Question

Y2=x3+8 at the point (-2,0)

Answer 1

Consider the elliptic curve y2=x3+A for any A∈Q. Given a rational point (x∗,y∗), we can then find another rational point (x,y) by considering the intersection of the tangent line through that point, see figure below (which is a slight modification of one found in Silverman's book mentioned below).

                             enter image description here

The slope of the curve at the point (x∗,y∗) is dydx=3x2∗2y∗ so the tangent line is given by

y=y∗+3x2∗2y∗(x−x∗)

This line intersects the curve at the point (x,y) determined by the qubic equation

(y∗+3x2∗2y∗(x−x∗))3=x2+A

Expanding the brackets above we see that x∗ is a double root and the sum of the roots is 9x4∗4y2∗. This gives us that the third root is

x=x4∗−8Ax∗4y2∗⟹y=8A2−20Ax3∗−x6∗8y3∗

This shows that if (x∗,y∗) is any rational point on the curve the so is (x,y) given by the formula above. As a small historical note I might add that this result was first derived by Claude Bachet already in 1621, see e.g. Rational points on elliptic curves by Joseph H. Silverman.

Applying this procedure with A=8 to the known rational points (x∗,y∗)=(2,4) and (x∗,y∗)=(1,3) gives us the points (−74,±138) and by continuing to apply the formula we can construct even more rational points:

(310732704,±5491823140608)

((892933489418780033326211856441766464,±994822284545339876177687617186315768712277487672882688)