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

Hence the result.

NOTE:
 * The relabelling, $k = n + j - 1$, used above simply serves to cast the interim result into the required final form. The symmetries of the $cos$ function and the existence of all odd multiples of $\pi/{2 n}$ between $0$ and $2 \pi$ mean that the $xy$ term can be added or subtracted; similarly, the periodicity of the $cos$ function also allows us to choose any function we want, e.g. ${2 k - 1}$, as the generator for odd numbers. In more concise terms, the sign and odd-number generator function used simply change the order in which each factor is calculated - the value of the final product is unaffected because multiplication is commutative.