# parity of $\tau $ function

If the prime factor^{} decomposition of a positive integer $n$ is

$n={p}_{1}^{{\alpha}_{1}}{p}_{2}^{{\alpha}_{2}}\mathrm{\cdots}{p}_{r}^{{\alpha}_{r}},$ | (1) |

then all positive divisors^{} of $n$ are of the form

$${p}_{1}^{{\nu}_{1}}{p}_{2}^{{\nu}_{2}}\mathrm{\cdots}{p}_{r}^{{\nu}_{r}}\mathit{\hspace{1em}}\text{where}\mathit{\hspace{1em}}0\le {\nu}_{i}\le {\alpha}_{i}\mathit{\hspace{1em}}(i=1,\mathrm{\hspace{0.17em}2},\mathrm{\dots},r).$$ |

Thus the total number of the divisors is

$\tau (n)=({\alpha}_{1}+1)({\alpha}_{2}+1)\mathrm{\cdots}({\alpha}_{r}+1).$ | (2) |

From this we see that in to $\tau (n)$ be an odd number^{}, every sum ${\alpha}_{i}+1$ shall be odd, i.e. every exponent^{} ${\alpha}_{i}$ in (1) must be even. It means that $n$ has an even number of each of its prime divisors ${p}_{i}$; so $n$ is a square of an integer, a perfect square^{}.

Consequently, the number of all positive divisors of an integer is always even, except if the integer is a perfect square.

Examples. 15 has four positive divisors 1, 3, 5, 15 and the square number 16 five divisors

1, 2, 4, 8, 16.

Title | parity of $\tau $ function |
---|---|

Canonical name | ParityOftauFunction |

Date of creation | 2013-03-22 18:55:43 |

Last modified on | 2013-03-22 18:55:43 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 6 |

Author | pahio (2872) |

Entry type | Feature |

Classification | msc 11A25 |

Related topic | TauFunction |