An arithmetic function $f$ has the property $$\sum_{d\mid n}f(d)=\begin{cases}0 & \text{ if n is divisible by a square of a prime} \\ n & \text{ otherwise}\end{cases}$$ Calculate $f(6300)$.
I have done the following :
From the Mobius inversion formula we get $$f(n)=\sum_{d\mid n}\mu (d)g(n/d)$$ where $g$ is the above defined function.
We have that $6300=2^2\cdot 3^2\cdot 5^2\cdot 7$
So as $d$ do we take all $54$ divisors? Or just the prime ones $2,3,5,7$ ?
$\sum_{d\mid n}f(d)=g(n) $
where $g(n)=\begin{cases} 0& \text{ if n is divisible by a square of a prime} \\ n & \text{ otherwise}\end{cases}$
Then $f=\mu\star g$. Since it's easy to see that $g$ is multiplicative, this implies that $f$ is also multiplicative.
Let $n=\Pi_{i=1}^{k} p_i^{a_i}$ . Then by multiplicative property of $f$ , we have
$\begin{align}f(n) &=\Pi_{i=1}^{k} \mu\star g(p_i^{a_i})\\&=\Pi_{i=1}^{k}\sum_{j=0}^{a_i}\mu({p_i^{a_{i}-j}) }{g(p_i^{j})} \end{align}$
Given $n=6300=2^2\cdot 3^2\cdot 5^2\cdot 7$
$\begin{align} f(6300)=&\sum_{j=0}^{2}\mu({2^{2-j}) }{g(2^{j})}\cdot \sum_{j=0}^{2}\mu({3^{2-j}) }{g(3^{j})}\cdot \sum_{j=0}^{2}\mu({5^{2-j}) }{g(5^{j})}\cdot\sum_{j=0}^{1}\mu({7^{1-j}) }{g(7^{j})} \end{align}$
$\begin{align} &\sum_{j=0}^{2}\mu({2^{2-j}) }{g(2^{j})}\\&=\mu(2)g(2)\\&=-2\end{align}$
$\begin{align} &\sum_{j=0}^{2}\mu({3^{2-j}) }{g(3^{j})}\\&=\mu(3)g(3)\\&=-3\end{align}$
$\begin{align} &\sum_{j=0}^{2}\mu({5^{2-j}) }{g(5^{j})}\\&=\mu(5)g(5)\\&=-5\end{align}$
$\begin{align} &\sum_{j=0}^{1}\mu({7^{1-j}) }{g(7^{j})}\\&=\mu(7)g(1)+\mu(1)g(7)\\&=-1+7\\&=6\end{align}$
$f(6300) =-2\cdot(-3) \cdot(-5) \cdot 6=-180$