Factors of Sum of Two Even Powers

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Theorem

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

Then:

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


Proof

From Factorisation of $z^n + a$:

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

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:

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

From Permutation of Indices of Product:

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

Hence:

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



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$\blacksquare$


Sources

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