Odd Powers Only

$(x+\dfrac{1}{x})^2=2$ $x^2+\dfrac{1}{x^2}=0$ $x^4=-1$ $x^2=\pm i$ $\dfrac{1}{x^2}=\mp i$ $\sum_{n=0}^{2015} (x^{2m+1}+\dfrac{1}{x^{2m+1}})^2=\sum_{n=0}^{2015} (x^{4m+2}+\dfrac{1}{x^{4m+2}}+2)$ $=\sum_{n=0}^{2015} ((x^4)^m*x^2+\dfrac{1}{(x^4)^m*x^2}+2)$ $=\sum_{n=0}^{2015} ((\pm(-1)^mi\mp(-1)^m i+2)=\sum_{n=0}^{2015} (2)=4032$

By using algebraic approach, we can rewrite the initial equation as followed:

$\left( x + \frac1x\right)^2 = 2$

$x^{2}$ + 2(x) $\frac{1}{x}$ + $\frac{1}{x^{2}}$ = 2

$x^{2}$ + 2 + $\frac{1}{x^{2}}$ = 2

$x^{2}$ = - $\frac{1}{x^{2}}$

Now, if we multiply both sides by x, we will get: $x^{3}$ = - $\frac{1}{x}$ . _(1)

Similarly, if we divide both sides by x, we will get: $x$ = - $\frac{1}{x^{3}}$ . _(2)

Adding (1) & (2), we will have a new equation: $x^{3}$ + $\frac{1}{x^{3}}$ = - $x$ - $\frac{1}{x}$ = -[ $x$ + $\frac{1}{x}$ ]

As a result, if we square both sides, $\left( x^{3} + \dfrac1{x^{3}}\right)^2$ = $\left( x + \frac1x\right)^2 = 2$ from the first constraint.

Then multiplying squared x to (1), we can raise the power to 5, $x^{5}$ = - $x$ . _(3)

And by dividing squared x to (2), $\frac{1}{x}$ = - $\frac{1}{x^{5}}$ _(4)

Adding (3) & (4), we will also get: $x^{5}$ + $\frac{1}{x^{5}}$ = - $x$ - $\frac{1}{x}$ = -[ $x$ + $\frac{1}{x}$ ]

Therefore, $\left( x^{5} + \dfrac1{x^{5}}\right)^2$ = $\left( x + \frac1x\right)^2 = 2$ .

Now, multiplying squared x to (3), $x^{7}$ = - $x^{3}$ _(5)

And, dividing squared x to (4), $\frac{1}{x^{3}}$ = - $\frac{1}{x^{7}}$ _(6)

From adding (5)&(6) then squaring like before, $\left( x^{7} + \dfrac1{x^{7}}\right)^2$ = $\left( x^{3} + \dfrac1{x^{3}}\right)^2$ = 2.

And the same process can go on and on: as long as the odd powers are retained, the squared term of the higher degree will be equalized to the lower degree one (specifically power minus 4) and will always consequently have a value of 2.

As a result, since there are 2016 squared terms in the summation, the total value will equal to

$2016 \times 2$ = 4032

$(x+\frac{1}{x})^2=2\Rightarrow x^2+x^{-2}=0\Rightarrow x^{-2}=-x^2.\\ \begin{aligned} (x^{2m+1}+\frac{1}{x^{2m+1}})^2&=x^{4m+2}+x^{-(4m+2)}+2\\ &= (x^2)^{2m+1}+(x^{-2})^{2m+1}+2\\ &=(x^2)^{2m+1}+(-x^2)^{2m+1}+2\\ &=(x^2)^{2m+1}-(x^2)^{2m+1}+2\\ &=2 \end{aligned} \\\sum_{m=0}^{2015}{(x^{2m+1}+\frac{1}{x^{2m+1}})^2}=\sum_{m=0}^{2015}{2}=2\times2016=4032$

Factoring the expression yields $\sum \left(x^{4m+2}+\frac1{x^{4m+2}}+2\right)=4032+\sum \left(x^{4m+2}+\frac1{x^{4m+2}}\right)$ To see that the last sum vanishes observe that $2=\left(x+\frac1x\right)^2=x^2+x^{-2}+2\quad\Rightarrow x^2+x^{-2}=0$ hence $x^4+1=0$ So $x$ is a primitive 8th root of unity. Now imagine the roots of unity and the rest follows. In detail: The inverse of a root of unity is its complex conjugate, so the sum $x^k+x^{-k}$ is just twice the realpart of $x$ so all in all we are done if the realparts in the sum at the beginning are zero. But $x^{4m+2}=(x^2)^{2m+1}=y^{2m+1}$ where $y$ is a 4th root of unity which has no real part so the whole expression has none as $2m+1$ is odd. NOTE: that this is way easier to understand (when looking at a unit circle with the roots of unity) than to express by words as I have tried here.