Factorization of $a^n \pm b^n$

Question

I am trying to figure out a way to factor $a^n \pm b^n$, but all I found is negative odd cases.

Is there any such way to factor using complex numbers?

Answer 1

You can factor $x^n-1$ with cyclotomic polynomials:

$$ x^n-1=\prod_{d\mid n}\Phi_d(x). $$

The product ranges over all divisors $d$ of $n$.

One formula for $\Phi_n(x)$ involves cancelling factors and Mobius inversion:

$$ \Phi_n(x)=\prod_{d\mid n}(x^d-1)^{\mu(n/d)}. $$

Here $\mu(k)$ is the Mobius function, defined to be $(-1)^{\omega}$ if $\omega$ is the number of distinct prime factors of $k$ and $k$ is squarefree, or is $0$ if $k$ is divisible by a nontrivial perfect square.

In practice there are lots of recursive formulas that let you calculate $\Phi_n(x)$.

Over $\mathbb{Q}$, each $\Phi_d(x)$ is an irreducible polynomial. Over $\mathbb{C}$ it factors as $\prod(x-\zeta)$ where $\zeta$ ranges over all primitive $d$th roots of unity, i.e. $\zeta=e^{2\pi ik/d}$ for all $1<k<d$ with $\gcd(k,d)=1$.

Thus, while the roots of $x^n-1$ are all $n$th roots of unity, we group the roots according to their order (the minimal exponent it takes to power to $1$) and these give the cyclotomic factors $\Phi_d(x)$ for each $d\mid n$. This is related to Galois theory.

To factor $x^n+1$, notice it is part of $(x^n+1)(x^n-1)=x^{2n}-1$, therefore the factors of $x^n+1$ are precisely the cyclotomic polynomials $\Phi_d(x)$ for divisors $d\mid 2n$ which are not divisors of $n$.

To factor $a^n\pm b^n$, you homogenize the factorization of $x^n\pm1$. For instance,

$$ x^3-1=(x-1)(x^2+x+1)\iff a^3-b^3=(a-b)(a^2+ab+b^2). $$