Prove $\sum_{k\mid n}{\mu(k)d(k)}=(-1)^{\omega{(n)}}$

Question

I have the following exercise.

Show that for all natural numbers $n$, the following equality holds $$\sum_{d|n}{\mu{(d)}d(d)}=(-1)^{\omega{(n)}}$$ Here, $\mu$ is the Möbius function, $d$ counts the number of divisors of $n$, and $\omega$ counts the number of distinct prime divisors of $n$.

I tried looking at a number like $n=10$ just to see what it looks like expanded. So since the divisors of $n$ are $1,2,5,$ and $10$, I can show that $$\sum_{d|10}{\mu{(d)}d(d)}=(-1)^{\omega{(10)}}=\mu(1)d(1)+\mu(2)d(2)+\mu(5)d(5)+\mu(10)d(10)=1$$ This gives me $$1\cdot 1+-1\cdot 2+-1\cdot 2 +1\cdot 4 $$ I'm thinking somehow since both $\mu$ and $d$ are multiplicative that we can rewrite though as $$\mu(1)d(1)+\mu(2)d(2)+\mu(5)d(5)+\mu(2)d(2)\mu(5)d(5)$$ $$=\mu(1)d(1)+\mu(5)d(5)+\mu(2)d(2)+\mu(2)d(2)\mu(5)d(5)$$ $$=\mu(1)d(1)+\mu(5)d(5)+\mu(2)d(2)(1+\mu(5)d(5))$$ $$=\mu(1)d(1)+\mu(5)d(5)+\mu(2)d(2)(\mu(1)d(1)+\mu(5)d(5))$$ $$=(1+\mu(2)d(2))(1+\mu(5)d(5))$$ $$=\sum_{d|2}{\mu(d)d(d)}\sum_{d|5}{\mu(d)d(d)}$$ So this original summation function is multiplicative. But this isn't helping me see the how to move forward. I know that the $\mu$ function is defined as $\mu(n)=(-1)^{\omega(n)}$ if $n$ is square free and $0$ if divisible by a square, so I think this plays a role somehow, but again, I'm feeling lost.

EDIT: Would looking at $n=\prod_{i=1}^k{p_i^{\alpha_i}}$ be a more useful approach to the problem? Knowing the larger summed function is multiplicative means I can focus my approach on the $p_i^{\alpha_i}$...

Answer 1

Look at the sum for a prime power $p^k$. The divisors of $p^k$ are $1,p,p^2,...,p^{k-1}$. All of them contain a square except $1$ and $p$. That means $\mu(p^a)=0$. So the sum is $$\mu(1)d(1)+\mu(p)d(p)\\=1\times 1+(-1)\times 2=1-2=-1$$ So each different prime, or its power, contributes a factor (-1).

Answer 2

Another, Combinatorial, way would be like $$\sum _{d|n} \mu (d) d(d)=\sum _{i=0}^{w(n)}\sum _{p_1<p_2 \ldots < p_i}\mu (p_1 \dots p_i)d (p_1 \dots p_i)=\sum _{i=0}^{w(n)}\binom{w(n)}{i}(-1)^i2^i=(1-2)^{w(n)}$$ where the $p_j$ are primes in the descomposition of $n$.