The 2000th Power

Algebra Level 3

Suppose that $x + \frac{1}{x} = \frac{1 + \sqrt{5}}{2}.$ Compute $x^{2000} + x^{-2000}$ .

The answer is 2.

2 solutions

Alan Yan
Feb 20, 2018

Let $\theta = 36^{\circ}$ . Observe that $\cos \theta = \frac{1 + \sqrt{5}}{4} = \frac{1}{2} \left ( x + \frac{1}{x} \right ) \implies x = e^{\theta i}.$ So, $x^{2000} + x^{-2000} = 2 \cos (2000\theta) = \boxed{2}$ .

@Alan Yan Could you show how you derived $\theta=36^{\circ}$ ?

John Frank - 3 years, 3 months ago

Here's a quick geometric proof:

Consider a regular pentagon $ABCDE$ inscribed in a circle. Let $AB = 1$ and $AC = d.$ From triangle $ABC,$ we can quickly derive $d = 2 \cos 36^{\circ}.$ Now, apply Ptolemy's Theorem to quadrilateral $ABCE$ :

$\begin{aligned} AB(CE) + AD(BE) &= AC(BE) \\ d + 1 &= d^2 \\ d^2 - d - 1 &= 0 \\ d &= \dfrac{1 + \sqrt{5}}{2} \\ 2 \cos 36^{\circ} &= \dfrac{1 + \sqrt{5}}{2} \\ \cos 36^{\circ} &= \dfrac{1 + \sqrt{5}}{4} \quad \blacksquare \end{aligned}$

Steven Yuan - 3 years, 3 months ago

E Koh
Nov 7, 2019

x + (1/x) = 2 cos (2pi/5) x^2000 + (1/x^2000) = 2 cos (2000pi/5) = 2

The 2000th Power

The answer is 2.

2 solutions

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