Exercise 4.3.8 ($\sum_{d \mid n} \mu(d) \phi(d) = 0$ if $n$ is even)

Question

If $n$ is any even integer, prove that $\sum_{d \mid n} \mu(d) \phi(d) = 0$.

Richard Ganaye · Accepted Answer

\begin{proof} 
We define for all positive integer $n$,
$$F(n) = \sum_{d \mid n} \mu(d) \phi(d).$$
We know that $\mu$ and $\phi$ are multiplicative functions, thus their product $\mu \phi$ is also multiplicative. By Theorem 4.8, $F$ is multiplicative.

If $n$ is even, we write $n = 2^\alpha m$, where $\alpha \geq 1$ and $m$ is odd. Then $F(n) = F(2^\alpha) F(m)$, and
\begin{align*}
F(2^\alpha) &= \sum_{d \mid 2^\alpha} \mu(d) \phi(d)\
&= \sum_{i=0}^\alpha \mu(2^i) \phi(2^i)&(d = 2^i)\
&= 1 - \phi(2)\
&= 0.
\end{align*}
Therefore $F(n) = F(2^\alpha) F(m) = 0.$

If $n$ is any even integer, then
  $$\sum_{d \mid n} \mu(d) \phi(d) = 0.$$
\end{proof}

Exercise 4.3.8 ($\sum_{d \mid n} \mu(d) \phi(d) = 0$ if $n$ is even)

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Exercise 4.3.8 ($\sum_{d \mid n} \mu(d) \phi(d) = 0$ if $n$ is even)

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