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If $x + \frac{1}{x} = 1$, then what is

$$ x^{2000} + \frac{1}{x^{2000}} = ?$$

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Assume that $x=e^{i\theta}$. Then you are trying to write $2\cos(2000\cdot\theta)$ in terms of $2\cos(\theta)$, so the answer is given by Chebyshev polynomials of the first kind:

$$ \left(x^{2000}+\frac{1}{x^{2000}}\right) = 2\cdot \widetilde{T}_{\!2000}\left(x+\frac{1}{x}\right) $$ where $\widetilde{T}_{n}(z) = T_n(z/2)$. In our very particular case, we may notice that $\frac{1}{x}+x=1$ implies $x=e^{\pm \pi i/3}$: since $2000\equiv 2\pmod{6}$, given $x+\frac{1}{x}=1$ we have: $$ x^{2000}+\frac{1}{x^{2000}}=x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2 = \color{red}{-1}.$$

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$$x^2-x+1=0\implies x^3+1=(x+1)(x^2-x+1)=0\iff x^3=-1$$

$$x^{2000}=(x^3)^{666}\cdot x^2=(-1)^{666}\cdot x^2=x^2$$

Finally use $a^2+b^2=(a+b)^2-2ab$ for $x^2+\dfrac1{x^2}$

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  • $\begingroup$ Thanks for writing this in 'elementary' terms $\endgroup$ – Simon S Oct 5 '15 at 13:13
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This is a not so elegant approach.

Let $a_n=x^n+\frac1{x^n}$ then $$a_n\left(x+\frac1x\right)=a_{n+1}+a_{n-1}$$ or $$a_{n+1}=a_{n}-a_{n-1}$$ We have $a_1=1$ and $a_2=-1$, hence solving recursively - plugging - we obtain \begin{align} a_{6n-4}=a_{6n-2}&=-1\\ a_{6n-3}&=-2\\ a_{6n}&=2\\ a_{6n-1}=a_{6n+1}&=1 \end{align} Now since $2000=6\times 334 -4$, we have that $a_{2000}=-1$.

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The first equation is equivalent to $x^2-x+1 = 0$ so $x = (1 \pm \sqrt{-3})/2 = \exp(\pm i \pi/3)$

Taking an exponential to a power is a simple multiplication

$(\exp(\pm i \pi/3)^m = (\exp(\pm i m\pi/3))$

If $n$ is an integer $\exp(i 2n\pi) = 1$ and $\exp(i (2n-1)\pi) = -1$.

Substituting back in gives $x^{2000} + x^{-2000} = e^{\pm i \pi (2/3 + 666 )} + e^{\mp i \pi(2/3+666)} = e^{\pm i \pi 2/3} + e^{\mp i \pi 2/3} = 2\cos(2 \pi/3) = -1$

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