A perfect welcome to 2016, Part 2

Aareyan has written a fine solution. For the sake of variety, let's look at another approach:

Let $c_m(n)$ be the sum of the $n$ th powers of the primitive $m$ th roots of unity. There is a useful formula, due to Otto Hölder: $c_m(n)=\frac{\phi(m)\mu(r)}{\phi(r)}$ where $r=\frac{m}{\gcd(m,n)}$ . To prove this formula, first observe that both sides are multiplicative functions in $m$ . The verification for a prime power $m=p^k$ is a bit tedious but straightforward.

In our case we have $n=480, m=2016$ and $d=21$ , so that $S=\frac{\phi(2016)\mu(21)}{\phi(21)}=48$ . The answer we seek is $\boxed{6}$ ... a perfect number... a perfect year.

nice problem. let $c_m(n)$ be the sum of the nth power of the mth roots of unity.if you dont know about them see this awesome wiki .

from the wiki we see that $c_m(n)=\begin{cases}0 &&,m\not\mid n\\m&&,m\mid n\end{cases}$ . let $r_m(n)$

be the nth powers of the primitive mth roots of unity.

note that every nth root of unity is a dth primitive root of unity for d dividing n.

for example,lets look at the 4th roots of unity: $1,-1,i,-i$ . the $i,-i$ are primitive 4th roots of unity, $-1$ is a primitive 2nd root of unity and finally $1$ is a primitive 1st root of unity.assuming this elucidated the earlier statement, lets move on.

we can say that the sum of the nth power of the mth roots of unity are the sum of the primitive dth roots of unity such that $d|n$ . or $c_m(n)=\sum_{d|n}(r_d(n))$ .

written in Dirichlet convolution form this is $c=r*u$ . convoluting both sides with $\mu$ we get $c*\mu=r*I=r$ or $r_m(n)=\sum_{d|m}c_d(n)\mu(\dfrac{m}{d})$ .

we remember $c_d(n)$ "dissapears" for $d\not\mid n$ .so basically we want d|n and d|m. from this we want a common divisor of n,m or we want $d|gcd(m,n)$ .

so the summation turns into $r_m(n)=\sum_{d|gcd(m,n)}d\mu(\dfrac{m}{d})$ .

put m,n=2016,480. their greatest common is $96=2^5*3$ .

remember that $\mu$ dissapears over non-squarefree integers. $2016=2^5*3^2*7$ . for $\dfrac{2015}{d}$ to be square free, power of 2 atleast 4 and power of 3 atleast 1. so the possible d are $2^4*3=48$ and $2^5*3=96$ .

put this in to get $-48+96=\boxed{48}$ . divide by 8 to get $\boxed{6}$ .