What is the number of polynomial factors of $a^n-b^n$?

Question

This is a number theoretical problem that I discovered myself. Let $f(n)$ be the number of factors of $a^n-b^n$ with integer coefficients when its completely factored. For example:

$f(1)=1$, because $a-b$ can't be factored further.
$f(2)=2$, because $a^2-b^2=(a-b)(a+b)$
$f(3)=2$, because $a^3-b^3=(a-b)(a^2+ab+b^2)$
$f(4)=3$, because $a^4-b^4=(a-b)(a+b)(a^2+b^2)$
$f(6)=4$, because $a^6-b^6=(a-b)(a+b)(a^2+ab+b^2)(a^2-ab+b^2)$

I inserted the sequence in OEIS, and I found that it was exactly $\tau(n)$, the number of divisiors of $n$. But how could I prove it? My idea was to show two things:

$f(p^n)=n+1$
$f(pq)=f(p)f(q)$ if $\gcd(p,q)=1$.

To prove the basic statement that $f(p)=2$, I tried to set $b=1$ and use complex numbers. I could prove some small cases, namely $p=3$ and $p=5$, by looking to the sum and products of the complex roots. But I couldn't generalize it.

I also tried to use the identity $\gcd(x^a-1, x^b-1)=x^{\gcd(a,b)}-1$, but this says nothing about other possible factors of lower degree.

Lastly I tried to involve group theory, because the number of subgroups of $C_n$ is $\tau(n)$ and the complex roots form a cyclic group as well. But I have no idea how to proceed.

The problem has been solved, but it has a nice corollary:

The number of factors of $a^{2n}-b^{2n}=(a^n-b^n)(a^n+b^n)$ is $\tau(2n)$, and the number of factors of $a^n-b^n$ is $\tau(n)$. So $a^n+b^n$ has $\tau(2n)-\tau(n)$ polynomial factors.

Similiarly, one can get:

The number of factors of $a^n-b^n$ is $\tau(n)$.
The number of factors of $a^n+b^n$ is $\tau(2n)-\tau(n)$.
The number of factors of $a^{2n}+a^nb^n+b^{2n}$ is $\tau(3n)-\tau(n)$.
The number of factors of $a^{2n}-a^nb^n+b^{2n}$ is $\tau(3n)-\tau(2n)+\tau(n)$.

@ajotatxe Yes, one through ten as well as a few around 100 and 200. — wythagoras, Jul 30 '15 at 15:36
@wythagoras Using Lucian's remark and the well-known fact that $$x^p - 1 = (x-1)(x^{p-1} + \ldots + 1) = (x-1)\cdot \phi_p(x)$$ where $p$ is prime should prove $f(p) = 2$. — AlexR, Jul 30 '15 at 15:39
@Lucian I see, they are interesting. I never really studied these things very well, I thought they were mainly related to $\varphi(n)$ and not $\tau(n)$. — wythagoras, Jul 30 '15 at 15:40
For the sum, this is related: http://math.stackexchange.com/questions/695266/factoring-xn-yn-over-the-integers — Watson, Aug 22 '16 at 16:53

Jyrki Lahtonen · Accepted Answer · 2015-12-05T13:20:02.430

As Lucian pointed out this follows immediately from the properties of cyclotomic polynomials. We have the factorization (into polynomials irreducible over $\Bbb{Q}$): $$ x^n-1=\prod_{d\mid n}\Phi_d(x). $$ Your observation follows from this because it yields the factorization: $$ a^n-b^n=b^n\left[\left(\frac ab\right)^n-1\right]=b^n\prod_{d\mid n}\Phi_d\left(\frac ab\right)= \prod_{d\mid n}b^{\deg \Phi_d}\Phi_d\left(\frac ab\right).$$ The multiplier $b^{\deg\Phi_d}$ exactly cancels the denominator created by plugging in $x=a/b$ to $\Phi_d(x)$.

The factors that you have found are exactly the polynomials $b^{\deg \Phi_d}\Phi_d(\frac ab)$ - one for each divisor of $n$.

What is the number of polynomial factors of $a^n-b^n$?

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