If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
 A: 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
$$x^{2000}+\frac{1}{x^{2000}} = ({x^{5}})^{400}+\frac{1}{(x^{5})^{400}} = (-1)^{400}++\frac{1}{(-1)^{400}} = 1+1 = 2$$
A: Here's another approach for the record.


*

*The equation $x+ \frac{1}{x}=\alpha$  where $\alpha \in [-2,2]$ can be solved as follows. Identify $\alpha$ is $2\cos (\theta)$, and observe that letting $z=e^{i\theta}$,from the definition of $\cos (\theta)$, we have 
$$z+ \frac{1}{z}=2\cos(\theta)=\alpha.$$

*Also, from the definition of $\cos$,  
$$z^k + \frac{1}{z^k} = 2\cos (k \theta).$$  


*Now back to our question. Let $\alpha = \frac{1+\sqrt{5}}{2}$. Then $\cos(\theta) = \frac{1+\sqrt{5}}{4}$. This gives us  $\theta = \pm \frac{\pi}{5} \mod 2\pi$. Therefore the answer to the problem is $2\cos (400\pi)=2$.  

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

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
A: 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 $\phi^2=\phi+1$
A: Let $\psi = \frac{1+\sqrt{5}}{2} $
$$x + \frac{1}{x} = \psi$$
$$x^2 + 1 = \psi x $$
$$x^2 - \psi x + 1 = 0$$
Use quadratic formula here to solve for $x$. Then plug that into the given expression.
A: Nice try of yours. You can further continue...:
$$
x^{2000} + \frac{1}{x^{2000}} = \left(x^{1000} + \frac{1}{x^{1000}}\right)^2 - 2 \\
x^{1000} + \frac{1}{x^{1000}} = \left(x^{500} + \frac{1}{x^{500}}\right)^2 - 2 \\
x^{500} + \frac{1}{x^{500}} = \left(x^{250} + \frac{1}{x^{250}}\right)^2 - 2 \\
\vdots \\
x^{n} + \frac{1}{x^n} = \left(x^{n/2} + \frac{1}{x^{n/2}}\right)^2 - 2 \\
$$
In the case of $2000 = 2^4 \cdot 5^3$. That is, when you are done with all the $2$s, you will be left with $5$:
$$
x^{125} + \frac{1}{x^{125}} = 
\left[\left(x^{25}\right)^5 + \left(\frac{1}{x^{25}}\right)^5\right] = 
$$
We do basically the same you did:
$$
(a+b)^5 = {{5}\choose{0}}a^5 + {{5}\choose{1}}a^4 b + \cdots \\
(a+a^{-1})^5 = {{5}\choose{0}}a^5 + {{5}\choose{5}}b^{-5} + \cdots \\
$$
Notice the intermediate terms:
$$
{{5}\choose{1}}a^{4} a^{-1} + 
{{5}\choose{2}}a^{3} a^{-2} + 
{{5}\choose{3}}a^{2} a^{-3} + 
{{5}\choose{4}}a^{1} a^{-4}
= 
{{5}\choose{1}}a^{3} + 
{{5}\choose{2}}a + 
{{5}\choose{3}}a^{-1} + 
{{5}\choose{4}}a^{-3}
$$
That is:
$$
\left[x^{5} + \frac{1}{x^{5}}\right]^5 = 
\left[x^{5} + \frac{1}{x^{5}}\right] + 
10\left[\left(x^{5}\right)^3 + \left(\frac{1}{x^{5}}\right)^3\right] + 
5\left[\left(x^{5}\right)^5 + \left(\frac{1}{x^{5}}\right)^5\right]
$$

You can keep doing so and simplifying to smaller and smaller terms. You can do this because:
$$
{{n}\choose{k}} = {{n}\choose{n-k}} = \frac{n!}{k!(n-k)!}
$$
That is, $(a + a^{-1})^n$ will have the terms $a^{-k} a^{n-k}$ and $a^k a^{-(n-k)}$ with the same coeficients. And this term forms: 
$$
a^k a^{-n+k} = a^{k-n+k} = a^{2k-n} \\
a^{-k} a^{n-k} = a^{-k+n-k} = a^{-2k+n} = a^{-(2k-n)}
$$
So, the coeficients opposite terms will be equal, then you will always be able to build terms like $x^k + 1/x^k$. Thus you can keep doing it until you are left only with the $x + 1/x$. It might take a while..... =D.
A: Just giving another way.
$$x+\frac 1x=\frac{1+\sqrt2}{2}\\ x^2+\frac{1}{x^2}=\frac{-1+\sqrt5}{2}\\ x^4+\frac{1}{x^4}=-\frac{1+\sqrt5}{2}\\ x^8+\frac{1}{x^8}=\frac{-1+\sqrt5}{2}$$ Hence $$\left(x^8+\frac{1}{x^8}\right)\left(x^2+\frac{1}{x^2}\right)=\left(\frac{-1+\sqrt5}{2}\right)^2\\ x^{10}+\frac{1}{x^{10}}=\left(\frac{-1+\sqrt5}{2}\right)^2-\left(-\frac{1+\sqrt5}{2}\right)=2\iff(x^{10}-1)^2=0$$
Thus $x^{10}=1$ from which the result  $2$.
