Skip to main content
\( \newcommand{\lt}{ < } \newcommand{\gt}{ > } \newcommand{\amp}{ & } \)

Section15.5Making More and More and More Points

Based on our previous work, we know that we should be able to do these things to get new rational points.

Subsection15.5.1Toward integer points

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.

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.

Rotten luck. But in some circumstances, this strategy works very well indeed. The following hyperbola is simple, just \(x^2-dy^2=1\).

So let's try it. What happens when we take the tangent line to the point \((3,2)\) as the solution to \(x^2-2y^2=1\)?

A new point, amazing! And if we plug that one in, another one. Hmm …

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!

Remark15.5.3

In the event, Pell did not have anything to do with them; 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).

Subsection15.5.2A surprising application

The particular equation \(x^2-2y^2=1\) was studied by Greeks such as Theon of Smyrna to shed light on \(\sqrt{2}\), though not in the generality we are. Why would this help?

Well, imagine that \((x,y)\) fulfill this equation. Then divide and rearrange the original equation to get \begin{equation*}\frac{x^2}{y^2}=2+\frac{1}{y^2}\end{equation*} If you can find a solution to this equation with a big \(y\), then \(\frac{x^2}{y^2}\) should be pretty close to \(2\), which means \(x/y\) itself is pretty close to \(\sqrt{2}\).

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

The easy one for \(d=2\) was \((3,2)\). And after all, \(\frac{3}{2}=1.5\) isn't too far from \(\sqrt{2}\approx 1.414\). There seems to be another point if we zoom out, but that would be a tedious way to compute them …

Example15.5.4

What if we double the point and take the tangent at \((3,2)\)? (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 \((17,12)\). (See Exercise 15.7.12.) Indeed, \(17^2-2\cdot 12^2=1\) and \(17/12\approx 1.417\), already correct to three significant digits. Those Greeks!