Ramble, Part 49: Fighting for Fermat

[stoutfellow]

Another of the developments that led to increasingly abstract algebra stemmed from the struggle to prove Fermat's Last Theorem. Unsurprisingly, this story also begins with Gauss.

In an earlier Ramble, I gave a geometric argument for a characterization of primitive Pythagorean triples, i.e., for triples of whole numbers a, b, c having no common factor greater than 1 and satisfying a²+b²=c². Assuming that b is even (and either a or b must be), there are whole numbers m, n such that a=m²-n², b=2mn, and c=m²+n². I'm going to give a purely algebraic argument for that characterization.

Let's begin by making the following observation. If a and b are two whole numbers with no common factors (in the jargon, they are relatively prime) and ab is a perfect square, then a and b must both be squares. Thus 4x9=36 is a square, and 4 and 9 are relatively prime; they are also squares. 3x12=36 is a square, and 3 and 12 are not squares, but they aren't relatively prime either - they're both divisible by 3.

Now, this isn't quite true. (-4)x(-9) is a square, and -4 and -9 are relatively prime, but they aren't squares. However, each of them is a square times -1. Now -1 is an unusual whole number, in that its reciprocal is also a whole number. (In fact, it is its own reciprocal.) Suppose u is a whole number whose reciprocal is a whole number; then if a and b are two relatively prime squares, au and b/u are also relatively prime, but they may not be squares - but their product is ab, which is a square. We call a number like u a unit; the units in the usual whole numbers are 1 and -1. In the situation above, then, a and b may not be squares, but they must be squares multiplied by units.

Now, suppose that a, b, c is a primitive Pythagorean triple. Then a²+b²=c². But a²+b²=(a+bi)(a-bi). Since a and b are relatively prime, we can show that a+bi and a-bi are also relatively prime. Hence, by the observation above, they must be squares. So say a+bi=(m+ni)². If we expand the right side, we get (m²-n²)+2mni; but the real parts (a and m²-n²) must be equal, and likewise for the imaginary parts b and 2mn. We get the characterization again.

What about units? Well, in this situation, we get some additional units, namely, i and -i. (a-bi)(i)=b+ai, swapping the roles of a and b; making the stipulation that b is even takes care of the problem.

There's a little more that needs to be done to make this argument work, though. Most important is this question: where are we? The original observation had to do with the whole numbers; but a+bi anda-bi are not whole numbers. Instead, they are Gaussian integers, numbers of the form a+bi where a and b are whole numbers. The Gaussian integers can be added, subtracted, and multiplied, and the results are themselves Gaussian integers. It is possible to talk about factors, and common factors, and being relatively prime; and the argument works.

What does this have to do with Fermat's Last Theorem? Well, if we can factor a²+b², can we factor a³+b³? aⁿ+bⁿ? The answer is yes, and this opens the door to an attack on that Theorem - an ultimately unsuccessful, but nonetheless fruitful one. That will be the subject of the next Ramble.

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Comments

[countrycousin.livejournal.com]

huh.

Didn't know about Gaussian integers as a term. No doubt I've been told and that plant wasn't watered sufficiently.

I suppose the multiplication table is basically 4 dimensional.

*wonders where to get a 4D spreadsheet*

*goes to get an aspirin*

[stoutfellow.livejournal.com]

You'd only run into the phrase "Gaussian integers" in a course in abstract algebra, if at all.

There's a broader definition of "algebraic integer" lurking in the background. A complex number is an algebraic integer if it is a zero of a polynomial all of whose coefficients are (ordinary) integers and whose leading coefficient is 1. Thus Sqrt[2] is an algebraic integer (x^2-2); i/2 is not (4x^2+1); and, perhaps surprisingly, (Sqrt[5]+1)/2 is (x^2-x-1).