How can I calculate this sum?
$\sum\limits_{d~|~(n_1, n_2)} \mu(d) \tau\left(\dfrac{n_1}{d}\right) \tau\left(\dfrac{n_2}{d}\right)$,
where $(n_1, n_2)$ is gcd of $n_1$ and $n_2$, $\mu$ is Mobius function and $\tau(n)$ is the number of positive divisors of $n$.
I tried to use Mobius inversion formula, but still can't manage to deal with the problem.
Any ideas? Thanks.
P.S. Note that sum in the above equation runs over positive divisors only.
First observe that $S=\sum\limits_{d~|~(n_1, n_2)} \mu(d) \tau\left(\dfrac{n_1}{d}\right) \tau\left(\dfrac{n_2}{d}\right)$ is multiplicative.
Let, $n_1=\prod_{i=1}^{k} p_i^{\alpha_i}\prod_{i=1}^{s}q_i^{\beta_i}$ and$n_2=\prod_{i=1}^{k} p_i^{\delta_i}\prod_{i=1}^{t}r_i^{\gamma_i}$ and $A =\{p_1, p_2,...p_k \}$
So, $S = \prod_{p \in A} ( -\tau\left(\frac{n_1}{p}\right)\tau\left(\frac{n_2}{p}\right) + \tau(n_1)\tau(n_2) )$.
Now, using the fact that $\tau(n)$ is multiplicative.
$S=(\tau(n_1) \tau(n_2))^k \prod_{i=1}^{k}\left( 1-\frac{\alpha_i\delta_i}{(\alpha_i +1)(\delta_i +1)}\right)$
Where k is the number of common prime divisors of $n_1$ and $n_2$