Let $\alpha, \beta,\gamma \in \mathbb{C}$ be the three roots of $x^3 + x+1$. For any $n\in\mathbb{N}$, let $a_n = \dfrac{(\alpha^n-1)(\beta^n-1)(\gamma^n-1)}{(\alpha-1)(\beta-1)(\gamma-1)}$. Prove that $a_n\ge 1$ for all $n\ge 1$ with equality iff $n=1,2,4,5$
Let $f(x)=x^3+x+1$ and let $S = \{\alpha,\beta,\gamma\}$. From Vieta's formulas, $\alpha + \beta + \gamma = 0, \alpha\beta + \alpha\gamma + \beta\gamma = 1, \alpha\beta\gamma = -1$. Also, by definition, $x^3+x+1 = (x-\alpha)(x-\beta)(x-\gamma)$. Clearly $a_1 = 1$.
$a_2 = -f(-1) = 1,$
$a_3 = \prod_{x\in S} (x^2 + x+1) = \prod_{x\in S} (x^2 -x^3) = \prod_{x\in S} x^2(1-x) = (\prod_{x\in S}x)^2 f(1) = 3.$
And $a_4 = \prod_{x\in S}(x^3+x^2+x+1) =(\prod_{x\in S}x)^2 = 1$.
Also, $a_5 = \prod_{x\in S}(x^4+x^2) = (\prod_{x\in S}x)^2 \cdot -f(-1) = 1$.
Now we just need to show that $a_n > 1$ for all $n\neq 1,2,4,5$, which is the part where I'm stuck. I think it might be possible to prove some sort of recurrence relation involving the $a_n$'s, but I'm not sure about the details.
