<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] Thank you all for those who answered. Different amazing solutions!
But I'd still share my final solution, this is to be done as fast as possible so there's no proving needed and algebraic/arithmetic manipulation is probably the desired/suggested solution (not that mine is the optimal one, I just find this easier).
$x^2 + 1 = -3x \Rightarrow x+ \dfrac{1}{x} = -3$
$x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2 = 9 -2 = 7$
$x^4 + \dfrac{1}{x^4} = (x^2 + \dfrac{1}{x^2})^2 - 2 = 49 - 2 = 47$
seeing the pattern, let $s_n = x^{2^n} + \dfrac{1}{x^{2^n}}$
What is needed:  $s_{11}$
$s_1 = -3$
$s_2 = 7$
$s_3 = 47$
$s_4 = 47^2 - 2 = 2207$
$\vdots$
Or simply, this can be done under mod 3:
$((((((((((-3)^2 - 2)^2 - 2)^2 - 2)^2 - 2)^2 -2)^2 -2)^2 -2)^2 -2)^2 - 2)^2 -2 $
The innermost is just $(-3)^2 - 2 \equiv 1~$  mod 3,
and then it can even be done mentally continuing the process $(1)^2 - 2 \equiv -1~$ mod 3, squaring: $(-1)^2$ and getting the same, $(-1)^2 - 2 \equiv -1~$  mod 3 .
Recognizing the pattern it's $-1 \equiv 2$ mod 3 all the way. So it isn't divisible by 3.
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