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

Let $n$ be a positive integer. Then the binomial ${x}^{n}-1$ has as many prime factors^{} (http://planetmath.org/PrimeFactorsOfXn1) with integer coefficients as the integer $n$ has positive divisors^{}, both numbers thus being $\tau (n)$ (http://planetmath.org/TauFunction).

Proof. If ${\mathrm{\Phi}}_{d}(x)$ generally means the $d$th cyclotomic polynomial^{}

$${\mathrm{\Phi}}_{d}(x):=(x-{\zeta}_{1})(x-{\zeta}_{2})\mathrm{\dots}(x-{\zeta}_{\phi (d)}),$$ |

where the ${\zeta}_{j}$s are the primitive $d$th roots of unity^{}, then the equation

$$\prod _{d|n,d>0}{\mathrm{\Phi}}_{d}(x)={x}^{n}-1$$ |

is true, because each $n$th root of unity is also a primitive (http://planetmath.org/RootOfUnity) $d$th root of unity for one and only one positive divisor of $n$. The cyclotomic factor polynomials ${\mathrm{\Phi}}_{d}(x)$ have integer coefficients and are irreducible (http://planetmath.org/IrreduciblePolynomial2). 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.

Title | factors of $n$ and ${x}^{n}-1$ |
---|---|

Canonical name | FactorsOfNAndXn1 |

Date of creation | 2013-03-22 16:35:05 |

Last modified on | 2013-03-22 16:35:05 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 7 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 11R60 |

Classification | msc 11C08 |

Classification | msc 11R18 |

Related topic | PrimeFactorsOfXn1 |