# 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 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 ParityOftauFunction 2013-03-22 18:55:43 2013-03-22 18:55:43 pahio (2872) pahio (2872) 6 pahio (2872) Feature msc 11A25 TauFunction