Fact 15.1.2.
All lines with rational slope through \((1,0)\) intersect the unit circle in a second rational point.
Section 15.1 Rational Points on Conics
Subsection 15.1.1 Rational points on the circle
Remember that in Section 3.4 we thought of Pythagorean triples as solutions to
\begin{equation*}
x^2+y^2=z^2\text{.}
\end{equation*}
Now, let’s divide the whole Pythagorean thing by \(z^2\text{:}\)
\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\text{.}
\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\text{,}
\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. This is not necessarily visible in Figure 15.1.1!
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.
Proof.
In fact, we can do even better than prove this; we can get a formula for the points.
First, any line with slope \(t\) has formula \(y=t(x-1)\text{.}\) We can then obtain all intersections with the circle \(x^2+y^2=1\) by plugging in \(y\text{,}\) so:
\begin{equation*}
x^2+(t(x-1))^2=1\Rightarrow x^2+t^2 x^2-2xt^2+t^2=1
\end{equation*}
We will skip the algebra (see Exercise 15.7.1) showing that the quadratic formula yields the two answers \(\frac{t^2\pm 1}{t^2+1}\text{.}\)
Note that \(\frac{t^2+1}{t^2+1}=1\) gives the point \((1,0)\) which we already knew. The other, new, point is \(\frac{t^2-1}{t^2+1}=x\text{;}\) plugging this in gives \(y=t\left(\frac{t^2-1}{t^2+1}-1\right)=\frac{-2t}{t^2+1}\text{.}\) In summary, every rational slope \(t\) gives us the point \(\left(\frac{t^2-1}{t^2+1},\frac{-2t}{t^2+1}\right)\text{.}\)
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!
Remark 15.1.3.
You could start the whole process with \((-1,0)\) or \((0,1)\text{,}\) use all lines through it with rational slopes, and get a different parametrization.
Subsection 15.1.2 Parametrization in general
But will this always work? Certainly not every curve gets rational points by intersecting rational slope lines with it.
Example 15.1.4.
Consider the curve given by \(y=x^3\) and the point \((0,0)\text{.}\) A rational slope line through that point would be \(y=\frac{p}{q}x\text{.}\) Substituting we get
\begin{equation*}
\frac{p}{q}x=x^3\Longrightarrow \frac{p}{q}x-x^3=0 \Longrightarrow x\left(\frac{p}{q}-x^2\right)=0
\end{equation*}
which clearly will have irrational \(x\)-coordinates for most choices of the slope \(p/q\text{.}\)
In the quadratic context it works, though! Here is an amazing fact we will not prove.
Fact 15.1.5.
Suppose you have a curve given by a quadratic equation with rational coefficients which contains at least one rational point. Then all lines with rational slope (including vertical 1 lines) through that point on the curve intersect the curve in only rational points, and all rational points on the curve are generated in this way.
Example 15.1.6.
Here’s an example with \(x^2+3y^2=1\text{.}\)
As in the proof of Fact 15.1.2, the line going through \((1,0)\) has equation \(y=t(x-1)\text{.}\) Here, the ellipse has equation \(x^2+3y^2=1\text{,}\) so that we must solve the equation
\begin{equation*}
x^2+3t^2(x-1)^2=1\Rightarrow x^2+3t^2x^2-6t^2x+3t^2-1=0
\end{equation*}
for \(x\) to find a parametrization of \(x\) in terms of \(t\text{.}\) Figure 15.1.7 might help visualize the process.
Solving this equation seems daunting. Here are two strategies (see Exercise 15.7.2 to try them).
- We already know that there is a solution \(x=1\text{,}\) 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 \(x=1\text{.}\)
In either case you should get
\begin{equation*}
x=\frac{3t^2-1}{3t^2+1},y=\frac{-2t}{3t^2+1}
\end{equation*}
Now you can find all kinds of interesting solutions like \(\left(\frac{11}{13},\frac{-4}{13}\right)\text{.}\)
Where does this go? One place these solutions lead is to integer solutions of three-variable equations. In the previous example, since \(x\) and \(y\) have a common denominator, we can just multiply through by the square of that denominator to get
\begin{equation*}
11^2+3(-4)^2=13^2\text{.}
\end{equation*}
One could consider this to be an integer point on the surface given by \(x^2+3y^2=z^2\text{,}\) which you may play around with in the following interact if you are online.
That is a rather non-obvious solution to this equation in three variables, to say the least, and only one of many that this method can help us find.
Subsection 15.1.3 When curves don’t have rational points
However, the rational slope 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.
Fact 15.1.8.
The circle \(x^2+y^2=15\) has no rational points.
Proof.
First, note this is a much stronger statement than what we already know, which is that this curve has no integer points (see Fact 13.1.1). The way to prove this is to correspond this to integer points on the surface \(x^2+y^2=15z^2\text{.}\)
Every rational point on the circle can be written using a common denominator as \((p/q,r/q)\) for some \(p,r,q\in\mathbb{Z}\text{,}\) where \(\gcd(p,q)=1=\gcd(r,q)\text{.}\) Then simply multiplying through by \(q\) gives integer points \((x,y,z)=(p,r,q)\) on the surface. (This isn’t a one-to-one correspondence, as the surface point \((0,0,0)\) shows.)
But now consider the whole equation \(p^2+r^2=15q^2\) modulo \(4\text{.}\) The reader should definitely check that there are no legitimate possibilities! (See Exercise 15.7.5; don’t forget that the rational points are written in lowest terms.)
As we can see experimentally in the interact below, 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 \(z=0\text{,}\) which would give a zero denominator on the circle. Here is a place where rational points are illuminated by questions of integer points rather than vice versa.
Let’s do another example.
Example 15.1.9.
Try to find rational points on the ellipse \(2x^2+3y^2=1\text{.}\)
Solution.
A rational point would correspond to integer points on \(2x^2+3y^2=z^2\text{.}\) You can try looking at it modulo four, but that goes nowhere. Instead, given the three as a coefficient, look at it modulo 3!
In this case it reduces to
\begin{equation*}
2\equiv (zx^{-1})^2\text{ (mod }3)
\end{equation*}
This is impossible since \([0],[1],[2]\) all square to \([0]\) or \([1]\) in \(\mathbb{Z}_3\text{.}\)
The point is that, at least sometimes, modular arithmetic and going back and forth between integer and rational points helps us find points, or prove there are no such points.
The long reason for this is projective space; the short and not-quite-rigorous reason is that \(\infty=1/0\) is a rational fraction, right? … Right?
