4
$\begingroup$

I'm trying to prove $\sum_{m|n} \mu(m)^2/\phi(m) = n/\phi(n)$

My first realization was that $\mu(m)^2 = 1$ iff $m=1$ or $m$ is a squareless factor of $n$ and otherwise is 0. Let $\{1,m_1,m_2, ... m_k\}$ be the set of squarefree factors of $n$.

So now we have a sum $\frac{1}{\phi(1)} +\frac{1}{\phi(m_1)} + \cdots +\frac{1}{\phi(m_k)}.$

How do I show the sum equals $n/\phi(n)$? Or am I going in the wrong direction?

  • 0
    Gah, so tired. Thank you again.2011-03-31

2 Answers 2

6

You may know the result that if $f$ is a multiplicative function, and $F(n)=\sum_{d\mid n} f(d)$, then $F$ is multiplicative.

Using that fact, you only need to verify your assertion for $n$ a prime power, where it is immediate.

  • 0
    One needs to show it holds for any power of a prime. However, the sum on the left, for a prime power, is quite simple, since the Moebius function terms are mostly 0.2011-03-31
2

Hint. For a different approach at establishing that your sum is what you want: $\displaystyle\phi(n) = n\!\!\!\!\!\prod_{p\mid n,\ p\text{ prime}}\!\!\!\!\left(1 - \frac{1}{p}\right).$