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