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Third point of elliptic curve $E: y^2 = x^3 + Ax + B$ given points $P_1=(x_1,y_1), P_2=(x_2, y_2)$ on $E$ (Weierstrass equation).

Assume $x_1 \neq x_2$.

I create the straight line $y = m(x-x_1) + y_1$ through $P_1$ and $P_2$. Then considering $(m(x-x_1) + y_1)^2 = x^3 + Ax + B$, we have $x^3 + Ax + B - (m(x-x_1) + y_1)^2 = 0$ (polynomial of degree $3$). I already know $2$ roots of this equation namely $x_1, x_2$. So it is possible to find a third root $x_3$ corresponding to a point $P^{'}_3$ on $E$ (factor the polynomial ...)

The book strongly indicate that $P^{'}_3 \neq P_2,P_1$ that is $x_3 \neq x_1,x_2$. But how can I see this is the case ?

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  • $\begingroup$ Rather than factoring the polynomial to find $x_3$, consider using Vieta's first formula: The coefficient of $x^2$ is $-m^2$, so the sum of the roots is $m^2$. Thus $x_3 = m^2 - x_1 - x_2$. $\endgroup$ – Maurice P Apr 12 '17 at 13:21
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No, that's wrong: $P_3$ can be equal to $P_1$ or $P_2$.

But $P_3$ is not the sum of $P_1$ and $P_2$: the sum of $P_1$ and $P_2$ is $Q = -P_3 = (x_3,-y_3)$. And it is $Q$ that can't equal $P_1$ or $P_2$.

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  • $\begingroup$ Thanks ! Can $y_3 = 0$ that is $Q$ equals $P_1$ or $P_2$ ? Also, $x_1$ or $x_2$ can be a multiple root of the polynomial, since $P_3$ can be equal to $P_1$ or $P_2$. $\endgroup$ – Mikkel Jul 27 '14 at 19:39
  • $\begingroup$ I see you write $Q$ can't equal $P_1$, $P_2$, but why can't $y_3 = 0$ ? $\endgroup$ – Mikkel Jul 27 '14 at 19:43
  • $\begingroup$ I never said that $Q$ couldn't equal $P_3$! And indeed this will happen if $P_1 = -P_2$. $\endgroup$ – TonyK Jul 27 '14 at 21:25
  • $\begingroup$ You write $P_3$ can be equal to $P_1$ or $P_2$ - and $Q$ can't equal $P_1$ or $P_2$. In the comment above you state $Q$ can equal $P_3$. This implies $Q$ can equal $P_1$ or $P_2$ ?? $\endgroup$ – Shuzheng Jul 28 '14 at 8:57
  • $\begingroup$ No, in the case that $P_3$ equals $P_1$ or $P_2$, then $Q$ can't equal $P_3$. If $P_3 = P_1$ or $P_3 = P_2$, then the line $P_1P_2$ must be a tangent to the curve at $P_3$, so the $y$-coordinate of $P_3$ can't be zero. (I am assuming here that neither $P_1$ nor $P_2$ is the Point at Infinity, since they were both presented as $(x,y)$-coordinates.) $\endgroup$ – TonyK Jul 28 '14 at 9:26

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