Let $\mu(p,d)$ denote the value of the Möbius function at the gcd of $p$ and $d$. Prove that for every prime $p$ we have
$$\sum_{d|n}\mu(d)\mu(p,d) = \begin{cases} 1 & \text{if $n=1,$} \\ 2 & \text{if $n=p^a,a \ge1,$}\\ 0 & \text{otherwise} \end{cases}$$
I'm not sure where to start on this question at all; I've always had trouble with these kinds of proofs, and working with primes. Could anyone help with this?
Case 1: $n=1$ In this case our sum simplifies to $$\mu(1)\mu(p,1)=1$$ The gcd of a prime number and $1$ is obviously $1$.
Case 2: $n=p^a,a\ge1$
We have $$\sum_{d|p^a}\mu(d)\mu(p,d).$$ Here the crutial term is $\mu(d)$. If $d\neq1$ and $d\neq p$, $\mu(d)$ is always equal to zero, since we can rewrite $d$ as $p^2p^b$, where $b\ge0$. If $d=1$ it is trivial to show that the term is equal to $1$. If $d=p$, $\mu(p)\cdot \mu(p)$ = $1$.
Case 3: $\mathrm{otherwise}$
For non-quadratics: Let $p_n,n\in\mathbb{N}$ a prime number. We start with the product $p_1\cdot p_2$. The divisors are $1,p_1,p_2$ and $p_1\cdot p_2$. We will define $\nu(p,d) := \mu(d)\mu(p,d)$. $$\nu(p,1)=+1,\quad \nu(p,p_1)=\mp1, \quad \nu(p,p_2) = -1, \quad \nu(p,p_1p_2)=\pm1$$
If we add another prime, then what happens? The number $p_1p_2p_3$ have the same divisors as $p_1p_2$ but also $p_1p_3$, $p_2p_3$, $p_3$ and $p_1p_2p_3$. So everytime we multiply our expression with another prime we have an increase in $\nu=1$ but the same decrease in $\nu=-1$.
For quadratics: Just an example would be the number $p_1^2p_2$. Divisors, $1,p_1,p_2,p_1p_2,p_1^2,p_1^2p_2$. Same argument as if it was a non-quadratic but $\nu(p,p_1^2)$ etc. equals zero.
Remark: You can formulate everything more rigorously and better. I had to do it in my head on my phone.