Special sum of Moebius function. Let $\mu$ be a Moebius function.

Find formula for 
  $$f(a,b)=\sum_{c: a|c , c|b}\mu (c)$$

For example it is known that $f(1,1)=1$ and $f(1,b)=0$ for $b>1$.
I tried to generalise that result using the same methods such us multiplicativity of $f(a,b)$ but i can't prove this.
Any help will be appreciated.
 A: Hint: If $a$ does not divide $b$, then the sum is over the empty set. Otherwise, the sum is equal to 
$$\sum_{t|\frac{b}a} \mu(at)$$
via $c=at$.
Any summand when $a$ and $t$ are not co-prime will be zero (as their product contains a square). Thus it is enough to sum over the the divisors $t$ that are coprime to $a$, which are the divisors of $b'$, which is $b$ divided by any prime powers in it's prime decomposition of a prime that divides both $a$ and $b$.
So we have 
$$\sum_{t|\frac{b}a} \mu(at) = \sum_{t|b'} \mu(at) = \sum_{t|b'} \mu(a)\mu(t)=\mu(a) \sum_{t|b'} \mu(t).$$
The sum on the right hand side is 0 unless $b'=1$.
A: $$f(a,b) = \sum_{an | b} \mu(an) $$
If $a \nmid b$ then $f(a,b)=0$. 
If $\mu(a) = 0$ then $\mu(an)=0$ and $f(a,b)=0$. 
In the remaining case $\frac{\mu(an)}{\mu(a)}$  is multiplicative  so let $b=am$ and $$g(m)=\frac{f(a,am)}{\mu(a)} = \sum_{ n | m} \frac{\mu(an)}{\mu(a)}$$
It is multiplicative, for $p$ prime and $k \ge 1$ $$g(p^k) = \sum_{l=0}^k \frac{\mu(ap^l)}{\mu(a)} = \cases{ 1 \text{ if } p | a \\ \sum_{l=0}^k \frac{\mu(a)\mu(p^l)}{\mu(a)} = \mu(1)+\mu(p) = 0 \ \text{ otherwise}}$$
Whence $$g(m) = \prod_{p^k \| m} g(p^k) = \prod_{p^k \| m} 1_{p | a} = 1_{rad(m) | a}$$
$$f(a,am) = \mu(a) 1_{rad(m)|a}$$
