Proving an identity of the Möbius function and Euler’s totient function product

Question

Could anyone kindly help me to prove that $$ \sum_{d|n} \mu(d) \varphi(d) = 0 $$ for all even integers $ n \geq 2 $, where $ \mu $ is the Möbius function and $ \varphi $ is Euler’s totient function? Thank you!

Answer 1

$$\sum_{d|n} \mu(d) \varphi(d) $$

$$=\prod_{\text{prime }p\mid n}[\mu(1)\phi(1)+\mu(p)\phi(p)+0]$$ (using the definition of Möbius function)

$$=\prod_{\text{prime }p\mid n}[1-(p-1)]$$

Answer 2

Since the $\mu$ function is zero over non-squarefree numbers, assuming that $\omega(n)=k$ and the primes dividing $n$ are $p_1,\ldots,p_k$, we have:

$$\sum_{d\mid n}\mu(d)\varphi(d) = \sum_{I\subset\{1,\ldots,k\}}(-1)^{|I|}\prod_{i\in I}(p_i-1)=(-1)^k\prod_{j=1}^{k}(p_j-1)\prod_{i=1}^{k}\left(1+\frac{-1}{p_i-1}\right)$$ so: $$\sum_{d\mid n}\mu(d)\varphi(d) = (-1)^{\omega(n)}\prod_{p\mid n}(p-2).$$ Obviously, if $n$ is even then the RHS is zero.