# Talk:Numbers whose count of divisors is prime

## The question is: Which numbers got odd count of divisors

If you look at prime decompostion of a number like n = p1^a*p2^b*...pn^z,

than the count of divisors is CoD = (a+1)*(b+1)*(c+1)*....(z+1).
That is only possible if all a..z are even.

That means all n has to be m*m.

Therefor only square numbers need to be tested.Horsth

- Brilliant, thanks! --PureFox (talk) 15:42, 11 July 2021 (UTC)
- Thanks, I was enjoying a weekend until 11pm on Sunday when a bunch of Italians spoiled it. Taking Horsth's analysis a step further, considering n with two Prime factors p
_{1}^{a}p_{2}^{b}then (a+1) and (b+1) are factors of CoD(n) which is therefore not prime. This makes the solution set the Cartesian product of n and g mapped to n^{g-1}where n is the set of primes and g is the set of of off odd primes. No need to factor, no need to determine primality.--Nigel Galloway (talk) 14:04, 13 July 2021 (UTC)