Find the third rational point on the curve: $y^2 = x^3 + 8$

Question

I am trying to find a third rational point on the curve $y^2 = x^3 + 8$

According to the my professor's solution, the idea is to find two rational points then solve for the third point.

These are the first two points:

$$(x,y) = (1, 3)~~~\text{and}~~~(x,y)=\left(-\frac{7}{4}, \frac{13}{8}\right)$$

The point $(1,3)$ is fairly straight forward and we can find it by trial and error, but we are also supposed to find the second point by trial and error. Is there a systematic way to go about this without having to try so many numbers?

Answer 1

Hint : $y^2 = (x+2)(x^2-2x+4)$ - What can you say about rational points now?

Answer 2

Consider the elliptic curve $y^2 = x^3 + A$ for any $A\in\mathbb{Q}$. 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).

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$

The slope of the curve at the point $(x_*,y_*)$ is $\frac{dy}{dx} =\frac{3x_*^2}{2y_*}$ so the tangent line is given by

$$y = y_* + \frac{3x_*^2}{2y_*}(x-x_*)$$

This line intersects the curve at the point $(x,y)$ determined by the qubic equation $$\left(y_* + \frac{3x_*^2}{2y_*}(x-x_*)\right)^3 = x^2 + A$$

Expanding the brackets above we see that $x_*$ is a double root and the sum of the roots is $\frac{9x_*^4}{4y_*^2}$. This gives us that the third root is

$$x = \frac{x_*^4-8Ax_*}{4y_*^2} \implies y = \frac{8A^2 -20Ax_*^3 - x_*^6}{8y_*^3}$$

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 $\left(-\frac{7}{4},\pm\frac{13}{8}\right)$ and by continuing to apply the formula we can construct even more rational points:

$$\left(\frac{31073}{2704}, \pm\frac{5491823}{140608}\right)$$

$$\left(\frac{(892933489418780033}{326211856441766464}, \pm\frac{994822284545339876177687617}{186315768712277487672882688}\right)$$

$$\ldots$$

The numbers gets quite big after a few itterations.

Answer 3

There is a systematic way. Draw a picture of $y^2 = x^3 + 8$ on graph paper.

http://www.printablepaper.net/category/graph

Per request: I am very, very old. When learning to draw graphs, I used a special kind of treeware called "PAPER," of a special type with little squares already printed on it. This special kind is called Graph Paper. If a printer is available, one may download a free pdf of graph paper, and print out a copy whenever one needs to draw a graph.

I made one. I bought a programmable calculator in 1983. I put in $\sqrt{x^3 + 8}$ and drew the following:

Note that I bought an inexpensive flatbed scanner for my home computer, maybe about 2012, which is how you get to see the picture.