Factors of Sum of Two Even Powers

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

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

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


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

where $\alpha$ is a complex $2 n$th roots of $-1$:

Then we have that:
 * $U_{2 n} = \set {\tuple {\alpha, \alpha^{2 n - 1} }, \tuple {\alpha^3, \alpha^{2 n - 3} }, \ldots, \tuple {\alpha^{2 k - 1}, \alpha^{2 n - 2 k + 1} }, \ldots, \tuple {\alpha^{n - 1}, \alpha^{n + 1} } }$

where $U_{2 n}$ denotes the complex $2 n$th roots of $-1$:
 * $U_{2 n} = \set {z \in \C: z^{2 n} = -1}$

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

Hence the result.