One might think this is all there is to say about points on the parabola. But before we go on, I want to point something very interesting out. Look at the following two setups. In one I create the line through two integer points on the conic, the other where I create the tangent line through one integer point.
Notice that in both cases you get another integer point! So we have two ways to get new (rational) points that often turn out to be integer points.
This is not a coincidence.
Indeed, often we will in fact get integer points. For our purposes, that means that we can try to create new points by either doing the slope-through-two-points thing (adding points) or the tangent-slope thing (doubling a point).
It doesn't always work, of course, as we are only guaranteed rational points. Below, I try this on the ellipse from before.
But in some circumstances, this strategy works very well indeed. The following hyperbola comes from something called Pell's equation; it is just a hyperbola \(x^2-dy^2=1\). (In the event, Pell did not have anything to do with them; it was all based on a misunderstanding. But names stick.)
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, a process very similar to this was already rigorously discussed by Brahmagupta. So it's a quite old idea!
Actually, \(x^2-2y^2=1\) was studied by Greeks such as Theon of Smyrna to shed light on \(\sqrt{2}\). What would this do? Well, imagine that \((x,y)\) fulfill this. Then divide and rearrange the original equation to get \[\frac{x^2}{y^2}=2+\frac{1}{y^2}\] so that if you can find a solution to this equation with a big \(y\), then \(\frac{x^2}{y^2}\) is pretty close to \(2\), which means \(x/y\) 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...
Going back up to the graphic, if we take the tangent at \((3,2)\) and move the same slope line to a “base” point (in this case, \((1,0)\)) then the next point we get is \((17,12)\).