<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 $x$ be a root of $x^2 + 3x + 1$.  Then $y = 1/x$ is also a root, since $$y^2 + 3y + 1 = (1/x)^2 + 3/x + 1 = \frac{1 + 3x + x^2}{x^2}$$ and $x \ne 0$.  Moreover, for each positive integer $n$,
$$x^{n+1} + y^{n+1} = (x+y)(x^n + y^n) - xy(x^{n-1} + y^{n-1}).$$
Since $x + y = -3$ and $xy = 1$, it follows that
$$x^{n+1} + y^{n+1} = -3(x^n + y^n) - (x^{n-1} + y^{n-1}).$$  Taken modulo $3$, we see that if $f_n = x^n + y^n$,
$$f_{n+1} \equiv -f_{n-1} \pmod 3.$$
So for instance, with $f_0 = x^0 + y^0 = 2$, we have
$$f_2 \equiv -f_0 \pmod 3, \\ f_4 \equiv -f_2 \equiv f_0 \pmod 3, \\ f_6 \equiv -f_4 \equiv -f_0 \pmod 3, \\ \vdots \\f_{2048} \equiv (-1)^{2048/2} f_0 = 2 \pmod 3.$$
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