Points on Elliptic Curves

Next suppose that $F$ is an irreducible cubic polynomial. The question of whether or not $F=0$ has a rational solution is still an open problem! We will not consider this problem further until we discuss the Birch and Swinnerton-Dyer conjecture.

Suppose that $F=0$ has a given rational solution. Then one can change coordinates so that the question of finding the rational solutions to $F=0$ is equivalent to the problem of finding all rational points on the elliptic curve

$$y^2 = x^3 + ax + b.$$

Recall that when $F$ has degree $2$ we can use a given rational point $P$ on the graph of $F=0$ to find all other rational points by intersecting a line through $P$ with the graph of $F=0$. The graph of $y^2 = x^3 + ax+b$ looks like

[egg and curvy line] or [curvier line]

Notice that if $P$ is a point on the graph of the curve, then a line through $P$ (usually) intersects the graph in exactly two other points. In general, these two other points usually do not have rational coordinates. However, if $P$ and $Q$ are rational points on the graph of $y^2 = x^3 + ax+b$ and $L$ is the line through $P$ and $Q$, then the third point of intersection with the graph will have rational coordinates. Explicitly, if $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ then the third point of intersection has coordinates¹

$$x_3 = \left(\frac{y_2 - y_1}{x_2-x_1}\right)^2 - x_1 - x_2, \qquad y_3 = \frac{y_2-y_1}{x_2-x_1} x_3 - \frac{y_2 x_1 - y_1 x_2}{x_2-x_1}.$$

Thus, given two points on $E$, we can find another. Also, given a single point, we can draw the tangent line to $E$ through that point and obtain a third point.