## 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

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 $x = y (mod p)$. We also have $x^p + (-z)^p = (-y)^p$, which by the same reasoning implies $x = -z (mod p)$. Thus

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 $I$.