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 $\alpha_k$ are the complex $2n$th roots of negative unity:

Then we have that:
 * $U_{2 n} = \set {\tuple {\alpha_0, \alpha_{2 n - 1} }, \tuple {\alpha_1, \alpha_{2 n - 2} }, \ldots, \tuple {\alpha_k, \alpha_{2 n - k - 1} }, \ldots, \tuple {\alpha_{n - 1}, \alpha_n } }$

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

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

However

Consider the permutation:


 * $\sigma = \begin{pmatrix}

1 & 2    & \cdots & k     & \cdots & n - 1 & n \\ n & n - 1 & \cdots & n - k & \cdots & 2    & 1 \end{pmatrix}$

From Permutation of Indices of Product:


 * $\displaystyle \prod_{\map R k} p_k = \prod_{\map R {\map \sigma k} } p_{\map \sigma k}$

Hence: