Factors of Difference of Two Even Powers

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Theorem

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

Then:

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


Proof

From Factorisation of $z^n - a$:

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

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

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


From Complex Roots of Unity occur in Conjugate Pairs:

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

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

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


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

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


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

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

Hence the result.

$\blacksquare$


Also see


Sources

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