Section15.1Rational Points on Conics

What we are going to do is to slowly try to make our way to finding integer solutions to some more difficult Diophantine equations, using an idea about rationals which simplifies Pythagorean triple geometry.

Subsection15.1.1Rational points on the circle

Remember that we thought of Pythagorean triples as solutions to \[x^2+y^2=z^2\; .\] Now, let's divide the whole Pythagorean thing by \(z^2\): \[\frac{x^2}{z^2}+\frac{y^2}{z^2}=1\Rightarrow \left(\frac{x}{z}\right)^2+\left(\frac{y}{z}\right)^2=1\, .\] Since we can always get any two rational numbers to get a common denominator, what that means is the Pythagorean problem is the same as finding all rational solutions to \[a^2+b^2=1\, ,\] which seems to be a very different problem. Let's investigate this.

The blue line intersects the circle \(x^2+y^2=1\) in the point \((1,0)\) and has rational slope denoted by slope. If you change the variable slope, then the line will change.

It is not a hard exercise to see that the line through two rational points on a curve will have rational slope, nor what its formula is, so that every rational point on the circle is gotten by intersecting \((1,0)\) with a line with rational slope.

It is a little harder to show that intersecting such a line with the circle always gives a rational point, but this is also true! It is also far more useful, as it gives us a technique to find all rational points and hence all Pythagorean triples.

Even the inputs \(t=0\) and \(t=\infty\) have an appropriate interpretation in this framework, believe it or not. Such a description of the (rational) points of the circle is called a parametrization. Plug in various \(t\) and see what you get!

You could start the whole process with \((-1,0)\) or \((0,1)\), use all lines through it with rational slopes, and get a different parametrization.

Subsection15.1.2Parametrization

Will this always work? Here is an amazing fact we will not prove.

Example15.1.3

Here's an example with \(x^2+3y^2=1\).

  • Once again, the line above goes through \((1,0)\) and so has equation \(y=t(x-1)\).
  • Again, the ellipse is \(x^2+3y^2=1\), so that \[x^2+3t^2(x-1)^2=1\Rightarrow x^2+3t^2x^2-6t^2x+3t^2-1=0\] is the equation we must solve for \(x\) to find a parametrization of \(x\) in terms of \(t\).
  • This seems daunting; however, we already know that there is a solution \(x=1\), so that \(x-1\) must be a factor of the expression!
    • So we could factor it out if we wished.
  • Alternately, we could use the quadratic formula and discard the solution \(x=1\). Do that now as an exercise!
  • At the end you should get \[x=\frac{3t^2-1}{3t^2+1},y=\frac{-2t}{3t^2+1}\]
  • Now you can find all kinds of interesting solutions like \(\left(\frac{11}{13},\frac{-4}{13}\right)\).

These solutions lead us to integer solutions of three-variable equations like \(x^2+3y^2=z^2\).

Since \(x\) and \(y\) have a common denominator, we can just multiply through by the square of that denominator to get a solution to this. E.g. \[11^2+3(-4)^2=13^2\] which is a rather non-obvious solution, to say the least, and only one of many that this method can help us find.

Subsection15.1.3When curves don't have rational points

However, this method does NOT always work. Namely, you need at least one rational point to start off with. And what if there isn't one that exists? It turns out that Diophantus already knew of some such curves.

As we can see, there are NO rational points on a circle of radius \(\sqrt{15}\) because there are no integer points on the corresponding surface other than ones with \(x,y=0\) – and those correspond to \(z=0\), which would give a zero denominator on the circle. Here is a place where rational points are helped by integer points instead of vice versa.

Example15.1.5

Another one to try is finding rational points on the ellipse \(2x^2+3y^2=1\).

  • This would correspond to \(2x^2+3y^2=z^2\).
  • Here, a different technique might help - look at it modulo 3!
  • In this case it reduces to \[2\equiv (zx^{-1})^2\text{ mod }(3)\]
  • This is impossible since \([0],[1],[2]\) all square to \([0]\) or \([1]\) in \(\mathbb{Z}_3\).

The point is that, at least sometimes, modular arithmetic and going back and forth between integer and rational points helps us both find points and prove there are no such points.