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