Let $x^2 + 3x +1 = 0$. Is $x^{2048} + \dfrac{1}{x^{2048}}$ divisible by 3? Let $x^2  + 3x +1 = 0$. Solve for $x^{2048} + \dfrac{1}{x^{2048}}$. Is it divisible by 3?
$x^2 + 1 = -3x \Rightarrow x+ \dfrac{1}{x} = -3$
$x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2 = 9 -2 = 7$
$x^4 + \dfrac{1}{x^4} = (x^2 + \dfrac{1}{x^2})^2 - 2 = 49 - 2 = 47$
seeing the pattern, let $s_n = x^{2^n} + \dfrac{1}{x^{2^n}}$
I need $s_{11}$
$s_1 = -3$
$s_2 = 7$
$s_3 = 47$
$s_4 = 47^2 - 2 = 2207$
$\vdots$
$((2207^2 -2)^2 - 2)^2 -2)^2 ... - 2)$
quite big
But I realized I didn't have to simplify it, I just have to check if
$((((((2207^2 - 2)^2 -2)^2 -2)^2 -2)^2 -2)^2 - 2)^2 -2$ is divisible by 3

To get a "closed form" solution, you can do the following:

*

*Simply evaluate roots, and write the answer as $\big(\frac{-3+\sqrt{5}}{2}\big)^{2048}+\big(\frac{-3-\sqrt{5}}{2}\big)^{2048}$


*Or, you could solve the recurrence relation as follows:
$s_n$ is defined in the question. $s_{n+1}=(s_n)^2-2$, where $s_0=-3$.
A solution is: $$s_n=2\cos(\cos^{-1}(-1.5)\cdot2^n)$$
Note the use of the identity $\cos(2\theta)=2\cos^2(\theta)-1$.
To compute $\cos^{-1}(-1.5)$, let $\cos(x+iy)=-1.5$, and proceed ($i$ is the imaginary unit). This gives a closed form.
