TITLE: Proof with Möbius function, divisor function and little omega function
QUESTION [1 upvotes]: For all natural numbers n, show 
$\sum_{k|n}$$\mu$(k)d(k) = (-1)$^{\omega(n)}$
where 
$\omega$(n) := $\sum_{p|n}$1
$\omega$(1) = 0
$\mu$(n) is the Möbius function and d(n) is the divisor function
So I know the $\omega$(.) function is the sum of all distinct primes of n, but this problem doesn't make sense to me. The RHS will always be $\pm$1 but the LHS seems like it could be other values since d(n) can be greater than 1. 
Any pointers or solutions would be much appreciated!

REPLY [1 votes]: In response to query by OP we observe that with $g(n)$ multiplicative
we have
$$\sum_{d|n} g(d)
= \prod_{p^v||n} (g(1)+g(p)+g(p^2)+\cdots+g(p^v)).$$
Here we have $g(n) = \mu(n) d(n)$ and with $\mu(n)$ zero when $n$
is not squarefree this becomes
$$\sum_{d|n} g(d)
= \prod_{p^v||n} (g(1)+g(p)).$$
Observe  that $g(1)  =  \mu(1)d(1) =  1$  and $g(p)  =  \mu(p) d(p)  =
(-1)\times 2$ so we get
$$\sum_{d|n} g(d)
= \prod_{p^v||n} (1+(-1)\times 2)
= \prod_{p^v||n} (-1)  = (-1)^{\omega(n)}$$
as claimed.
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