<s>[INST] 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
 [/INST] Let $a_n = x^n + \dfrac{1}{x^n}\implies a_1 = 3, a_2 = 7, a_{n+1}= x^{n+1}+\dfrac{1}{x^{n+1}}= \left(x^n+\dfrac{1}{x^n}\right)\left(x+\dfrac{1}{x}\right) - \left(x^{n-1}+\dfrac{1}{x^{n-1}}\right)= 3a_n-a_{n-1}\implies a_{n+1} - 3a_n + a_{n-1} = 0$. From here you can use characteristic equation for difference equation and find the general term $a_n$ and from it you can find a closed form for $a_{2048}$.
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