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# factors of $n$ and $x^{n}-1$

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.

## Mathematics Subject Classification

11R60*no label found*11C08

*no label found*11R18

*no label found*

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