Show $\sum_{d|n} d\mu(d) = ((-1)^{ω(n)}ϕ(n)s(n))/n$

Question

I shall define some functions first. For $n$ being any positive integer,

$ω(n) =$ no. of distinct primes $p$ which divides $n$.

$\mu(n) = (-1)^{ω(n)}$ if $n$ is square-free; $= 0$ if not.

$s(n) =$ the largest square-free factor of $n = \prod_{p|n} p$.

How can we show that $$\sum_{d|n} d\mu(d) = ((-1)^{ω(n)}ϕ(n)s(n))/n?$$

Many thanks.

Answer 1

With $p|n$ the primes dividing $n$ we get for the LHS

$$\prod_{p|n} (1+\mu(p)p) = \prod_{p|n} (1-p).$$

We get for the RHS

$$(-1)^{\omega(n)} \times n \times \prod_{p|n} \left(1-\frac{1}{p}\right) \times \prod_{p|n} p \times \frac{1}{n} \\ = (-1)^{\omega(n)} \prod_{p|n} \left(1-\frac{1}{p}\right) \times \prod_{p|n} p \\ = (-1)^{\omega(n)} \prod_{p|n} \left(p-1\right) \\ = \prod_{p|n} \left(1-p\right).$$

This is the claim.