Factors of Sum of Two Even Powers

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Theorem

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

Then:

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


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 $-1$:

\(\displaystyle \alpha\) \(=\) \(\displaystyle e^{i \pi / \paren {2 n} }\) from Roots of Complex Number
\(\displaystyle \) \(=\) \(\displaystyle \map \cos {\dfrac {\pi} {2 n} } + i \map \sin {\dfrac {\pi} {2 n} }\)


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:

\(\displaystyle \paren {x + \alpha^k y} \paren {x + \alpha^{2 n - k} y}\) \(=\) \(\displaystyle \paren {x + \alpha^k y} \paren {x + \overline {\alpha^k} y}\) Complex Roots of Unity occur in Conjugate Pairs
\(\displaystyle \) \(=\) \(\displaystyle x^2 + x y \paren {\alpha^k + \overline {\alpha^k} } + \alpha^k y \overline {\alpha^k} y\)
\(\displaystyle \) \(=\) \(\displaystyle x^2 x y \paren {\alpha^k + \overline {\alpha^k} } + \cmod {\alpha^k}^2 y^2\) Modulus in Terms of Conjugate
\(\displaystyle \) \(=\) \(\displaystyle x^2 + x y \paren {\alpha^k + \overline {\alpha^k} } + y^2\) Modulus of Complex Root of Unity equals 1
\(\displaystyle \) \(=\) \(\displaystyle x^2 + x y \paren {\cos \dfrac {\paren {2 k + 1} \pi} {2 n} + i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} + \cos \dfrac {\paren {2 k + 1} \pi} {2 n} - i \sin \dfrac {\paren {2 k + 1} \pi} {2 n} } + y^2\) Definition of $\alpha$
\(\displaystyle \) \(=\) \(\displaystyle x^2 + 2 x y \cos \dfrac {\paren {2 k + 1} \pi} {2 n} + x y\) simplification

Hence the result.

$\blacksquare$


Sources