Fermat’s Last Theorem

A lot of algebraic number theory developed in connection to Fermat’s last theorem. The theorem itself is of no theoretical significance, but the theories developed in order to help prove it are tremendous. Kummer’s original work on ideal factorization was partly developed in order to try rescuing a proof that assumed all number fields had unique factorization. If you’ve read the thoroughly misleading book Fermat’s Enigma, you’ll know about the Taniyama-Shimura theorem, which is theoretically important (for a more technical explanation, go here; you’ll never want to read a book by Simon Singh again).

If you’ve followed my posts about ideal theory and class numbers, you’ll have the theoretical background for a very basic treatment of Kummer’s proof. Unfortunately, Kummer’s proof only applies to regular primes. A prime p is regular if the class number of the pth cyclotomic field is not divisible by p, where the nth cyclotomic field is a number field generated by all nth roots of unity. In that case, a trick similar to what I did with the equation x^2 + 5 = y^3 can work.

But for now, I’ll only deal with the most basic case of Fermat’s theorem, the one with n = 3. The case with n = 4 is elementary, but the proof requires some annoying trick with integers satisfying x^2 + y^2 = z^2 that I’d rather not deal with.

First, suppose that x, y, and z are numbers in Z[w], where w = (1 + SQRT(-3))/2, such that x^3 + y^3 = uz^3, where u is any unit. The ring Z[w] is a PID, so factorization is unique. Further, its field of fractions is a purely imaginary quadratic field, Q(SQRT(-3)), so its only units are the roots of unity, {1, w, w – 1, -1, –w, 1 – w}.

We can assume that x, y, and z have no factors in common. If they do, we can divide by their greatest common divisor and get a smaller solution.

Now, we can factor x^3 + y^3 as (x + y)(x + (w – 1)y)(x – wy). Any prime dividing the first two factors divides their difference, (w – 2)y. If it divides y then it also divides x, contradicting the assumption that x and y are coprime. So it divides w – 2, which means it’s associated to it since it’s irreducible. In that case it also divides the third factor, since the difference between it and the first is (1 + w)y, where w – 2 = (1 + w)(w – 1) since w^2 = w – 1.

In fact, we can assume w – 2 is the greatest common divisor of the three factors, rather than 1. To do that, note that the only cubic residues mod 3 in Z[w] are -1, 0, and 1. If w – 2 doesn’t divide z, then 3 doesn’t divide z^3, and x^3 + y^3 != 0 mod 3. Then one of x or y is divisible by w – 2, say x, and uz^3 = (+/-)1 mod 3. But uz^3 = (+/-)u mod 3, so u = (+/-)1. As 1 and -1 are cubes, this implies that x^3 + y^3 = z‘^3, where z‘ is z or –z. Then we can switch the equation to y^3 + (-z‘)^3 = (-x)^3, and x is divisible by w – 2.

Now, note that x + y = (1 – w)(x + (w – 1)y) + w(x – wy), and that (x + y)/(2 – w), (x + (w – 1)y)/(2 – w), and (x – wy)/(2 – w) are coprime. The last three factors’ product is u(z/(2 – w))^3, so by unique factorization, each of them is a unit times a cube. If c^3 ~ (x + y)/(2 – w), b^3 ~ (x + (w – 1)y)/(2 – w), and a^3 ~ (x – wy)/(2 – w), then for some units v and v‘, we get va^3 + v‘b^3 = c^3.

If v = (+/-)v‘, then we can just divide throughout by v and get a smaller solution to x^3 + y^3 = uz^3, which will yield infinite descent, a contradiction. We can perform a similar trick if v or v‘ is (+/-)1. So assume neither is true. In that case, vv‘ = (+/-)1, so vv‘a^3b^3c^3(2 – w)^3 is a cube. That’s enough to show uz^3 is a cube, so we can suppose u = 1.

Suppose z is not divisible by 3, and consider the residues of x^3, y^3, and z^3 mod 9. We have z = (+/-)(w + 1) mod 3, so z^3 = (+/-)(6w – 3) mod 9, where 2w – 1 = SQRT(-3). For cubic residues mod 9 that aren’t divisible by 3, note that d^3 and (d + 3e)^3 are the same mod 9, so it’s enough to look at residue classes mod 3, the only two of which that are cubic are 1 and -1. In other words, x^3 + y^3 = (+/-)1 (+/-)1 != (+/-)6w – 3 mod 9. This is a contradiction, so z is in fact divisible by 3.

Finally, since z is divisible by 3, z^3 is divisible by 27, so abc is divisible by (2 – w)^3. Therefore at least one of a, b, and c is divisible by 2 – w. If it’s c, then we get that mod 3, (+/-)v (+/-)v‘ = 0, so v = (+/-)v‘. If it’s a, then we get that v‘ = (+/-)1. If it’s b, we get that v = (+/-)1.

In any of the three cases, we get a suitable equation of the form a^3 + b^3 = u‘c^3, where u‘ is a unit, and |abc| < |z| < |xyz|. But that process we can repeat ad infinitum, which is a contradiction. So the original assumption that there’s a solution to x^3 + y^3 = uz^3 is wrong.

Clearly, Fermat’s last theorem for n = 3 follows from that, since we only need to restrict to integers and ignore u.