2015 again

$x^4 = 1 \implies \begin{cases} x^3+x^2+x+1&=0 \\ 1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} & = 0 \end{cases}$

$\begin{aligned} \sum_{n=0}^{2015} \left(x^n + \frac{1}{x^n} \right) & = \sum_{n=0}^{2015} x^n + \sum_{n=0}^{2015} \frac{1}{x^n} \\ & = \sum_{k=0}^{503} x^{4k}(1+x+x^2+x^3) + \sum_{k=0}^{503} \frac{1}{x^{4k}} \left( 1 + \frac{1}{x} + \frac{1}{x^2} + \frac{1}{x^3} \right) \\ & = \sum_{k=0}^{503} x^{4k}(\color{#D61F06}{0}) + \sum_{k=0}^{503} \frac{1}{x^{4k}} \left( \color{#D61F06}{0} \right) \\ & = \boxed{0} \end{aligned}$

We have that $x^{4} - 1 = (x^{2} - 1)(x^{2} + 1) = (x - 1)(x + 1)(x + i)(x - i).$

So the roots distinct from $\pm 1$ are $\pm i.$ Now $i^{n} = i^{n+4}$ and $(-i)^{n} = (-i)^{n+4},$ so for each of these values we have that

$\displaystyle\sum_{n=0}^{2015} \left(x^{n} + \dfrac{1}{x^{n}}\right) = 504 \sum_{n=0}^{3} \left(x^{n} + \dfrac{1}{x^{n}}\right),$

since the given sum has $2016 = 4*504$ terms, and thus $504$ "cycles" of the same sum. Now

$\displaystyle\sum_{n=0}^{3} \left(i^{n} + \dfrac{1}{i^{n}}\right) =$

$\left(1 + \dfrac{1}{1}\right) + \left(i + \dfrac{1}{i}\right) + \left(-1 + \dfrac{1}{-1}\right) + \left(-i + \dfrac{1}{-i}\right) =$

$2 + \dfrac{i^{2} + 1}{i} + (-2) + \dfrac{i^{2} + 1}{-i} = 0.$

Similarly for $x = -i.$ Thus the original sum is $\boxed{0}$ as well.

Let $x=e^{i\theta}$

$x^4=1$

$e^{4i\theta}=e^{2ik\pi}$

$\theta=\frac{k\pi}{2}$

By De Movire's Theorem, $x^{n}+x^{-n}=e^{\frac{ink\pi}{2}} +e^{\frac{-ink\pi}{2}}=2\cos{\frac{nk\pi}{2}}$

The sum thus reduces to

$2\sum_{n=0}^{2015}\cos{\frac{nk\pi}{2}}$

Notice that $\cos{\frac{0\pi}{2}}+\cos{\frac{\pi}{2}}+\cos{\frac{2\pi}{2}}+\cos{\frac{3\pi}{2}}=0$

The cosine function is periodic, with a period of $\pi$ , hence we have 503 "cycles" of the same sum of zero, since $2015= 4(503)+3$ .

Thus, $2\sum_{n=0}^{2015}\cos{\frac{nk\pi}{2}}$ = $2[503×0+\cos{\frac{0\pi}{2}}+\cos{\frac{\pi}{2}}+\cos{\frac{2\pi}{2}}]$ $=2(0+1+0-1)=\boxed{0}$