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
In the interact above, the blue line intersects the circle in the point 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 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 intersect the unit circle in a second rational point.
Proof.
In fact, we can do even better than prove this; we can get a formula for the points.
First, any line with slope has formula . We can then obtain all intersections with the circle by plugging in , so:
We will skip the algebra (see Exercise 15.7.1) showing that the quadratic formula yields the two answers .
Note that gives the point which we already knew. The other, new, point is ; plugging this in gives . In summary, every rational slope gives us the point .
Even the inputs and have an appropriate interpretation in this framework. Such a description of the (rational) points of the circle is called a parametrization. Plug in various and see what you get!
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 verticalβ1β
The long reason for this is projective space; the short and not-quite-rigorous reason is that is a rational fraction, right? β¦ Right?
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.
Where does this go? One place these solutions lead is to integer solutions of three-variable equations. In the previous example, since and have a common denominator, we can just multiply through by the square of that denominator to get
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 isnβt one that exists? It turns out that Diophantus already knew of some such curves.
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 .
Every rational point on the circle can be written using a common denominator as for some , where we cancel any common divisor of all three numbers. Then simply multiplying through by gives integer points on the surface. (This isnβt a one-to-one correspondence, as the surface point shows.)
But now consider the whole equation modulo . The reader should definitely check that there are no legitimate possibilities! (See Exercise 15.7.5; donβt 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 because there are no integer points on the corresponding surface other than ones with β and those correspond to , 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.
A rational point would correspond to integer points on . You can try looking at it modulo four, but that goes nowhere. Instead, given the three as a coefficient, look at it modulo 3!
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.