<s>[INST] How to find $x^{2000}+x^{-2000}$ when $x + x^{-1} = \frac{1}{2}(1 + \sqrt{5})$ 
Let $x+x^{-1}=\dfrac{1+\sqrt{5}}{2}$. Find $x^{2000}+x^{-2000}$.

How many nice methods do you know for solving this problem? Thank you everyone.
My method: because $x+\dfrac{1}{x}=2\cos{\dfrac{2\pi}{5}}$, so $$x^{2000}+\dfrac{1}{x^{2000}}=2\cos{\dfrac{2000\pi}{5}}=2.$$
Can you think of other nice methods? Or this problem has not used Euler's theorem: $(\cos{x}+i\sin{x})^n=\cos{nx}+i\sin{nx}$
 [/INST] $$x+x^{-1}=\frac{\sqrt5+1}2$$
$$\implies 2x^2-x+2=\sqrt5x$$
$$\text{On squaring,} (2x^2-x+2)^2=5x^2$$
$$\implies 1-x+x^2-x^3+x^4=0$$ 
$$\text{ or,    }x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2\cdot x\cdot\frac1x=\left(\frac{\sqrt5+1}2\right)^2-2=\frac{\sqrt5-1}2$$
$$\implies x+\frac1x-\left(x^2+\frac1{x^2}\right)=1\implies 1-x+x^2-x^3+x^4=0$$
which is a Geometric Series with common ratio $=-x$ and the first term being $=1$
$$\implies 1-x+x^2-x^3+x^4=\frac{1+x^5}{1+x}\implies 1+x^5=0$$
$$\text{ or } 1+x^5=(1+x)(1-x+x^2-x^3+x^4)=0$$
$\implies x^5=-1\implies x^{10n}=(x^5)^{2n}=(-1)^{2n}=1$
Put $n=200,-200$
</s>