If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $?

Note that $a = \frac{1+\sqrt{5}}{2}$ satisfies the equation $a^2 - a - 1 = 0$

Substituting $a=x+\frac{1}{x}$ gives:

$$0 = \left(x+\frac{1}{x}\right)^2 - \left(x+\frac{1}{x}\right) - 1 = x^2 - x + 1 - \frac{1}{x} + \frac{1}{x^2}$$

$$\iff \quad x^4 - x^3 + x^2 - x + 1 = 0$$

Multiplying by $x+1$ results in $x^5+1=0$, so $x$ is a complex $5^{th}$ root of $-1$ therefore $x^5 = -1$.

Then $x^{2000}+\frac{1}{x^{2000}} = \big(x^{5}\big)^{400} + \cfrac{1}{\big(x^{5}\big)^{400}} = (-1)^{400} + \cfrac{1}{(-1)^{400}} = 1 + 1 = 2$.

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P.S. For a heavy-handed "solution" to the tune of "how to crack a nut with a sledgehammer", let Wolfram Alpha do all the work:

resultant[resultant[x^2 - a x + 1, a^2 - a - 1, a ], x^4000 - b x^2000 + 1, x] = (b-2)^4

If $a_n=x^n+\frac{1}{x^n}$ then $a_n=a_1\cdot a_{n-1}+a_{n-2}=\phi\cdot a_{n-1}-a_{n-2}$ which is a second-order linear recurrence, where $\phi=\frac{1+\sqrt 5}{2}$. The initial conditions are $a_1=\phi$ and $a_2=a_1^2-2=\phi^2-2=\phi-1$, since $\phi^2=\phi+1$

Edit: Found another solution, removed old answer (it was incorrect anyway)

You have $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$. By simple algebraic manipulation you can get the $$x^4-x^3+x^2-x+1 = 0$$ Now notice that $x^4 = x^3-x^2+x-1$ and multiplying both sides by $x$ you get $x^5 = x^4-x^3+x^2-x=-1$.

Therefore $$x^{2000}+\frac{1}{x^{2000}} = ({x^{5}})^{400}+\frac{1}{(x^{5})^{400}} = (-1)^{400}++\frac{1}{(-1)^{400}} = 1+1 = 2$$