=x^3+x^2-25x+29$ and

$$P_1=\left (\frac{61}{4}, \frac{-469}{8}\right ), P_2=\left ( \frac{-335}{81}, \frac{-6868}{729}\right ) , P_3=\left ( 21, 96\right )$$ I have to show that these points are $\mathbb{Z}-$linearly dependent and indeed that

$$-3P_1-2P_2+6P_3=0$$

To calculate the point $3P_1$, I tried to find firstly $2P_1$ :

$$\lambda=\frac{3x_1^2+2x_1-25}{2y_1}, v=\frac{-x_1^3-25x_1+2\cdot 29}{2y_1}$$

$$2P=(\lambda^2-1-x_1-x_2, -\lambda \cdot x_3-v)$$

$$P_1=\left ( \frac{61}{4}, \frac{-469}{8} \right ) : $$

$$\lambda=\frac{3(\frac{61}{4})^2+2\frac{61}{4}-25}{-2\frac{469}{8}}=\frac{3\frac{61^2}{16}+\frac{61}{2}-25}{-\frac{469}{4}}=\frac{3 \cdot 61^2+ 8 \cdot 61-16 \cdot 25}{- 4 \cdot 469}=-\frac{11251}{1876}, \\ v=\frac{-(\frac{61}{4})^3-25\frac{61}{4}+2\cdot 29}{-2\frac{469}{8}}=\frac{-\frac{61^3}{64}-25\frac{61}{4}+58}{-\frac{469}{4}}=\frac{-61^3-25 \cdot 16 \cdot 61+ 64 \cdot 58}{- 16 \cdot 469}=\frac{-247669}{-7504}=\frac{247669}{7504}$$

$$2P_1=(x_3, y_3) \\ x_3=\lambda^2-1-2 \frac{61}{4}=\frac{11251^2}{1876^2}-1-\frac{61}{2}=\frac{15724657}{3519376}, \\ y_3= -\lambda \cdot x_3-v=\frac{11251}{1876} \cdot \frac{15724657}{3519376}-\frac{247669}{7504}=-\frac{40991967729}{6602349376}$$

Is it right? Or have I done something wrong? (Thinking)

Is it the only way to show that $-3P_1-2P_2+6P_3=0$ ? (Thinking)

 

[jvicens]

Hello there! Merry Christmas to you too! I can confirm that your calculations for $2P_1$ are correct. However, there are other ways to show that $-3P_1-2P_2+6P_3=0$.

One way is to use the fact that if $P_1, P_2, P_3$ are points on an elliptic curve $E$, then $P_1+P_2+P_3=0$ if and only if $P_1, P_2, P_3$ are $\mathbb{Z}$-linearly dependent.

Another way is to use the group law on elliptic curves. Since $P_1, P_2, P_3$ are points on the same curve, we can add them together and see if we get the identity point $O$. We can also use scalar multiplication to see if $-3P_1-2P_2+6P_3$ is equal to $O$.

Overall, your method and the other methods are all valid ways to show that $-3P_1-2P_2+6P_3=0$. So don't worry, you did not do anything wrong. Keep up the good work and have a great day! (Bigsmile)