Elliptic Curves

In the last installment, we found the rational points on a singular cubic curve. What about non-singular cubics? Non-singular cubic curves are called elliptic curves. (Actually, to be elliptic, we need to add "points at infinity" and insist that they be non-singular as well, but we will finesse this point.) The study of elliptic curves is an incredibly rich branch of mathematics. Wiles' proof of Fermat's Last Theorem was actually a proof of a certain fact about elliptic curves.

Suppose you are given two rational points on an elliptic curve. How could you construct a third point? Draw a line through the two points and intersect it with the elliptic curve. The resulting equation will be cubic, but you already know that it has two rational roots (corresponding to the original points) hence the third root must be rational as well.


What if you were only given one point? Borrowing an idea from first-semester calculus, we could draw the tangent line through the point and use that line to intersect with the curve.

For example, we know by Fermat's Last Theorem, that there are no non-trivial solutions to the equation a^3 + b^3 = c^3, but what about the variant:

a^3 + b^3 = 7c^3?
It is easy to find the small solution: a=2, b=-1, c=1. We will use this to find other solutions. Letting x=a/c, y=b/c, we have the elliptic curve x^3 + y^3 = 7 with rational point (2,-1). We want to find the equation of the tangent line to the curve at this point. Implicitly differentiating we find

3x^2 + 3y^2(dy/dx) = 0, so
dy/dx = - x^2/y^2 = - 4

and the equation of the tangent line is y = -4x + 7. To find the intersection of this line with the cubic we substitute to obtain:

x^3 + (-4x + 7)^3 = 7
63x^3 - 336 x^2 + 588x -336 = 0 3x^3 - 16x^2 + 28x -16 = 0.

But we already know that 2 is a root of this equation. In fact, it must be a double root, since the line is tangent to the cubic at that point, so (x - 2)^2 must be a factor of the left-hand side:

(x - 2)^2(3x - 4) = 0

so x = 4/3 (and hence y = -4(4/3) + 7 = 5/3) are the coordinates of our third point and give the solution a = 4, b = 5, c = 3 to our original equation.

We could find more solutions by repeating the construction above with our new point (4/3,5/3), but we want to illustrate the method with two distinct points. Unfortunately, if we choose (4/3,5/3) and (2,-1) we will just get (2,-1) back. (Why?) Instead, we note that the symmetry of the equation tells us that (-1,2) is also a solution. Now the equation of the line through (4/3,5/3) and (-1,2) is y = - (1/7)x + 13/7 and substituting yields:

x^3 + (-1/7x + 13/7)^3 = 7 and (after clearing denominators)
114x^3 + 13x^2 - 169x -68 = 0

But we know that -1 and 4/3 must be roots and hence that (x + 1)(3x - 4) must be a factor of the left-hand side:

(x + 1)(3x - 4)(38x + 17) = 0

So x = - 17/38 (and hence y = -(1/7)(-17/38) + 13/7 = 73/38) are the coordinates of our third point and give the solution a = -17, b = 73, c = 38 to our original equation.

We can repeat these processes to obtain an infinite number of (primitive) solutions to a^3 + b^3 = 7c^3.

Exercise: Find a small solution to a^3 + b^3 + c^3 = 6abc and then use the techniques of this section to find other solutions. [Note the symmetry of the equation.]