Why is the sum across divisors of the Mobius function times the Euler function equal to zero when n is even? I want to show that when $n$ is even, $$\sum_{d|n} \mu(d)\varphi(d) = 0.$$ I've played around with it for a while but I can't seem to get it.
 A: Put $$f\left(n\right)=\sum_{d\mid n}\mu\left(d\right)\phi\left(d\right)
 $$ we can note that $f\left(n\right)$ is a multiplicative function, so we can consider the Euler product $$f\left(n\right)=\prod_{p\mid n}\left(1+F\left(p\right)\right)$$ where $F\left(p\right)=f\left(p\right)+f\left(p^{2}\right)+\dots$ but $\mu\left(k\right)\neq0$ only if $k$ is a square-free number, then $$f\left(n\right)=\prod_{p\mid n}\left(1+\mu\left(p\right)\phi\left(p\right)\right)=\prod_{p\mid n}\left(1-\left(p-1\right)\right)$$ and if $n$ is a even number the product is obviously $0$.
A: I give an answer equivalent to the other answer, but not passing through Euler products.
Since both $\mu$ and $\varphi$ are multiplicative, you can consider the sums
$$ \sum_{d \mid p^k} \mu(d) \varphi(d).$$
For even numbers, you can consider the case when $p = 2$. And since $\mu(n) = 0$ if $n$ is a square, we have that for any $k \geq 1$,
$$ \sum_{d \mid 2^k} \mu(d) \varphi(d) = \mu(1)\varphi(1) + \mu(2)\varphi(2) = 1 - 1 = 0.$$
A: Let $S= \{d :d|n \land \mu (d)\ne 0\}. $ Let $S(1)$ be the set of odd members of $S. $ Let $S(2)=S\backslash S(1). $ When $n$ is even we have $S(2)= \{2d :d\in S_1\}. $  Observe that if $d\in S(1)$ then $\phi (2d)=\phi (d)$ and $\mu (2d)=-\mu (d) .  $ Hence, when $n$ is even we have $$\sum_{d|n}\mu (d)\phi (d)=    \sum_{d\in S}\mu (d)\phi (d)=$$   $$=\sum_{d\in S(1)}\mu (d)\phi (d)+\sum_{e\in S(2)}\mu (e)\phi (e)=$$   $$=\sum_{d\in S(1)}\mu (d)\phi(d)+\sum_{d\in S(1)}\mu (2d)\phi (2d)=$$ $$=\sum_{d\in S(1)}\mu (d)\phi (d)+\sum_{d\in S(1)}(-\mu(d)\phi (d))=0 .$$
