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# parity of $\tau$ function

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

$\displaystyle n\;=\;p_{1}^{{\alpha_{1}}}p_{2}^{{\alpha_{2}}}\cdots p_{r}^{{% \alpha_{r}}},$ | (1) |

then all positive divisors of $n$ are of the form

$p_{1}^{{\nu_{1}}}p_{2}^{{\nu_{2}}}\cdots p_{r}^{{\nu_{r}}}\quad\mbox{where}% \quad 0\leq\nu_{i}\leq\alpha_{i}\quad(i=1,\,2,\,\ldots,\,r).$ |

Thus the total number of the divisors is

$\displaystyle\tau(n)\;=\;(\alpha_{1}\!+\!1)(\alpha_{2}\!+\!1)\cdots(\alpha_{r}% \!+\!1).$ | (2) |

From this we see that in order 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.

Keywords:

square of an integer

Related:

TauFunction

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Feature

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## Mathematics Subject Classification

11A25*no label found*

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