Well suppose you have a function such that . Then by the inversion formula you have , which is precisely .
You actually don't even need the hypothesis that is multiplicative.
Prove that if f is multiplicative and the sum of f(d) (where the sum extends over all positive divisors of n) = n for all n, then f = Phi ( where the phi function is defined as the the number of positive integers not exceeding n that are relatively prime to n). As related to the Mobius Inversion Formula.
Can anyone get me going on this proof?
Hi. I'm working on a few other questions as well. If you be so kind, I'd appreciate your input on this problem.
Suppose n>0, p/n, p is prime, p^2/n, Show that there exists integers a and b such that n=a^3b^2.
Now, it seems to me that all exponents of the prime factors of n have to be at least power of 2 which means that the odd exponents have power at least 3 which would satisfy the above statement. But I'm not sure how to show this symbolically - perhaps the explanation is enough?
Actually, you do not need to know Möbius formula for this.
If you think of it, having: for a given function , determines for all positive integers . (In fact is linear combination of for )
(THINK INDUCTIVELY)
Now since on the other post we proved that satisfies it follows that for all positive integers .
PS: But I have to say that proving the assertion above is as long (if not longer) than proving Möbius inversion formula. - although it seems a pretty obvious fact-