parity of τ function

If the prime factorMathworldPlanetmath decomposition of a positive integer n is

n=p1α1p2α2prαr, (1)

then all positive divisorsMathworldPlanetmathPlanetmath of n are of the form

p1ν1p2ν2prνrwhere0νiαi(i=1, 2,,r).

Thus the total number of the divisors is

τ(n)=(α1+1)(α2+1)(αr+1). (2)

From this we see that in to τ(n) be an odd numberMathworldPlanetmathPlanetmath, every sum αi+1 shall be odd, i.e. every exponentPlanetmathPlanetmath αi in (1) must be even.  It means that n has an even number of each of its prime divisors pi; so n is a square of an integer, a perfect squareMathworldPlanetmath.

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 τ 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