The points are $\mathbb{Z}$-linearly dependent

Question

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?

Answer 1

I think you have got the elliptic curve wrong as well as the relation among the points. The correct curve seems to be $y^2 = x^3 + x^2 - 25x + 39$, which contains the points $P_1$, $P_2$ and $P_3$. The correct relation is $3P_1+2P_2+6P_3 = 0$. It can be checked by (tedious) computation by hand, or more easily using a CAS as suggested by Jyrki Lahtonen.

Sage is a free to use at cloud.sagemath.org. A session with your curve that verifies the relation runs like this.

sage: E = EllipticCurve(x^3+x^2-25*x+39 == y^2)

sage: P1 = E.lift_x(61/4)

sage: P2 = E.lift_x(-335/81)

sage: P3 = E.lift_x(21)

sage: 3*P1+2*P2+6*P3

(0 : 1 : 0)

If you intend to check by hand beware of the very large coefficients. For instance

sage: 6*P3

(17631797546863867480163645661711294049/2834578067615933833996300908324147456 : -6090252960717733600018 1399672827762453069546262535228527/4772353810493036247904139120367622993558177805319376896 : 1)

Answer 2

Thanks to Jesper Petersen for correcting the equation. Once we have the right coefficients and coordinates, the identity can be verified by writing the $P_i$ in terms of small generators, avoiding the calculation of huge numbers such as the 50+ digit denominator that arises in the direct verification.

According to mwrank the group of rational points has rank $2$, and the points $G_1$, $G_2$ with $(x,y)$ coordinates $(1,4)$ and $(-5,8)$ generate the group modulo its torsion subgroup. According to gp the torsion subgroup has order $2$, with generator $T=(3,0)$. Hence the group of rational points is the direct sum of the torsion group $\{0,T\}$ with ${\bf Z}G_1 \oplus {\bf Z}G_2$. Now it's a matter of recognizing $P_1,P_2,P_3$ in this group, and it turns out that $$ P_1 = 2 G_2,\quad P_2 = 3 G_1,\quad P_3 = T + G_1 + G_2. $$ The desired identity follows. (Note that having found $G_1$, $G_2$, and $T$ one doesn't actually need to prove that they generate the full group of rational points to prove that the particular combinations $P_1,P_2,P_3$ satisfy $3P_1+2P_2=6P_3$.)