# 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

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

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

 $\ds \alpha_k$ $=$ $\ds e^{i \paren {2 k + 1} \pi / {2 n} }$ from Roots of Complex Number $\ds$ $=$ $\ds \map \cos {\dfrac {\paren {2 k + 1} \pi} {2 n} } + i \, \map \sin {\dfrac {\paren {2 k + 1} \pi} {2 n} }$ $k \in \set {0, 1, 2, \ldots, 2 n - 1}$

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:

 $\ds p_k$ $=$ $\ds \paren {x - \alpha_k y} \paren {x - \alpha_{2 n - k - 1} y}$ $\ds$ $=$ $\ds \paren {x - \alpha_k y} \paren {x - \overline {\alpha_k} y}$ Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs $\ds$ $=$ $\ds x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + \alpha_k y \overline {\alpha_k} y$ $\ds$ $=$ $\ds x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + \cmod {\alpha_k}^2 y^2$ Modulus in Terms of Conjugate $\ds$ $=$ $\ds x^2 - x y \paren {\alpha_k + \overline {\alpha_k} } + y^2$ Modulus of Complex Root of Negative Unity equals 1 $\ds$ $=$ $\ds x^2 - x y \paren {\cos \dfrac {\paren {2 k + 1} \pi} {2 n} + i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} + \cos \dfrac {\paren {2 k + 1} \pi} {2 n} - i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} } + y^2$ Definition of $\alpha_k$ $\ds$ $=$ $\ds x^2 - 2 x y \cos \dfrac {\paren {2 k + 1} \pi} {2 n} + y^2$ simplification

However

 $\ds p_{n - k}$ $=$ $\ds x^2 - 2 x y \cos \dfrac {\paren {2 n - 2 k + 1} \pi} {2 n} + y^2$ $\ds$ $=$ $\ds x^2 - 2 x y \cos \dfrac {\paren {-2 k + 1} \pi + 2 n \pi} {2 n} + y^2$ $\ds$ $=$ $\ds x^2 + 2 x y \cos \dfrac {\paren {-2 k + 1} \pi} {2 n} + y^2$ Cosine of Angle plus Straight Angle: $\map \cos {x + \pi} = -\cos x$ $\ds$ $=$ $\ds x^2 + 2 x y \cos \dfrac {\paren {2 k - 1} \pi} {2 n} + y^2$ Cosine Function is Even: $\map \cos {- x} = \cos x$

Consider the permutation:

$\sigma = \begin{pmatrix} 1 & 2 & \cdots & k & \cdots & n - 1 & n \\ n & n - 1 & \cdots & n - k & \cdots & 2 & 1 \end{pmatrix}$
$\displaystyle \prod_{\map R k} p_k = \prod_{\map R {\map \sigma k} } p_{\map \sigma k}$

Hence:

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

$\blacksquare$