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: $(\cos{x}+i\sin{x})^n=\cos{nx}+i\sin{nx}$
 A: Just one note.
If $x \in \mathbb{R}$, then $x+x^{-1} \geqslant 2$. So, there are no $x$, such that $x+x^{-1}=\varphi=\dfrac{1+\sqrt{5}}{2}$.
If $x \in \mathbb{C}$, then denote $x = m (\cos \alpha + i\sin \alpha)$, where $m,\alpha \in \mathbb{R}$. 
So, $x^{-1} = m^{-1}(\cos\alpha - i\sin \alpha)$, and 
$$
x+x^{-1} = (m+m^{-1})\cos \alpha + i (m - m^{-1})\sin\alpha \implies
$$
$
m = m^{-1}=1, \cos \alpha = \varphi/2$.
Yes, if $x\in \mathbb{C}$, then there is sense.
A: $$x+x^{-1}=\frac{\sqrt5+1}2$$
$$\implies 2x^2-x+2=\sqrt5x$$
$$\text{On squaring,} (2x^2-x+2)^2=5x^2$$
$$\implies 1-x+x^2-x^3+x^4=0$$ 
$$\text{ or,    }x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2\cdot x\cdot\frac1x=\left(\frac{\sqrt5+1}2\right)^2-2=\frac{\sqrt5-1}2$$
$$\implies x+\frac1x-\left(x^2+\frac1{x^2}\right)=1\implies 1-x+x^2-x^3+x^4=0$$
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 $n=200,-200$
A: Just adding a different way of looking at it (fully endorsing Steve's excellent answer based on factoring the tenth cyclotomic polynomial over $\mathbb{Q}[\sqrt5]$).
Let $t=x+\dfrac1x$. Let us denote $t_n=x^n+\dfrac1{x^n}$. We calculate
$$
t^3=x^3+3x+\frac3x+\frac1{x^3}=t_3+3t,
$$
$$
t^5=x^5+5x^3+10x+10\frac1x+5\frac1{x^3}+\frac1{x^5}=t_5+5t_3+10t.
$$
From these we can easily solve
$$
t_3=t^3-3t,\qquad \text{and}\qquad t_5=t^5-5t_3-10t=t^5-5t^3+5t.
$$
This time $t=(1+\sqrt5)/2$, so $t^3=2+\sqrt5$ and $t^5=(11+5\sqrt5)/2$, and thus from the above formula we get
$$
t_5=\cdots=-2.
$$
From the equation (a quadratic in the unknown $x^5$)
$$
x^5+\frac{1}{x^5}=t_5=-2
$$
we get $x^5=-1$ as the only solution. The OP's result follows easily.
Anyway, what I wanted to add is that the so called Dickson polynomials give us (among other things) formulas for $t_n$ as a degree $n$ polynomial of $t$. They occasionally come in handy with problems like this.
A: 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 $x^{10}=1$.
A: Depending on what one wants to do with this, one can treat the process as an isoseries, based on $a(n+1) = k \cdot a(n) - a(n-1)$.  There are algorithms that can generate these numbers for any form of multiplication, including the modular form (ie $(a\cdot b ) mod c$.
The following iterations show the value for finding $a(37)$.  The first two columns are actual integers as shown.   37 is odd, which produces a '1'.  This means that we keep the values  in O0 and O1.  The value of 37 divided by 2 gives 18, an even number.  We keep the valuse in E0 and E1.
The values in brackets are the values of $a(n)$, shown as (n). The third column is powers of 2, by $a(2n) = a(n)^2 - 2$.  The values in columns 4 onwards show the powers, in steps of the powers.  We have, eg $a(5) = $a(2) \cdot a(3) - a(1)$.
$a(0)=2 = x^0 + x^{-0}$, and $a(1) = a$.
                  P2   E0     O0     E1    O1
       37   1    (1)   (0)   (1)    (2)    (3)
       18   0    (2)   (1)   (3)    (5)
        9   1    (4)   (1)   (5)    (9)   (13)
        4   0    (8)   (5)   (13)   (21)
        2   0    (16)  (5)   (21)   (37)
        1   1    (32)  (5)   (37)  

