## Fermat’s Last Theorem

Next we attempt to show $$x^n + y^n = z^n$$ has no solutions over the integers for $$n > 2$$. 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 $$p > 3$$, 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 $$p > 3$$ 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 💡