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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?

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    $\phi$ is multiplicative, but it is not *completely* multiplicative. You seem to be assuming that the squarefree fractors of $n$ are pairwise relatively prime and multiply out to $n$; this is not necessarily the case. Take $n=60$; then your set contains $1$, $2$, $3$, $5$, $6$, $10$, and $15$. But $\phi(1)\phi(2)\phi(3)\phi(5)\phi(6)\phi(10)\phi(15)\neq\phi(60)$.2011-03-31
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    You're correct. I was going in the wrong direction. I'll edit it.2011-03-31
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    This is still incorrect: you are assuming that if $m_i$ is a squarefree divisor of $n$, then $\gcd(m_i,n/m_i) = 1$. Again: take $n=60$, and $m_i$ the squarefree divisor $6$. Then $n/m_i = 10$ is not relatively prime with $m_i=6$, and it is false that $\phi(n/m_i)\phi(m_i) = \phi(n)$. $\phi(10) = 4$, $\phi(6)=2$, and $\phi(60) = 16\neq 8=\phi(6)\phi(10)$.2011-03-31
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    Gah, so tired. Thank you again.2011-03-31

2 Answers 2

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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.

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    Thank you. Now I simply show the sum holds for each prime, and then I can show the sum holds for every other number through its factorization.2011-03-31
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    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
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Hint. For a different approach at establishing that your sum is what you want: $$\displaystyle\phi(n) = n\!\!\!\!\!\prod_{p\mid n,\ p\text{ prime}}\!\!\!\!\left(1 - \frac{1}{p}\right).$$