Factors of Difference 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

From Factorisation of $z^n - a$:

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

where $\alpha$ is a primitive complex $2 n + 1$th roots of unity, for example:

\(\ds \alpha\) \(=\) \(\ds e^{2 i \pi / \paren {2 n + 1} }\)
\(\ds \) \(=\) \(\ds \cos \dfrac {2 \pi} {2 n + 1} + i \sin \dfrac {2 \pi} {2 n + 1}\)


From Complex Roots of Unity occur in Conjugate Pairs:

$U_{2 n + 1} = \set {1, \tuple {\alpha, \alpha^{2 n} }, \tuple {\alpha^2, \alpha^{2 n - 1} }, \ldots, \tuple {\alpha^k, \alpha^{2 n - k + 1} }, \ldots, \tuple {\alpha^n, \alpha^{n + 1} } }$

where $U_{2 n + 1}$ denotes the complex $2 n + 1$th roots of unity:

$U_{2 n + 1} = \set {z \in \C: z^{2 n + 1} = 1}$


The case $k = 0$ is taken care of by setting $\alpha^0 = 1$, from whence we have the factor $x - y$.


Taking the product of each of the remaining factors of $x^{2 n + 1} - y^{2 n + 1}$ in pairs:

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

Hence the result.

$\blacksquare$


Also see


Sources

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