statements including Mobius Inversion Formula Find a simple formula for


*

*$f(n) = \sum \limits_{x|n} \mu(x) \tau(x)$

*$f(n) = \sum \limits_{x|n} \sigma(x) \mu(n/x)$ which I believe is equal to n. 

*$f(n) = \sum \limits_{x|n} \mu(x) \mu(n/x)$.


When we know that $\tau(n) = \sum \limits_{x|n} 1$ and $\sigma(n) = \sum \limits_{x|n} d$.We also know that the given functions are multiplicative and I was thinking of using Mobius Inversion Formula, but I wasn't sure on expressing them clearly.
 A: Here, the important thing is if $g$ and $h$ are multiplicative arithmetic functions i.e. $g$ (and $h$) are functions from $\mathbb{N}^*\rightarrow \mathbb{R}$ such that $g(nm)=g(n)g(m)$ whenever $gcd(n,m)=1$ then the functions $f_1$ and $f_2$ defined as :
$$f_1(n):=\sum_{d|n}g(d)h(n/d)\text{ and }f_2(n):=\sum_{d|n}g(d)h(d)  $$
are multiplicative arithmetic functions as well. Now it is easy to see that if $f$ is a multiplicative arithmetic function and $n=p_1^{a_1}...p_r^{a_r}$ then :
$$f(n)=f(p_1^{a_1})...f(p_r^{a_r})$$
So that, in order to compute $f(n)$ in each of your cases, you just have to do it for $p^a$ where $p$ is a prime number and $a\geq 1$. 

$f(n):=\sum_{d|n}\mu(d)\tau(d) $

Let $n=p^a$ we know that $\mu(p^k)=1$ if $k=0$, $-1$ if $k=1$ and $0$ if $k\geq 2$, furthermore $\tau(p^k)=k+1$ hence :
$$f(p^a)=\sum_{k=0}^a\mu(p^k)\tau(p^k)=1-2=-1 $$
Finally if $n=p_1^{a_1}...p_r^{a_r}$ with $a_i\geq 1$ then :
$$f(n)=(-1)^r\text{ where } r\text{ is the number of distinct primes dividing } n $$ 
I let you do the same for the others.

$f(n):=\sum_{d|n}\sigma(d)\mu(n/d) $ then $f(n)=n$ (as you said).

And for the last one (now corrected) :

$f(n):=\sum_{d|n}\mu(d)\mu(n/d) $

Then $f(n)=0$ if $n$ is divisible by a cube. If $n$ is not divisible by a cube then :
$$f(n)=(-2)^r\text{ where } r\text{ is the number of prime divisors of } n\text{ whose square does not divide } n $$
The solution is the following, like before, I only need to tell $f(p^a)$ where $p$ is a prime number and $a\geq 1$. You have :
$$f(p^a)=\sum_{k=0}^a\mu(p^k)\mu(p^{a-k}) $$
But you know that $\mu(p^k)=0$ if $k\geq 2$, so :
$$f(p^a)=\mu(p^0)\mu(p^a)+\mu(p^1)\mu(p^{a-1})=\mu(p^a)-\mu(p^{a-1}) $$
If $a>2$ then $f(p^a)=0-0=0$.
If $a=2$ then $f(p^2)=\mu(p^2)-/mu(p^{1})=0-(-1)=1$.
If $a=1$ then $f(p)=\mu(p^1)-\mu(p^0)=-1-1=-2$.
Using the multiplicativity of $f$ you get the result.
A: As an  addendum to  the excellent answer  already provided  suppose we
evaluate these using Dirichlet series and Euler products.
First. 
Here we are interested in
$$\sum_{d|n} \mu(d)\tau(d).$$
Note that
$$\sum_{n\ge 1} \frac{1}{n^s} \mu(n)\tau(n)
= \prod_p \left(1-\frac{2}{p^s}\right).$$
Therefore
$$\sum_{n\ge 1} \frac{1}{n^s}\sum_{d|n} \mu(d)\tau(d)
= \zeta(s) \prod_p \left(1-\frac{2}{p^s}\right)
= \prod_p \frac{1-2/p^s}{1-1/p^s}
\\ = \prod_p \left(1-\frac{1}{1-1/p^s}\right)
= \prod_p 
\left(-\frac{1}{p^s}-\frac{1}{p^{2s}}-\frac{1}{p^{3s}}\cdots\right).$$
This is zero for $n=1$ and $(-1)^{\omega(n)}$ otherwise.
Second.
Suppose we are interested in
$$\sum_{d|n} \sigma(d)\mu(n/d).$$
Observe that
$$\sum_{n\ge 1} \frac{1}{n^s} \sigma(n)
= \zeta(s) \sum_{n\ge 1} \frac{n}{n^s}
= \zeta(s)\zeta(s-1).$$
Recall furthermore that by definition of the Moebius function we have
$$\sum_{n\ge 1} \frac{1}{n^s} \mu(n)
= \prod_p \left(1-\frac{1}{p^s}\right) = \frac{1}{\zeta(s)}.$$
Therefore
$$\sum_{n\ge 1} \frac{1}{n^s} \sum_{d|n} \sigma(d)\mu(n/d)
= \zeta(s)\zeta(s-1) \frac{1}{\zeta(s)} = \zeta(s-1).$$
Hence this function is equal to $n.$
Third.
Suppose we are interested in
$$\sum_{d|n} \mu(d)\mu(n/d).$$
We have
$$\sum_{n\ge 1} \frac{1}{n^s} \sum_{d|n} \mu(d)\mu(n/d)
= \frac{1}{\zeta(s)^2}
= \prod_p \left(1-2/p^s+1/p^{2s}\right).$$
This  is  zero   if  $n$  is  divisible  by   a  cube,  and  otherwise
$$(-1)^{q(n)}$$  where $q(n)$  is  the number  of  prime factors  with
exponent one in the prime factorization of $n.$
