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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.

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1 Answer 1

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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, denote the three roots of $x^3 + x+1$ by $\,a,b,c\,$ where

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

Use Vieta's formulas to get

$$ e_3 := abc= -1, \quad e_2 = ab+ac+bc = 1, \quad e_1 := a+b+c = 0. \tag2 $$

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

$$ (a-1)(b-1)(c-1) = e_3-e_2+e_1-1 = -3. \tag3 $$

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

Your guess

I think it might be possible to prove some sort of recurrence relation involving the an's

is correct, but I don't think that approach is needed to prove the desired result.

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