NTIC Making More and More and More Points

Section 15.5 Making More and More and More Points

Recall from Fact 15.1.5 that the following two strategies should give new rational points on a conic section. We will give these strategies names.

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Algorithm 15.5.1. Getting New Rational Points.

Two ways to obtain new rational points on a conic from rational points you already have are:

Connect two points with a secant line, and then make a line with the same slope but through another (rational) point. We call this adding points.
Find the tangent line through a point, and then make a line with the same slope but through another point. We call this doubling a point.

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Fact 15.5.2.

The set of rational points on a conic section is an Abelian group. Assuming you have a point selected as an identity element, the group operation on two points $P$ and $Q$ is given by the first, “adding points”, operation Algorithm 15.5.1. That is, you connect $P$ and $Q$ by a secant line of slope $m\text{,}$ and then connect the identity to a fourth point $P+Q$ with a line of slope $m\text{.}$ Adding a point $P$ to itself uses the slope of the tangent line at $P\text{,}$ the second, “doubling points”, operation in Algorithm 15.5.1.

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Subsection 15.5.1 Toward integer points

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More germane to our investigation, our limited experience in the previous section suggests these processes may often give you integer points. This is not a coincidence; in general, we should try to add or double points to get (new) integer points.

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As we are only guaranteed rational points, this doesn't always work. Below, I try this on the ellipse from the beginning of Section 15.4.

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Figure 15.5.3. Trying to find more integer points on an ellipse

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Rotten luck. But in some circumstances, this strategy works very well indeed. Figure 15.5.4 gives an example of the simple family of hyperbolas

$x^2-dy^2=1$ where

$d=2\text{.}$

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So let's try the strategy of Algorithm 15.5.1. What happens when we take the tangent line to the curve

$x^2-2y^2=1$ at the point

$(3,2)\text{,}$ and then create a new line with the same slope through

$(1,0)\text{?}$ See Figure 15.5.5.

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Figure The hyperbola $x^2-2y^2=1$ with more points

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It intersects in a new integer point, amazing! And if we repeat the process with the new point, we get another one – use the interact to see. Hmm …

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d=2
var('x,y')
@interact
def _(x_0=3,y_0=2,lattice=False,auto_update=False):
    g(x,y)=x^2-d*y^2
    x_1,y_1=x_0^2+2*y_0^2,2*x_0*y_0
    plot1 = implicit_plot(g-1,(x_0-4,x_1+4),(x_0-4,x_1+4), plot_points = 200)
    grid_pts = [[i,j] for i in [x_0-4..x_1+4] for j in [x_0-4..x_1+4]]
    plot_grid_pts = points(grid_pts, rgbcolor=(0,0,0), pointsize=2)
    lattice_pts = [coords for coords in grid_pts if (coords[0]^2-d*coords[1]^2==1)]
    plot_lattice_pts = points(lattice_pts, rgbcolor = (0,0,1),pointsize=20)
    line1 = plot((x_0/(2*y_0))*(x-x_0)+y_0,x_0-4,x_1+4, color='red')
    line2 = plot((x_0/(2*y_0))*(x-1),x_0-4,x_1+4, color='red', linestyle='--')
    if lattice:
        show(plot1 + plot_grid_pts + plot_lattice_pts + line1 + line2, figsize = [5,5], xmin = x_0-4, xmax = x_1+4, ymin = y_0-4, ymax = y_1+4, aspect_ratio=1)
    else:
        show(plot1+plot_lattice_pts+line1+line2, figsize = [5,5], xmin = x_0-4, xmax = x_1+4, ymin = y_0-4, ymax = y_1+4, aspect_ratio=1)
    pretty_print(html("The new points are $x_1=%s$ and $y_1=%s$"%(x_1,y_1)))

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As it turns out, this is quite an old idea. Finding integer solutions to this hyperbola is called solving Pell's equation, and has been studied in this form since the seventeenth century. But a process very similar to this was already rigorously discussed by Brahmagupta centuries before that!

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Historical remark 15.5.6. Brahmagupta.

Brahmagupta is one of the earliest Indian mathematicians we have records from, though as was typical for mathematicians around the world for over a millennium, he was the head of an astronomical observatory. In addition to working on Pell's equation (see for example Wikipedia), we saw earlier the Brahmagupta-Fibonacci identity, and he also had prescient results in approximation and geometry.

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Historical remark 15.5.7. Stigler's Law.

In the event, Pell did not have anything to do with these equations; it was all based on a misunderstanding. But names stick. In mathematics this phenomenon of not naming things after the actual discoverer is sometimes called Boyer's law, more generally Stigler's law of eponymy (which are themselves self-referential).

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Subsection 15.5.2 A surprising application

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The particular equation

$x^2-2y^2=1$ was studied by Greeks such as Theon of Smyrna (though not in this generality) to shed light on

$\sqrt{2}\text{.}$ Why would solutions to this equation help?

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Well, imagine that

$(x,y)$ fulfill the equation. Then divide and rearrange the original equation to get

$\begin{equation*} \frac{x^2}{y^2}=2+\frac{1}{y^2} \end{equation*}$

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If you can find a solution to this equation with a big

$y\text{,}$ then

$\frac{x^2}{y^2}$ should be pretty close to

$2\text{,}$ which means

$x/y$ itself is pretty close to

$\sqrt{2}\text{.}$

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Let's see this in action. We already tried to find integer points on the curve in the following interact.

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var('x,y')
@interact
def _(viewsize=slider(10,20,1),d=2):
    f(x,y)=x^2-d*y^2
    plot1 = implicit_plot(f-1, (-viewsize,viewsize), (-viewsize,viewsize), plot_points = 200)
    grid_pts = [[i,j] for i in [-viewsize..viewsize] for j in [-viewsize..viewsize]]
    plot_grid_pts = points(grid_pts,rgbcolor=(0,0,0),pointsize=2)
    lattice_pts = [coords for coords in grid_pts if (coords[0]^2-d*coords[1]^2==1)]
    plot_lattice_pts = points(lattice_pts, rgbcolor = (0,0,1),pointsize=20)
    show(plot1+plot_grid_pts+plot_lattice_pts, figsize = [5,5], xmin = -viewsize, xmax = viewsize, ymin = -viewsize, ymax = viewsize, aspect_ratio=1)
    pretty_print(html("Points on the curve $x^2-%sy^2=1$"%d))

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The easy one for

$d=2$ was

$(3,2)\text{.}$ And after all,

$\frac{3}{2}=1.5$ isn't too far from

$\sqrt{2}\approx 1.414\text{.}$ There seems to be another point if we zoom out, but that would be a tedious way to compute them …

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Example 15.5.8.

What if we double the point and take the tangent at $(3,2)\text{?}$ (See Algorithm 15.5.1.) Then we take that slope, and make a new line through the “base” point (in this case, $(1,0)$ ).

Then the next point we get is $(17,12)\text{.}$ (See Exercise 15.7.14.) Indeed, $17^2-2\cdot 12^2=1$ and $17/12\approx 1.417\text{,}$ already correct to three significant digits. Those Greeks!