More fun in 2015, Part 2

Prasun has given a beautiful conceptual proof. Here, for the same of variety, is a more pedestrian approach.

Let $w=e^{2\pi{i}/2015}$ . Then the 2015th roots of unity are $w^k$ for $k=0,1,...,2014$ . Now $\sum_{k=0}^{2014}(w^k)^{195}=\sum_{k=0}^{2014}(w^{195})^{k}=\frac{1-(w^{195})^{2015}}{1-w^{195}}=\frac{1-(w^{2015})^{195}}{1-w^{195}}=0$

More generally, if $m$ and $n$ are two positive integers such that $n\not|{m}$ , then the sum of the mth powers of all the nth roots of unity is 0. If $n|m$ then the sum is $n$ , of course.

We can use this observation to solve this

The $2015^{\textrm{th}}$ roots of unity are actually the roots of the monic degree $2015$ polynomial $P(x)=x^{2015}-1$ .

Denote by $e_i$ the $i^{\textrm{th}}$ elementary symmetric polynomial made up by the roots of $P(x)$ and denote by $P_i$ the $i^{\textrm{th}}$ power sum of the roots of $P(x)$ .

By Vieta's formulas , we have,

$e_i=0~\forall~i\in\Bbb{Z^+_{\leq 2014}}\quad\textrm{and}\quad e_{2015}=(-1)^{2015}\cdot\frac{(-1)}{1}=1$

Now, we simply use Newton's Identities for finding the value of $P_{195}$ which is our required answer.

$\begin{aligned}~P_{195}&=\left(\sum_{i=1}^{194}(-1)^{i-1}e_iP_{195-i}\right)+195e_{195}\\&=\left(\sum_{i=1}^{194}(0)\right)+195\times 0=\boxed{0}\end{aligned}$

An interesting conclusion that can be obtained using the general form of Newton's Identities is,

$P_i=0~\forall~i\in\Bbb{Z^+_{\leq 2014}}\quad\textrm{and}\quad P_{2015}=2015e_{2015}=2015$

We also get the following reduced recurrence,

$P_i=e_{2015}P_{i-2015}=P_{i-2015}~\forall~i\in\Bbb{Z_{\geq 2016}}$

Using this recurrence and the values obtained till now, we have the following result,

$P_i=\begin{cases}\begin{aligned}~0\quad\forall~&i\not\equiv 0\pmod{2015}\\ 2015\quad\forall~&i\equiv 0\pmod{2015}\end{aligned}\end{cases}~\forall~i\in\Bbb{Z^+}$

sum of nth (where n>=2) roots of unity is always zero.