Cyclotomic integers: Why do we have $x^n+y^n=(x+y)(x+\zeta y)\dots (x+\zeta ^{n-1}y)$? Why do we have the factorization $$x^n+y^n=(x+y)(x+\zeta y)\dots (x+\zeta ^{n-1}y)$$ for $\zeta$ a primitive $n^{\text{th}}$ root of unity where $n$ is an odd prime?
 A: Hints: use the cyclotomic polymomial
$$z^n-1=(z-1)(z-\zeta)\cdot\ldots\cdot(z-\zeta^{n-1})\implies\; \text{substitute}$$
$$z=-\frac xy\implies\left(-\frac xy\right)^n-1=\left(-\frac xy-1 \right)\left(-\frac xy-\zeta\right)\cdot\ldots\cdot\left(-\frac xy-\zeta^{n-1}\right)$$
Now just multiply the above equality by $\;y^n\;$ and, of course, use that $\,n\,$ is odd....
A: If $y = 0$, the factorisation reads $x^n = \underbrace{x\,.x \dots x}_{n\ \text{times}}$.
Consider $y\neq 0$. If $\zeta \in \mathbb{C}$ is a primitive $n^{\text{th}}$ root of unity then $\zeta^n = 1$ and $\zeta^k \neq 1$ for $k = 1, \dots, n-1$. In particular, $\zeta, \zeta^2, \dots, \zeta^{n-1}, \zeta^n=1$ are all distinct. Now consider $x^n + y^n$ as a polynomial in $x$ over the complex numbers (treat $y$ as a fixed constant). Then by the Fundamental Theorem of Algebra, $x^n + y^n$ has precisely $n$ zeroes. For each $k = 1, \dots, n$, $x = -\zeta^ky$ is a zero as 
\begin{align*}
x^n + y^n &= (-\zeta^ky)^n + y^n\\ 
&= (-1)^n(\zeta^k)^ny^n + y^n\\ 
&= -(\zeta^n)^ky^n + y^n\qquad \text{as $n$ is odd, $(-1)^n = -1$}\\ 
&= -y^n + y^n\\ 
&= 0.
\end{align*} 
As $y \neq 0$, $x = -\zeta^ky$ for $k=1, \dots, n$ are all distinct, so they are all of the zeroes of $x^n + y^n$. By the factor theorem, each zero corresponds to a linear factor, so we have 
\begin{align*}
x^n + y^n &= a(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y)(x+\zeta^ny)\\
&= a(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y)(x+y)\\
&= a(x+y)(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y)
\end{align*}
for some $a \in \mathbb{C}$. Comparing the coefficients of $x^n$, we deduce that $a = 1$ and arrive at the desired factorisation
$$x^n+y^n = (x+y)(x+\zeta y)(x+\zeta^2 y)\dots(x+\zeta^{n-1}y).$$
A: The equation $x^n-y^n = (x-y)(x-\zeta y)(x-\zeta^2 y) ... (x-\zeta^{n-1}y)$ actually applies to all values of $n$, not just when $n$ is prime.  
When $x=y=1$, this equation produces the product of chords of rational angles $2\pi x/n$, the product of which is simply $n$.  
This bit is tangential, but useful to know, and derives directly from this.
One couples this, with the fact that the union of the span of a finite set, closed to multiplication, and the set of rational numbers $F$ is the same as that of $Z$.  The $x^n-y^n$ factorises to a factor for each divisor of $n$, and the product of values $1-\zeta^m$, where $0<m<n$ and $hcf(m, n)=1$, then this factor is either $p$ where $n$ is some power of $p$, or 1 otherwise.  
This particular fact can be used to exclude, for example, that an angle with a chord like  $4-\sqrt{5}$ can not be a rational angle.  It is very useful, for example, in limiting the number of regular polyhedra to the known set.
