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The following relation is well-known: For odd $n$, \begin{align} x^{n} - y^{n} = \prod_{k = 0}^{n-1} ( \zeta_n^{k} x - \zeta_{n}^{-k} y) \qquad \zeta_n =e^{2\pi i/n}. \end{align} Is there a similar relation for even $n$? Is there a similar relation involving roots-of-unity to factor the polynomial $x^{n} - y^{m}$ for arbitrary positive integers $n, m \geqslant 1$? Of course, not all of these polynomials are reducible, but I expect this to be visible in the putative product, if one exists.

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    $\begingroup$ For even power $2m$, you have that $x^{2m} - y^{2m} = (x^m - y^m)(x^m + y^m)$. Then if $m$ happens to be even, you can repeat the process on the left-hand factor until it goes odd. What to do with all the factors with $+$ in them, I don't know. $\endgroup$ – Arthur Feb 26 '13 at 23:21
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For all $n$ we have

$$x^n - y^n = \prod_{k=0}^{n-1} (x - \zeta_n^k y).$$

It follows that if $d = \gcd(n, m)$ then

$$x^n - y^m = \prod_{k=0}^{d-1} (x^{n/d} - \zeta_n^k y^{m/d}).$$

If $\gcd(n, m) = 1$ then $x^n - y^m$ is irreducible (so the above is a factorization into irreducible factors).

Proof. If $f(x, y)$ is an irreducible factor then so is $f(\zeta_n^i x, \zeta_m^j y)$ for all $i, j$. The product of all distinct (up to scalar multiple) polynomials of this form is invariant under $x \mapsto \zeta_n x$ and $y \mapsto \zeta_m y$, hence is a polynomial in $x^n$ and $y^m$. Since $x^n - y^m$ is a linear polynomial in $x^n$ and $y^m$, this product must be $x^n - y^m$, so $x^n - y^m$ factors as a product of polynomials all of which have the same degree in $x$ and the same degree in $y$. But since $\gcd(n, m) = 1$, this is possible if and only if $f$ has degree $n$ in $x$ and degree $m$ in $y$. $\Box$

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