More fun in 2015, Part 46

Abdelhamid has submitted a fine solution. For the sake of variety, let me show a slightly different approach.

Let $c_m(n)$ be the sum of all the $n$ th powers of the primitive $m$ th roots of unity. Let $b_m(n)$ be the sum of the $n$ th powers of all $m$ th roots of unity. We know that $b_m(n)=m$ if $m|n$ and $b_m(n)=0$ otherwise. Since every $m$ th root of unity is a primitive $d$ th root of unity for some divisor $d$ of $m$ , we find that $b_m(n)=\sum_{d|m}c_d(n)$ . Mobius inversion gives $c_m(n)=\sum_{d|m}b_d(n)\mu(m/d)=\sum_{d|\gcd(m,n)}d\times\mu(m/d)$

In particular, since $\gcd(2015,1729)=13$ , we have $c_{2015}(1729)=\sum_{d|13}d\times\mu(2015/d)=-1+13=\boxed{12}$

Prime decomposition of $2015 = 5 \times 13 \times 31$ and $1729 = 7 \times 13 \times 19$

Let $\zeta _{q}=e^{\frac {2\pi i}{q}}$

We can write $w$ a primitive root uniquely as (from $C_{2015}$ and $C_{5} \times C_{13} \times C_{31}$ isomorphism) $w = \zeta _{5}^{a} \zeta _{13}^{b} \zeta _{31}^{c} \quad for \quad 1 \le a \le 4 \quad 1 \le b \le 12 \quad 1 \le c \le 30$

Then
$S = \sum w^{1729} = (\sum_{1 \le a \le 4} \zeta _{5}^{1729a}) (\sum_{1 \le b \le 12} \zeta _{13}^{1729b}) (\sum_{1 \le c \le 30} \zeta _{31}^{1729c})$

Let $\eta _{q}(n)=\sum _{k=1}^{q -1}\zeta _{q}^{kn}$

Then: $S = \eta _{5}(1729) \times \eta _{13}(1729) \times \eta _{31}(1729))$

It follows from the identity $x^q - 1 = (x - 1)(x^{q-1} + x^{q-2} + ... + x + 1)$ that for q prime

$\eta _{q}(n)={\begin{cases}-1&\;{\mbox{ if }}q\nmid n\\q-1&\;{\mbox{ if }}q\mid n\\\end{cases}}$

$S = -1 \times 12 \times -1 = 12$