Number Theory Euler totient function Prove that
$$\sum_{d\mid n}(-1)^{n/d}\varphi(d)=\begin{cases}-n&2\nmid n\\0&2\mid n\end{cases}$$
I have came across the above question.
I have done the following:


*

*$n$ is odd then so is $n/d$ which would result $(-1)^{n/d}$ $=-1$
I am assuming you would have to use Gauss: 
$$\sum_{d/n} φ(d) = n$$
$===>$ $$\sum_{d/n} (-1)^{n/d}φ(d) = -\sum_{d/n}φ(d)=-n$$


*

*$n$ is even then $n=2^{a}m$, where we can say $m$ is odd.


I am not sure what else to do from here.
Any help would be appreciated.
 A: If $n = 2^k$ for some $k \geq 1$, then:
\begin{align}
\sum_{j = 0}^k (-1)^{2^{k-j}} \phi(2^j) &= \sum_{j=0}^{k-1} \phi(2^j) - \phi(2^k)
\\
&= \sum_{j=0}^{k} \phi(2^j) - 2\phi(2^k)
\\
&= 2^k - 2 \phi(2^k)
\\
&= 2^k - 2\cdot 2^{k-1} = 0
\end{align}
Finally note that the sum is multiplicative.
A: One easy method is to notice that 
$$ a_n := \sum_{d\mid n}(-1)^{n/d}\varphi(d) \tag1 $$
is the Dirichlet convolution of two multiplicative sequences,
$\,(-1)^n\,$ and $\, \varphi(n).\,$
The Dirichlet generating function series of the first is
$$ f(s) := \sum_{n>0} (-1)^n/n^s = -1/1^s + 1/2^s - 1/3^s - \dots = -\frac{1-2/2^s}{1-1/2^s}\prod_{p>2}\frac1{1-1/p^s} \tag2 $$
and of the second is
$$ g(s) := \sum_{n>0} \varphi(n)/n^s = 1/1^s + 1/2^s + 2/3^s + 2/4^s + \dots =
\prod_{p} \frac{1-1/p^s}{1-p/p^s}. \tag3 $$
The Dirichlet convolution of the two sequences has Dirichlet generating function series
$$ f(s)\,g(s) = -1/1^s - 3/3^s - 5/5^s - \dots =  -\prod_{p>2} \frac1{1-p/p^s} \tag4 $$
because the $p=2$ factor in the second g.f. cancels the factor in the first.
A: Consider fractions $0<q\leq 1$ with denominator $n$, where $n$ is even.  If $d|n$ then the number of such fractions with denominator $d$ when written in lowest terms is $\varphi(d)$.
Now if $n/d$ is even, what do you know about the numerators of these $\varphi(d)$ fractions when written with denominator $n$?  Conversely if $n/d$ is odd what does that tell you about the numerators?
Your sum is just adding up the number of possible numerators where $n/d$ is even, and subtracting those that are odd.
