Let $n$ be a positive integer. Then the binomial $x^n\!-\!1$ has as many prime factors with integer coefficients as the integer $n$ has positive divisors, both numbers thus being $\tau(n)$ .

Proof. If $\Phi_d(x)$ generally means the $d$ th cyclotomic polynomial $$\Phi_d(x) := (x-\zeta_1)(x-\zeta_2)\ldots(x-\zeta_{\varphi(d)}),$$ where the $\zeta_j$ s are the primitive $d$ th roots of unity, then the equation $$\prod_{d|n,\,\,d>0}\!\Phi_d(x) = x^n\!-\!1$$ is true, because each $n$ th root of unity is also a primitive $d$ th root of unity for one and only one positive divisor of $n$ . The cyclotomic factor polynomials $\Phi_d(x)$ have integer coefficients and are irreducible. Thus the number of them is same as the number $\tau(n)$ of positive divisors of $n$ .

For illustrating the proof, let $n = 6$ (divisors 1, 2, 3, 6); think the sixth roots of unity: $\zeta^0$ , $\zeta^1$ , $\zeta^2$ , $\zeta^3$ , $\zeta^4$ , $\zeta^5$ (where $\zeta = e^{i\pi/3} = \frac{1+i\sqrt{3}}{2}$ ). From them, $\zeta^0 = 1$ is the primitive 1st root, $\zeta^3$ the primitive 2nd root, $\zeta^2$ and $\zeta^4$ the primitive 3rd roots, $\zeta^1$ and $\zeta^5$ the primitive 6th roots of unity.