Hello everyone,
I have some questions about the function d(n) which gives the number of positive divisors of n including n itself.
1. How can it be proven that d(n) is odd IFF n is a perfect square? (Kind of like proving that d(n) is odd iff n = k^2 for some integer k.
2. How is the following statement true: The product of all of the positive divisors of n (including n itself) is n^[d(n)/2] . Is there a proof for this ?
Any help is appreciated!
For all parts, note that if then , the idea would be the following:
For each divisor there's a corresponding one , the only divisor that is not counted here is in case of this being possible (that is, n is a square).
Thus if is not a square the divisors come in pairs, hence the number of divisors in that case is even, and in the case of n being a square it is odd (we add one).
For the second part use the above idea and note that -and now try to use it to calculate the product of the divisors-