How to find $x^{2000}+x^{-2000}$ when $x + x^{-1} = \frac{1}{2}(1 + \sqrt{5})$

Solution 1:

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 $x^{10}=1$.

Solution 2:

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.