# Factors of Difference of Two Even Powers

## Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:

$x^{2 n} - y^{2 n} = \paren {x - y} \paren {x + y} \displaystyle \prod_{k \mathop = 1}^{n - 1} \paren {x^2 - 2 x y \cos \dfrac {k \pi} n + y^2}$

## Proof

$z^{2 n} - y^{2 n} = \displaystyle \prod_{k \mathop = 0}^{2 n - 1} \paren {x - \alpha^k y}$

where $\alpha$ is a primitive complex $2 n$th roots of unity, for example:

 $\displaystyle \alpha$ $=$ $\displaystyle e^{2 i \pi / \paren {2 n} }$ $\displaystyle$ $=$ $\displaystyle \cos \dfrac {2 \pi} {2 n} + i \sin \dfrac {2 \pi} {2 n}$ $\displaystyle$ $=$ $\displaystyle \cos \dfrac \pi n + i \sin \dfrac \pi n$
$U_{2 n} = \set {1, \tuple {\alpha, \alpha^{2 n - 1} }, \tuple {\alpha^2, \alpha^{2 n - 2} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k} }, \ldots, \tuple {\alpha^{n - 1}, \alpha^{n + 1} }, -1}$

where $U_{2 n}$ denotes the complex $2 n$th roots of unity:

$U_{2 n} = \set {z \in \C: z^{2 n} = 1}$

The case $k = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $x - y$.

The case $k = n$ is taken care of by setting $\alpha^k = -1$, from whence we have the factor $x + y$.

Taking the product of each of the remaining factors of $x^{2 n} - y^{2 n}$ in pairs:

 $\displaystyle \paren {x - \alpha^k y} \paren {x - \alpha^{2 n - k} y}$ $=$ $\displaystyle \paren {x - \alpha^k y} \paren {x - \overline {\alpha^k} y}$ Complex Roots of Unity occur in Conjugate Pairs $\displaystyle$ $=$ $\displaystyle x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + \alpha^k y \overline {\alpha^k} y$ $\displaystyle$ $=$ $\displaystyle x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + \cmod {\alpha^k}^2 y^2$ Modulus in Terms of Conjugate $\displaystyle$ $=$ $\displaystyle x^2 - x y \paren {\alpha^k + \overline {\alpha^k} } + y^2$ Modulus of Complex Root of Unity equals 1 $\displaystyle$ $=$ $\displaystyle x^2 - x y \paren {\cos \dfrac {k \pi} n + i \sin \dfrac {k \pi} n + \cos \dfrac {k \pi} n - i \sin \dfrac {k \pi} n} + y^2$ Definition of $\alpha$ $\displaystyle$ $=$ $\displaystyle x^2 - 2 x y \cos \dfrac {k \pi} n + x y$ simplification

Hence the result.

$\blacksquare$