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Show that $n$ is square-free if and only if $\sum _{d\mid n} \mu (d)^2\varphi (d) = n$.

Attempt:

The only non-zero terms in the sum are the terms where $d = p_{i_1}\cdots p_{i_l}$ where $p_{i_1},\cdots ,p_{i_l}$ are distinct primes.

Then the sum becomes $\sum_{d\mid n}\varphi(d)$ where $d = p_{i_1}\cdots p_{i_l}$.

Now is there a way to show that this sum equals $n$?

Klaus
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Ash
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  • See this post –  Mar 11 '16 at 16:33
  • There are two things to prove: (i) If $n$ is square-free then the sum is $n$ and (ii) if the sum is $n$ then $n$ is square-free. It is best to prove these separately. For the first, there are many proofs. Maybe most mechanical, if you already have the result, is to use the fact that $\sum_{d\mid n} \varphi(d)$ is multiplicative. – André Nicolas Mar 11 '16 at 16:34
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    If $n$ is a perfect square or a multiple of a perfect square, then at least one $\mu(d)^2 = 0$, and hence $\mu(d)^2 \phi(d) = 0$. If $n = p^2$ (where $p$ is some positive prime), then $\mu(p)^2 \phi(p) = p - 1$, and the whole sum is $p$, which clearly falls short of $p^2$. Also note that squaring the Möbius function is essentially the same as taking its absolute value. – Robert Soupe Mar 11 '16 at 16:45

2 Answers2

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Taking the trouble to typeset this in mathematical notation we have that

$$\sum_{d|n} \mu(d)^2 \phi(d) = \sum_{S\subseteq Q} \prod_{p\in S} (p-1).$$

where $Q$ is the set of prime factors of $n.$ This equals

$$\left.\prod_{p\in Q} (1+(p-1)x)\right|_{x=1}$$

by the combinatorial subset construction.

Simplifying we get $$\prod_{p\in Q} p.$$

So the function in question is the product of the prime divisors of $n.$

Now for the foreward direction obviously if $n$ is squarefree then $\prod_{p\in Q} p = n.$ On the other hand if $\prod_{p\in Q} p = n$ then $n$ must be squarefree because if it isn't we have $\prod_{p\in Q} p \lt n,$ a contradiction.

Marko Riedel
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I'd just like to record an alternative solution.

Note that $$\sum_{d\mid n} \varphi(d) =n,$$ and $\mu(d)^2$ is either 0 or 1, so $$\sum_{d\mid n} \mu(d)^2\varphi(d) =n$$ if and only if for no $d\mid n$ we have $\mu(d)=0$. This is true if and only if no divisor $d$ of $n$ is divisible by the square of a prime if and only if all divisors of $n$ are square-free. Lastly, this is true if and only if $n$ is square-free.

Hence $$\sum_{d\mid n} \mu(d)^2\varphi(d) =n$$ if and only if $n$ is square-free as desired.

jgon
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