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


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?

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    $\begingroup$ First: what a horrible exercise (because of the horrible numbers, not because of the wonderful subject)! Second, why do you ask whether is that right or not? Substitute the result in the original equation and see if you get an equality! $\endgroup$ – Timbuc Dec 23 '14 at 12:31
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    $\begingroup$ @Timbuc Since I got such huge numbers, I thought that maybe I shouldn't substitute all the variables. So, is that the only way? $\endgroup$ – evinda Dec 23 '14 at 12:37
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    $\begingroup$ Well, the group law is rather simple to apply, and the problem is usually with the computation of the ugly numbers that sometimes come up, so I can't see any easier, more direct way to check your work than substituting. $\endgroup$ – Timbuc Dec 23 '14 at 12:39
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    $\begingroup$ I would suspect that anyone assigning this kind of homework expects you to work with a suitable CAS. You can do partial verification by doing all the arithmetic modulo a suitable prime (good reduction is needed). If you check enough many primes you can probably make it quite convincing! You do need to calculate several modular inverses though. $\endgroup$ – Jyrki Lahtonen Dec 23 '14 at 12:44
  • $\begingroup$ @JyrkiLahtonen Could you explain further what I am supposed to do? $\endgroup$ – evinda Dec 23 '14 at 13:23

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)

  • $\begingroup$ So, the only ways to compute these points are by hand, or using a CAS? $\endgroup$ – evinda Jan 3 '15 at 14:06
  • $\begingroup$ I think the best approach is given by the nice answer from Noam Elkies, which should enable you to solve the problem by hand. $\endgroup$ – Jesper Petersen Jan 3 '15 at 19:18
  • $\begingroup$ Jesper Petersen Could you take a look at my post to check if that's what I have tried is right? math.stackexchange.com/questions/1101953/… $\endgroup$ – evinda Feb 8 '15 at 8:48

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 $3P_1+2P_2=6P_3$.)

  • $\begingroup$ Could you explain further to me what I have to do instead of calculating these huge numbers by hand? $\endgroup$ – evinda Jan 3 '15 at 12:10
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    $\begingroup$ With the explanation from Noam Elkies, all you need to do is the following: 1) realize that $G_1$, $G_2$ and $T$ are points on E. 2) Verify that $P_1 = 2G_2$, $P_2 = 3G_1$, $P_3 = T+G_1+G_2$ and $2T = 0$ and therefore $6T = 0$, which should be feasible to do by hand because of the small coefficients. 3) Realize that you then immediately have that $3P_1 + 2P_2 = 6P_3$ $\endgroup$ – Jesper Petersen Jan 3 '15 at 19:16

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