Remember that in Section 3.4 we thought of Pythagorean triples as solutions to \begin{equation*}x^2+y^2=z^2\; .\end{equation*} Now, let's divide the whole Pythagorean thing by \(z^2\): \begin{equation*}\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\, .\end{equation*} Since we can always get any two rational numbers to have a common denominator, what that means is the Pythagorean problem is the same as finding all rational solutions to the simpler formula \begin{equation*}a^2+b^2=1\, ,\end{equation*} which seems to be a very different problem. Let's investigate this.

In the interact above, 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. Such a description of the (rational) points of the circle is called a parametrization. Plug in various \(t\) and see what you get!

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

Where does this go? These solutions lead us to integer solutions of three-variable equations. In this case, it gives solutions to ones 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. \begin{equation*}11^2+3(-4)^2=13^2\end{equation*} which is a rather non-obvious solution, to say the least, and only one of many that this method can help us find.

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.

Let's do another example.