Find $\alpha^{2016} + \beta^{2016} + \alpha^{2014} + \beta^{2014} \over \alpha^{2015} + \beta^{2015}$ for zeroes of the polynomial $x^2+3x +1$

Question

We have a polynomial $x^2 + 3x + 1$. There are 2 zeroes of it,

$\alpha$ and $\beta$

Now, what we need to find out is(What I couldn't) is as follows:

$$ \alpha^{2016} + \beta^{2016} + \alpha^{2014} + \beta^{2014} \over \alpha^{2015} + \beta^{2015}$$

Any ideas on how to do it?

Thanks a lot! -bone

Answer 1

If $x^2+3x+1=0$ has a root $\alpha,$ then $\alpha^2+1=-3\alpha$

In the numerator, $\displaystyle \alpha^{2016}+\alpha^{2014}=\alpha^{2014}(\alpha^2+1)=\alpha^{2014}(-3\alpha)=-3\alpha^{2015}$

Similarly, $\beta$ is another root of $x^2+3x+1=0, $

$\displaystyle\beta^{2016}+\beta^{2014}=\cdots=-3\beta^{2015}$

So, the numerator becomes $\cdots$

Answer 2

By general theory on recurrence equations, the sequence $(\alpha^n+\beta^n)$ is the unique solution of the recurrence equation $$ a_{n+2}=-3a_{n+1}-a_n,\qquad a_0=2,\quad a_1=-3 $$ In particular, $$ a_{n+2}+a_{n}=-3a_{n+1} $$ so $$ \frac{a_{n+2}+a_{n}}{a_{n+1}}=-3 $$