Prove that $\sum_{d|n, d \geq 1}{|\mu(d)|} = 2^{\omega(n)}$

Question

Let n ∈ Z with n > 0 and let ω(n) denote the number of distinct prime numbers dividing n.

Prove that

$$\sum_{d|n, d \geq 1}{|\mu(d)|} = 2^{\omega(n)}$$

I'm not sure how to approach this problem. All I've got so far is that

$$|\mu(d)| =\begin{cases} 0,& \text{if $n$ is not square free}\\ 1,& \text{if $n=1$ or if $n$ is square free} \end{cases} $$

Can anyone give me a hint about how to start/approach this proof?

Let $$n = \prod_{\kappa = 1}^k p_{\kappa}^{a_{\kappa}}.$$ So $\omega(n) = k$. Which divisors of $n$ are squarefree? — Daniel Fischer, Jan 21 '17 at 19:23
More generally $g(n) = \sum_{d | n} f(d) h(n/d)$ is multiplicative whenever $f(n),h(n)$ are multiplicative. Here $f(n) = |\mu(n)|$ and $g(n) =1$ are multiplicative, so it is enough to check the relation for $n = p^k$ a prime power. — reuns, Jan 21 '17 at 20:15

ajotatxe · Accepted Answer · 2017-01-21T19:33:03.913

A function $f:\Bbb N\to \Bbb C$ is multiplicative if $f(mn)=f(m)f(n)$ whenever $\gcd(m,n)=1$. The function $|\mu|$ is multiplicative, as you can easily check. Also we have that if $f$ is multiplicative then $$g(n)=\sum_{d\mid n}f(d)$$ is also multiplicative, because if $\gcd(m,n)=1$ then any divisor of $mn$ can be expressed uniquely as a product $ab$ where $a\mid m$ and $b\mid n$, and then $$g(mn)=\sum_{d\mid mn}f(d)=\sum_{a\mid m,b\mid n}f(ab)=\sum_{a\mid m}\sum_{b\mid n}f(a)f(b)=\sum_{a\mid m}f(a)\sum_{b\mid n}f(b)=g(m)g(n)$$

So you can prove the equality for powers of primes and then use multiplicity.

Prove that $\sum_{d|n, d \geq 1}{|\mu(d)|} = 2^{\omega(n)}$

1 Answers1