# Prove that $a_n\ge 1$ for all $n\ge 1$ with equality iff $n=1,2,4,5$

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.

The question is

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 $$n=1,2,4,5$$

First, I will use $$\,a,b,c\,$$ for the three roots of $$x^3 + x+1$$ where

$$a\approx -0.682,\;\;b\approx 0.341 - 1.161i,\;\;c\approx 0.341 + 1.161i. \tag1$$

Use Vieta's formulas to get

$$abc= -1, \quad ab+ac+bc = 1, \quad a+b+c = 0. \tag2$$

This implies that the denominator of $$\,a_n\,$$ is

$$(a-1)(b-1)(c-1) = (-1)-(1)+(0)-1 = -3. \tag3$$

The numerator expands to

$$(abc)^n -(ab)^n - (ac)^n - (bc)^n + a^n + b^n + c^n - 1. \tag4$$

Notice that

$$0<-a <|ab| = |ac|<1 = -abc < |b| = |c| < bc \approx 1.465. \tag5$$

Define the sequence

$$b_n := \frac{(bc)^n-b^n-c^n}3, \tag6$$

and thus

$$a_n - b_n = \frac{1 - (-1)^n + (ab)^n + (ac)^n - a^n}3. \tag7$$

This implies that the absolute error is bounded by $$\, |a_n - b_n| < 2 \;\;\text{ if }\;\; n>0.\,$$ But, now $$\,b_n > 3\,$$ if $$\,n>5\,$$ which implies that $$\,a_n > 1\,$$ if $$\,n>5.\,$$

Notice that $$\,\lim_{n\to\infty} a_{2n} - b_{2n} = 0\,$$ and $$\,\lim_{n\to\infty} a_{2n+1} - b_{2n+1} = 2/3.\,$$