<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] Here is an algebraic way avoiding trig functions: note that your number $x$ satisfies
$$x^2-(\frac{1+\sqrt{5}}{2})x+1=0 \quad \implies \quad \text{by multiplying by the conjugate} \quad x^4-x^3+x^2-x+1=0$$ and then use the factorization
$$x^{10}-1=(x^6+x^5-x-1)(x^4-x^3+x^2-x+1)$$ to see that $x^{10}=1$.
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