Prove that for every positive integer , .
The notation here is denoting Mobius inversion, with
(the number of distinct primes dividing ), and
(the number of positive divisors of )
I'm pretty stuck on this one. Any help would be much appreciated!
Prove that for every positive integer , .
The notation here is denoting Mobius inversion, with
(the number of distinct primes dividing ), and
(the number of positive divisors of )
I'm pretty stuck on this one. Any help would be much appreciated!
we know that for any integer which is not square-free. so in your sum you only need to look at those divisors of which are square-free, i.e. those which are a product of
distinct primes. now there exist exactly divisors of which are product of distinct primes. also if an integer is a product of distinct primes, then
thus:
Another approach: multiplicative implies multiplicative.
So is multiplicative.
Thus: where the product goes over the primes dividing ( is the greatest integer such that )
-since if n is not square free- hence:
(*) is multiplicative if and only if whenever