TITLE: Way to show $x^n + y^n = z^n$ factorises as $(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) = z^n$
QUESTION [5 upvotes]: For odd $n$ the Fermat equation $x^n + y^n = z^n$ factorises as $$(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) = z^n,$$
where $\zeta = e^{2 \pi i/n}$. I tried seeing this was true by multiplying the factors out by hand but got into a mess. Is there a neat way of showing this (e.g. is there a useful identity that can be used involving products of the $\zeta^i$ etc.?

REPLY [4 votes]: Fix $x \neq 0$. Let 
$$P(y)=(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) -x^n-y^n$$
Then $P(-x)=P(-\zeta x)=P(-\zeta^2 x)=..=P(-\zeta^{n-1} x)=0$ and $P(0)=x^n-x^n=0$.
Thus $P$ is a polynomial of degree at most $n$ which has at least $n+1$ roots. This proves that
$$(x + y)(x + \zeta y) \cdots (x + \zeta^{n-1}y) =x^n+y^n$$
for all $y$ and all $x \neq 0$. 
The case $x=0$ is trivial.