#### Fact 15.1.2.

All lines with rational slope through \((1,0)\) intersect the unit circle in a second rational point.

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.

All lines with rational slope through \((1,0)\) intersect the unit circle in a second rational point.

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!

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.

But will this always work? Certainly not every curve gets rational points by intersecting rational slope lines with it.

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.

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.

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?

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.

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.

The circle \(x^2+y^2=15\) has no rational points.

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 rational points on the circle 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 we cancel any common divisor of all three numbers. 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.

Try to find rational points on the ellipse \(2x^2+3y^2=1\text{.}\)

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.