stoutfellow

This is kind of a digression, I'm afraid. I don't know whether Fermat himself actually used the technique I'm about to describe, but he certainly was interested in Pythagorean triples, and I think it's the kind of thing he would have enjoyed. It's one of my favorite bits of mathematics, and a less-than-obvious illustration of the way geometry can be used to solve algebraic - in fact, number-theoretic - questions.

We start with some definitions. A "Pythagorean triple" is a triple a,b,c of whole numbers satisfying the equation a²+b²=c². To avoid duplications and uninteresting cases, let's assume that a,b,c are positive. Now, if you multiply the triple by another whole number d, you get another Pythagorean triple ad, bd, cd. Conversely, if the triple a,b,c has a common divisor d, you can set a'=a/d, b'=b/d, c'=c/d; then a',b',c' is a Pythagorean triple. If d is actually the greatest common divisor of a,b,c, then a',b',c' have g.c.d. 1; in other words, they form a "primitive Pythagorean triple".

The point of all that is that, if you want to know what all the Pythagorean triples are, it's enough to find all the primitive Pythagorean triples; the rest are just multiples of those. Now, let's take it a step further. If a,b,c is a primitive Pythagorean triple, set r=a/c, s=b/c; then r²+s²=1. In other words, the point (r,s) lies on the circle x²+y²=1. Conversely, if (r,s) is a point on that circle, where r,s are positive rational numbers, then you can set c to be the least common denominator of r and s, and set a=rc, b=sc; a,b,c is a primitive Pythagorean triple.

In sum: there is a one-to-one correspondence between primitive Pythagorean triples and points with rational coordinates on the circle x²+y²=1 which lie in the first quadrant (i.e., both coordinates are positive). To save repetition, let's call a point "rational" if its coordinates are rational.

Now, let N be the "North Pole" of the circle, the point (0,1). If P is a first-quadrant rational point on the circle, draw the line NP. This will strike the x-axis at a point (t,0) - and t will be rational and greater than 1. Conversely, if t is rational and greater than 1, the line connecting N to (t,0) will strike the circle at a second point P - and P will be rational, and in the first quadrant. That is: there is a one-to-one correspondence between first-quadrant rational points on that circle and rational numbers greater than 1.

Let's work this out in detail. The line connecting N to (t,0) has the equation x+ty=t. Let's set x=t-ty in the equation of the circle: (t-ty)²+y²=1. Expanding and simplifying, we get (t²+1)y²-2t²y+t²-1=0. This is a quadratic equation in y, and we know that 1 is one of the roots. (Remember, N is on the line and on the circle!) A little computation shows that the other root is y=(t²-1)/(t²+1). Substituting this back into x=t-ty, we get x=2t/(t²+1).

Now, since we're assuming that t is rational, let's set t=m/n, where m,n are positive whole numbers with g.c.d. 1. (Since t is greater than 1, m is greater than n.) The point (2t/(t²+1),(t²-1)/(t²+1) corresponds to the Pythagorean triple 2mn,m²-n²,m²+n². This triple won't be primitive if m,n are both odd; in that case, we have to divide by 2. With that proviso, every primitive Pythagorean triple must have this form.

Some examples: if m=2 and n=1, we get the familiar triple 4,3,5. If m=3 and n=2, we get 12,5,13. Let's look at a case where m,n are both odd: m=3 and n=1 gives us 6,8,10, and dividing by 2 gives us 3,4,5 - which is just a rearrangement of the triple given by m=2, n=1. This will always happen - that is, the triple given by m,n both odd will be a rearrangement of the triple given by a different choice m',n', where m',n' are not both odd, so that case may as well be ignored. Every primitive Pythagorean triple with a even has the given form, for some choice of m,n not both odd.

This technique can be adapted to handle other similar problems, like finding all choices of whole numbers a,b,c satisfying, say, a²+ab-2b²=c²; you'd use the hyperbola x²+2xy-2y²=1 instead of the circle, but the same line of attack works. I don't know about you, but I think that's neat.

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