All of Pythagoras's Friends

March, 2018


A lot of the reason I read a particular textbook has to do with its portability. I've got a lot of moving around to do over the course of the day. If I want to kill the commute time, reading has always been my go-to. Reading on the bus is kinda nice, but if you've got a text that's a little too large, it can be prohibitive. Not to mention the fact that you've got to carry it around all day. Fortunately, there are a plentitude of texts that fit right in your pocket! The one I picked off the shelf today is called Introduction to Elliptic Curves and Modular Forms, by a guy named Neal Koblitz. He's a cryptographer at the university of Washington, and an adjunct faculty member here at Waterloo.

I don't actually know any number theory, though. So this book has been an adventure. I've had to brush up on some definitions, sit and ponder a few things. The first problem I worked out today had to do with primitive Pythagorean triples, which is a 3-tuple of intergers (X,Y,Z) such that X2+Y2=Z2, with none of the intergers sharing a common factor. We suppose that a>b without loss of generality, and say that they're not both odd, and relatively prime. The problem is to show that, given the relations

X=a2b2,Y=2ab,Z=a2+b2

form a triple so-described, and that all primitive Pythagorean triples take this form. Showing that this is formulation of (X,Y,Z) is a triple is simple, all that's needed is to recognize that these values fulfill the Pythagorean theorem. Showing that they form all primitive solutions is a little trickier. Letting both a and b be even, we see that Y and Z are no longer coprime. Similarly, if both a and b are odd, all (X,Y,Z) become even, and thus no two of them are coprime. Thus, we need one of (a,b) to be odd, and the other even.

The reason we know all primitive triples are found this way is the most interesting part. The easiest way to see it is to draw a picture:

Unit Circle with chord from
                            (-1,0) to (u,v) in first quadrant

Re-paramaterizing from X=a2b2,Y=2ab,Z=a2+b2 to u=XZ,v=YZ, we see that following the Pythagorean theorem leads us to know that u2+v2=1, naturally leading us to a parametrization of the unit circle. Cycling through the intergers for a and b gives us all possible Pythagorean triples.