
Mobius Inversion Formula
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?

Well suppose you have a function $\displaystyle f$ such that $\displaystyle n=\sum_{d\mid n}f(d)$. Then by the inversion formula you have $\displaystyle f(n)=\sum_{d \mid n}\mu(d)\frac{n}{d}$, which is precisely $\displaystyle \phi(n)$.
You actually don't even need the hypothesis that $\displaystyle f$ is multiplicative.

I don't see how it is the same as phi(n). I understand the mechanics of the mobius inversion formula  but I'm not seeing the relationship to the phi function. Can you help me connect them?

Sure! Apply the inversion formula to $\displaystyle n=\sum_{d\mid n}\phi(d)$. What do you get? (Giggle)

I get Phi=Summation of Mfunction(n/d) (d) which is also equal to n? Also, how do I access the math symbols?

To learn how to use LaTeX, click here.
If you apply the inversion formula you get $\displaystyle \sum_{d\mid n}\mu(d)\frac n d$ (or $\displaystyle \sum_{d\mid n}\mu(\frac n d)d$, which is the same). Now compare this with above...

You're the best(Rofl) I can't express how thankful I am that you are willing to help me.

primes
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: $\displaystyle \sum_{dn}f(d)=g(n)$ for a given function $\displaystyle g$, determines $\displaystyle f$ for all positive integers $\displaystyle n$. (In fact $\displaystyle f(n)$ is linear combination of $\displaystyle g(d)$ for $\displaystyle dn$)
(THINK INDUCTIVELY)
Now since on the other post we proved that $\displaystyle \phi$ satisfies $\displaystyle \sum_{dn}\phi(d)=n$ it follows that $\displaystyle f(n)=\phi(n)$ for all positive integers $\displaystyle n$.
PS: But I have to say that proving the assertion above is as long (if not longer) than proving Möbius inversion formula. (Rofl)  although it seems a pretty obvious fact
