Factors of Difference of Two Even Powers

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

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

Proof
From Factorisation of $z^n - a$:


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

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

From Complex Roots of Unity occur in Conjugate Pairs:
 * $U_{2 n} = \set {1, \tuple {\alpha, \alpha^{2 n - 1} }, \tuple {\alpha^2, \alpha^{2 n - 2} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k} }, \ldots, \tuple {\alpha^{n - 1}, \alpha^{n + 1} }, -1}$

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:

Hence the result.