Let , () be the roots of the quadratic equation x^2-x-4=0.

Question

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

Answer: 16

Accepted Answer

Since $\alpha,\beta$ satisfy $x^2=x+4$, we get $P_{n+1}=P_n+4P_{n-1}$ (Newton-type recurrence).

So $P_{15}-P_{14}=4P_{13}$ and $P_{16}-P_{15}=4P_{14}$.

$$\frac{P_{15}P_{16}-P_{14}P_{16}-P_{15}^2+P_{14}P_{15}}{P_{13}P_{14}} = \frac{P_{16}(P_{15}-P_{14})-P_{15}(P_{15}-P_{14})}{P_{13}P_{14}} = \frac{(P_{15}-P_{14})(P_{16}-P_{15})}{P_{13}P_{14}}$$
$$=\frac{(4P_{13})(4P_{14})}{P_{13}P_{14}}=16.$$

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