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Talk:Numbers whose count of divisors is prime

From Rosetta Code

The question is: Which numbers got odd count of divisors[edit]

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)
I think, Nigel enjoys weekend :-) Horsth 15:47, 11 July 2021 (UTC)
Yes, he's usually onto these like a flash :) --PureFox (talk) 16:17, 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 p1ap2b 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 ng-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)