More fun in 2015, Part 12

The Möbius function is just what we needed, it is defined as the sum of the primitive n-th roots of unity. It is $\mu(n)$ , and if $n$ has repeated prime factors, then $\mu(n)=0$ ; if $n$ has an odd number of distinct prime factors, then $\mu(n)=-1$ ; and if $n$ has an even number of distinct prime factors, then $\mu(n)=1$ .

We know that $2015=5\times13\times31$ , it has $3$ distinct prime factors, hence $\mu(2015)=\boxed{-1}$ .

For the bonus question, $2009=7^2 \times 41$ , hence $\mu(2009)=0$ .