Factors of Sum of Two Odd Powers

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Theorem

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

Then:

\(\ds x^{2 n + 1} + y^{2 n + 1}\) \(=\) \(\ds \paren {x + y} \prod_{k \mathop = 1}^n \paren {x^2 + 2 x y \cos \dfrac {2 \pi k} {2 n + 1} + y^2}\)
\(\ds \) \(=\) \(\ds \paren {x + y} \paren {x^2 + 2 x y \cos \dfrac {2 \pi} {2 n + 1} + y^2} \paren {x^2 + 2 x y \cos \dfrac {4 \pi} {2 n + 1} + y^2} \dotsm \paren {x^2 + 2 x y \cos \dfrac {2 n \pi} {2 n + 1} + y^2}\)


Proof

\(\ds x^{2 n + 1} + y^{2 n + 1}\) \(=\) \(\ds x^{2 n + 1} - \paren {-\paren {y^{2 n + 1} } }\)
\(\ds \) \(=\) \(\ds x^{2 n + 1} - \paren {-y}^{2 n + 1}\)
\(\ds \) \(=\) \(\ds \paren {x - \paren {-y} } \prod_{k \mathop = 1}^n \paren {x^2 - 2 x \paren {-y} \cos \dfrac {2 \pi k} {2 n + 1} + \paren {-y}^2}\) Factors of Difference of Two Odd Powers
\(\ds \) \(=\) \(\ds \paren {x + y} \prod_{k \mathop = 1}^n \paren {x^2 + 2 x y \cos \dfrac {2 \pi k} {2 n + 1} + y^2}\) simplification

$\blacksquare$


Sources

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