$a$,$b$ and $c$ are roots of the equation $x^3-x^2-x-1=0$

Question

The roots of the equation $x^3-x^2-x-1=0$ are $a$,$b$ and $c$.

if $n \gt 21 $ and $n \in \mathbb{N}$ The find the possible values of $$E=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$$ in $[0 \: 2]$ are?

Since $x^3=x^2+x+1$ from the graphs of $x^3$ and $x^2+x+1$ its clear that they meet at one point. So number of real roots of $x^3-x^2-x-1=0$ is one which is positive and remaining two are complex conjugates. So let the roots be $a$ and $b=re^{i\theta}$ and $c=re^{-i\theta}$ But product of the roots is $$ar^2=1$$ hence $$r=\frac{1}{\sqrt{a}}$$

Now assuming $b=z$ and $c=\bar{z}$ then

$$E=\frac{a^n-z^n}{a-z}+\frac{z^n-\bar{z}^n}{z-\bar{z}}+\frac{\bar{z}^n-a^n}{\bar{z}-a}$$ so

$$E=2 \Re\left(\frac{a^n-z^n}{a-z}\right)+r^{n-1}\frac{\sin(n\theta)}{\sin\theta}$$

I am not able to proceed from here

Answer 1

HINT: If

$E_n=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$

then, because of $x^3=x^2+x+1$ you have $x^n=x^{n-1}+x^{n-2}+x^{n-3}$ and:

$E_n=E_{n-1} + E_{n-2} + E_{n-3}$.

It is easy to find out the first values of this recurrence.

Answer 2

Isn't this a trick question?

Let $$A_{n}=\frac{a^n-b^n}{a-b}$$

Then note $$a^{n+3}=a^{n+2}+a^{n+1}+a^{n}$$$$b^{n+3}=b^{n+2}+b^{n+1}+b^{n}$$ Thus subtract the two, and divide by $a-b$.

This gives us $$A_{n+3}=A_{n+2}+A_{n+1}+A_{n}$$

In a similar fashion, if $$B_{n}=\frac{b^n-c^n}{b-c}$$$$C_{n}=\frac{c^n-a^n}{c-a}$$

Then $$B_{n+3}=B_{n+2}+B_{n+1}+B_{n}$$$$C_{n+3}=C_{n+2}+C_{n+1}+C_{n}$$

Thus if $$E_n=\frac{a^n-b^n}{a-b}+\frac{b^n-c^n}{b-c}+\frac{c^n-a^n}{c-a}$$

Then $$E_{n+3}=E_{n+2}+E_{n+1}+E_{n}$$ since $E_{n}=A_{n}+B_{n}+C_{n}$.

Now, $E_0=0, E_1=3, E_2=2$ and use our above recurrence.