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).

Question

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 ?

Answer 1

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 $P_2$.