Rational points on the unit circle

A rational point on a plane curve is a point on the curve with rational coordinates. For example, $(3/5,4/5)$ is a rational point on the circle $C$ with equation $x^2+y^2=1$.

As one can guess from the example just given, rational points on $C$ are closely related to Pythagorean triples, i.e., the positive integer solutions to $a^2+b^2=c^2$. In fact, if $a$, $b$, $c$ are any integers satisfying $a^2+b^2=c^2$ and $c \not=0$, then $(a/c,b/c)$ will be a rational point on $C$.

Conversely, if $(x,y)$ is a rational point on $C$. then by choosing a common denominator for $x$ and $y$ one can write $x=a/c$ and $y=b/c$ for some integers $a,b,c$ with $c \not=0$, and the relation $x^2+y^2=1$ implies $a^2+b^2=c^2$. If moreover $x$ and $y$ are nonzero, then $a,b,c$ will all be nonzero, and $(\vert a\vert,\vert b\vert,\vert c\vert)$ will be a Pythagorean triple.

It would be nice to have a description of all the rational points on $C$, because then we would have a description of all the Pythagorean triples. Our goal now is to find such a description using geometry!

Consider the following construction. Start with the rational point $P=(-1,0)$ on $C$. Fix a rational number $t$. Draw the line $L_t$ with slope $t$ passing through $P$. This line will intersect the circle at a second point $Q_t$ (which depends on the number $t$).

By ``pure thought'' (no calculation), one can see that $Q_t$ must have rational coordinates, because its $x$-coordinate will arise as the solution to a quadratic equation which already has one rational root, namely the $x$-coordinate of $P$, and then the $y$-coordinate of $Q_t$ also will be rational (either by the same argument with $y$-coordinates, or by using the equation of $L_t$).

For the incredulous, here is a full calculation of $Q_t$. The equation of $L_t$ (in point-slope form) is $y=t(x+1)$. To intersect this with $C$, which is $x^2+y^2=1$, substitute $y=t(x+1)$ to obtain

$$x^2 + t^2 (x+1)^2 = 1$$

$$(x^2-1) + t^2 (x+1)^2 = 0$$

$$(x+1) \left[ (x-1) + t^2 (x+1) \right] = 0.$$

It was not just luck that the quadratic polynomial in $x$ factored: the point is that it had to have $x+1$ as a factor, because we already knew that there was a point with $x$-coordinate $-1$ in the intersection $L_t \cap C$, namely $P=(-1,0)$. Anyway, discarding the root $x=-1$ and solving for the other possible $x$-coordinate, we see that $Q_t$ has $x$-coordinate $(1-t^2)/(1+t^2)$, and $y$-coordinate

$$y=t(x+1)=t\left[ \frac{1-t^2}{1+t^2} + 1 \right] = \frac{2t}{t^2-1}$$

so

$$Q_t = \left( \frac{1-t^2}{1+t^2}, \frac{2t}{t^2-1} \right).$$

Since $t$ is rational, $Q_t$ has rational coordinates.

We now claim that every rational point on the circle $C$ other than $P=(-1,0)$ arises as $Q_t$ for exactly one rational number $t$. In other words, we obtain a parameterization of all the rational points on $C$ (except $P$). Recall that ${\mathbb{Q}}$ denotes the set of all rational numbers.

Theorem 1   The map

$${\mathbb{Q}} \rightarrow \{ x^2+y^2=1 (-1,0) \}$$

$$t \mapsto Q_t=\left( \frac{1-t^2}{1+t^2}, \frac{2t}{t^2-1} \right)$$

is a bijection (one-to-one correspondence).

Proof. There is a natural candidate for the inverse map, namely, the map

$$\{ x^2+y^2=1 (-1,0) \} \rightarrow {\mathbb{Q}}$$

$$(r,s) \mapsto \frac{s}{r+1}$$

sending a rational point $Q=(r,s)$ on $x^2+y^2=1$ other than $(-1,0)$ to the slope of the line $\overleftrightarrow{PQ}$.

To show that the two maps are indeed inverse bijections, it suffices to show that the composition of the two maps in either order is the identity map.

Given $t \in {\mathbb{Q}}$, if we construct $Q_t$, and then take the slope of the line $\overleftrightarrow{PQ_t}$, we get $t$ back, by definition of $Q_t$.

On the other hand, if we start with a rational point $Q \not=P$ on $C$, compute the slope $t$ of the line $L=\overleftrightarrow{PQ}$, and then construct $Q_t$, then $Q_t=Q$ because $Q$ is the intersection point other than $P$ of $C$ with the line $L$ through $P$ with slope $t$. This completes the proof. $\qedsymbol$

If $m$ and $n$ are positive integers with $m>n$, and we take $t=n/m$, then we obtain the point

$$Q_t = \left( \frac{m^2-n^2}{m^2+n^2} , \frac{2mn}{m^2+n^2} \right),$$

so $(m^2-n^2,2mn,m^2+n^2)$ is a Pythagorean triple.