Find the value of an expression related to the roots of a quadratic polynomial

Question

Let $\alpha,\beta\:\:(\alpha>\beta)$ be roots of the quadratic equation $x^2-x-4=0$. If $$P_n=\alpha^{n}-\beta^{n}$$ and $n\in\mathbb{N}$ then find the value of $$\frac{P_{15}P_{16}-P_{14}P_{16}-P_{15}^2+P_{14}P_{15}}{P_{13}P_{14} }$$

I decided to use the Newton sum formulas but because of the negative sign in $-\beta$, I wasn't able to do so. I also multiplied the whole equation by $\alpha^8$ and $\beta^8$ one-by-one but in vain. Any help is greatly appreciated.

Answer 1

Hints.

1. $$ \frac{P_{15}P_{16}-P_{14}P_{16}-P_{15}^2+P_{14}P_{15}}{P_{13}P_{14}}= \frac{(P_{15}-P_{14})(P_{16}-P_{15})}{P_{13}P_{14}} $$

2. $$ P_{n+1}-P_n=4P_{n-1}. $$

Answer 2

Using the quadratic equation, $\alpha^n=a_n\alpha+ b_n$ and $\beta^n=a_n\beta+b_n,$ whith $a_{n+1}=a_n+b_n$ and $b_{n+1}=4a_n,$ hence $a_{n+2}=a_{n+1}+4a_n.$

$P_n=a_nP_1$, so $$\begin{align}\frac{P_{15}P_{16}-P_{14}P_{16}-P_{15}^2+P_{14}P_{15}}{P_{13}P_{14}}&=\frac{a_{15}a_{16}-a_{14}a_{16}-a_{15}^2+a_{14}a_{15}}{a_{13}a_{14}} \\&=\frac{a_{15}(a_{15}+4a_{14})-a_{14}(a_{15}+4a_{14})-a_{15}^2+a_{14}a_{15}}{a_{13}a_{14}} \\&=\frac{4(a_{15}-a_{14})}{a_{13}} \\&=16 \end{align} $$