The elliptic curve $y^2=x^3+3x+4$ has points O,(-1,0) and (0,2). Find five more points with rational coordinates.
The answer to this example gives: (0,-2) (5,-12) (5,12) (71/25,744/125) and (71/25,-744/125)
It seems to me that by changing the y coordinates sign you can automatically get another point, is this true?
Also I known how to calculate points P + Q, but in this case I don't see the mod explicitly expressed, do I have to solve for this?
Also is there a way to know how many points exist? Is it infinite or not?
(0,2)+(5,12) $λ=12-2/5-0=10/5$
$x_r=λ^2-x_p-x_q$
$x_r=λ^2-0-5$
$y_r=λ(0-x_r)-2$
They're using $\lambda$ as the slope of the line $\vec{PQ}$.
Namely, $\lambda = \frac{y_Q - y_P}{x_Q - x_P}$.
Now the idea is: take the line $y - y_P = \lambda (x - x_P)$ and find a third point (other than $P$ and $Q$) on this line intersected with the curve.
To do this, plug in $x = x_P + t$ and $y = y_P + \lambda t$:
$(y_P + \lambda t)^2 = (x_P + t)^3 + 3 (x_P + t) + 4$.
One solution is $t = 0$ corresponding to $P$ and one solution is $t = x_Q - x_P$ corresponding to $Q$. The third solution to this cubic gives you the third point, i.e. the point $S$ such that $P+Q+S = O$. Then $P+Q = R$, where $R = -S$ is the point such that $R + S + O = O$.
In this case, this is just $S$ with its $y$-coordinate swapped (since a vertical line passes through $S$ and this, with its "third point of intersection" at infinity, i.e. $O$).
It looks like you've been given formulas to find $R$, so you could also just use those. The proof those formulas work is by my comments above plus some algebraic manipulation.