Fee, $\phi$ , Fo, Fum

$x + \frac{1}{x} = \phi$

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

$x^2 + \frac{1}{x^2} + 2 = \phi + 1$ (By Properties of $\phi$ )

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

$x^2 - x + 1 - \frac{1}{x} + \frac{1}{x^2} = 0$

$x^4 - x^3 + x^2 - x + 1 = 0$

$\dfrac{x^5 + 1}{x + 1} = 0$ , Note that $( x \not = -1 )$

$x^5 + 1 = 0$

$x^5 = -1$

$x^{10} = 1$

Therefore $\Rightarrow x^{2000} + x^{-2000} = (x^{10})^{200} + (x^{10})^{-200} = 1^{200} + 1^{-200} = 1 + 1 = \boxed{2}$

Misal x = e^(i.a), maka 1/x = e^(-i.a)

e^(i.a) = cosa + i.sina e^(-i.a) = cosa - i.sina --------------------------- + e^(i.a)+e^(-i.a) = 2cosa (1+V5)/2 = 2cosa Maka a = 72 derajat. x^2000 + 1/x^2000 = e^(2000i.a) + e^(-2000i.a) = 2cos(2000a) = 2cos(2000x72) = 2cos0 = 2

I wish you understand :)

$$ Let $x=e^{i\theta}$ , then $x^{-1}=e^{-i\theta}$ . Euler's formula states that, for any real number $\theta$ and $n$ , $$ $e^{in\theta}=\cos n\theta+i\sin n\theta$ and $e^{-in\theta}=\cos n\theta-i\sin n\theta.$ $$ Then $\begin{aligned} x+x^{-1}&=\frac{1+\sqrt{5}}{2}\\ e^{i\theta}+e^{-i\theta}&=\frac{1+\sqrt{5}}{2}\\ \cos\theta+i\sin\theta+\cos\theta-i\sin\theta&=\frac{1+\sqrt{5}}{2}\\ 2\cos\theta&=\frac{1+\sqrt{5}}{2}\\ \theta&=\cos^{-1}\left(\frac{1+\sqrt{5}}{4}\right)\\ &=\frac{2\pi}{5}\text{rad}\\ &=72^\circ. \end{aligned}$ $$ Therefore, for $n=2000$ , we obtain $$ $\begin{aligned} x^{2000}+x^{-2000}&=e^{2000i\theta}+e^{-2000i\theta}\\ &=2\cos(2000\theta)\\ &=2\cos(2000\cdot72^\circ)\\ &=2\cos(400\cdot360^\circ)\\ &=2\cos360^\circ\\ &=\boxed{2} \end{aligned}$ $$

Note : If you like this solution, please vote up Danang's solution. Thanks... :)

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

$x+x^{-1}=\frac{1+\sqrt{5}}{2} \\ x^{2}+x^{-2}=(x+x^{-1})^2-2=\frac{-1+\sqrt{5}}{2} \\ x^{3}+x^{-3}=(x+x^{-1})^3-3\cdot(x+x^{-1})=\frac{1-\sqrt{5}}{2} \\ (x^{2}+x^{-2})(x^{3}+x^{-3})=x^{5}+x^{-5}+(x+x^{-1})=\frac{-1+\sqrt{5}}{2}\cdot\frac{1-\sqrt{5}}{2}=\frac{-3+\sqrt{5}}{2}\\ x^{5}+x^{-5}=-2\\ x^{10}+1=-2\cdot x^{5}\\ x^{5}=-1\\ (x^{5})^{400}+(x^{5})^{-400}=1+1=\boxed{2} .\\$

$let \\x=e^{i\theta}\\x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}\\ae^{i\theta}+ae^{-i\theta}=\frac{1+\sqrt{5}}{2}\\comparing \ real \ and \ imaginary \ parts: a=1; \cos\theta=\frac{1+\sqrt{5}}{4}\\\theta=\frac{36\pi}{180}\\x^{2000}+x^{-2000}=e^{i\frac{36\pi}{180}2000}+e^{-i\frac{36\pi}{180}2000}=e^{i2(200)\pi}+e^{-i2(200)\pi}=1+1=2\\because\ e^{\pm i2n\pi} = 1\ where \ n=0,1,2,3,.........$

2 assume (x^2000+x^-2000)=y therfore (x^2000+x^-2000)^2=y^2(eqn1) and (x^2000+x^-2000 + 2)=y+2 adding 2 to both side this equal to (x^2000+x^-2000)^2=y+2(eqn2) using 1 and 2 eqn hence y=2

{(root5+1)/2}^2-1 is a golden ratio so we can solve it with this mystery

As, A.M > G.M therefore, [(a + b)/2] > (a.b)^(1/2) , [x^2000 + x^(-2000)] / 2 > [x^2000 . x^(-2000)]^(1/2) [x^2000 + x ^(-2000)] / 2 > 1 x^2000 + x^(-2000) > 2 Ans = 2.