Fermat’s Last Theorem

Next we attempt to show \( x^n + y^n = z^n \) has no solutions over the integers for \(n \gt 2\). There is a truly remarkable proof of this, but unfortunately this site is too small to contain it. We can make a start though.

Using the result on Pythagorean triples, it can be shown that the case \(n = 4\) has no solutions, so we need only consider when \(n\) is some odd prime \(p\).

There are two cases. The first, case 1, is that \(p\) divides none of \(x, y, z\). Case 2 is that \(p\) divides exactly one of \(x, y, z\). We only consider the first case. The second is messier, but the results are similar.

For \(p=3\), considering the equation modulo 9 immediately shows there are no case 1 solutions.

For \(p \gt 3\), we may attempt to mimic the process that used to find Pythagorean triples, as Kummer did. First we rewrite the equations as

\[ (x+y)(x+y \omega)...(x + y \omega^{p-1}) = z^p \]

where \(\omega = e^{2 \pi i / p}\).

Then if \(\mathbb{Z}[\omega]\) is a UFD, one can show that \(x + y \omega = u a^p\) for some unit \(u\) and \(a \in \mathbb{Z}[\omega]\). This in turn implies \(x = y \pmod p\). We also have \(x^p + (-z)^p = (-y)^p\), which by the same reasoning implies \(x = -z \pmod p\). Thus

\[ 2x^p = x^p + y^p = z^p = -x^p \pmod p \]

so \(p\) divides \(3x^p\), a contradiction since \(p \gt 3\) and \(p\) does not divide \(x\) because we are in case 1.

Unfortunately \(\mathbb{Z}[\omega]\) is not always a UFD. However, it turns out it is always a Dedekind domain, which means its ideals factor uniquely into prime ideals. Using this fact, one can show that the ideal \(\langle x + y \omega \rangle = I^p\) for some ideal \(I\).

If \(p\) is regular, defined below, this implies that \(I\) is also a principal ideal thus \(\langle x + y \omega \rangle = I^p = \langle a \rangle ^p = \langle a^p \rangle\) for some element \(a\). In other words, \(x + y \omega = u a^p \) for some unit \(u\), and we can continue as before to prove case 1 of Fermat’s Last Theorem for such primes.

Define a relation \(~\) on the set of ideals of \(\mathbb{Z}[\omega]\) as follows. For ideals \(A, B\), say \(A ~ B\) if \(a A = b B\) for some \(a, b \in \mathbb{Z}[\omega]\). It turns out this is an equivalence relation, and furthermore, there are finitely many equivalence classes. The number of classes is the class number, and denoted by \(h\).

We call a prime \(p\) regular if \(p\) does not divide \(h\).

The ideal classes form a group under multiplication. The identity element is the class of all principal ideals. This explains why regularity is important. If \(p\) is regular, then its ideal class group cannot contain an element of order \(p\), and hence if \(I^p\) is principal, then so is \(I\).


Ben Lynn blynn@cs.stanford.edu 💡