1. Given quadratic and its roots

The roots $\alpha,\beta$ of $x^2-x-4=0$ satisfy $\alpha+\beta=1, \qquad \alpha\beta=-4.$

Also, $P_n=\alpha^n-\beta^n.$

We need to evaluate $\frac{P_{15}P_{16}-P_{14}P_{16}-P_{15}^2+P_{14}P_{15}}{P_{13}P_{14}}.$

2. Factor the numerator

Group terms: \begin{align*} &P_{15}P_{16}-P_{14}P_{16}-P_{15}^2+P_{14}P_{15} \ &= P_{16}(P_{15}-P_{14})-P_{15}(P_{15}-P_{14}) \ &= (P_{15}-P_{14})(P_{16}-P_{15}). \end{align*}

So the expression becomes $\frac{(P_{15}-P_{14})(P_{16}-P_{15})}{P_{13}P_{14}}.$

3. Find a recurrence for $P_n$

Since $\alpha,\beta$ satisfy $r^2=r+4,$ we have for any $n$, $\alpha^{n+2}=\alpha^{n+1}+4\alpha^n, \qquad \beta^{n+2}=\beta^{n+1}+4\beta^n.$ Subtracting, $P_{n+2}=P_{n+1}+4P_n.$

Now define $Q_n=P_n-P_{n-1}.$ Then \begin{align*} Q_n &= P_n-P_{n-1} \ &= (P_{n-1}+4P_{n-2})-P_{n-1} \ &= 4P_{n-2}. \end{align*}

Hence, $P_{15}-P_{14}=4P_{13}, \qquad P_{16}-P_{15}=4P_{14}.$

4. Substitute into the expression

Therefore, \begin{align*} \frac{(P_{15}-P_{14})(P_{16}-P_{15})}{P_{13}P_{14}} &= \frac{(4P_{13})(4P_{14})}{P_{13}P_{14}} \ &= 16. \end{align*}

5. Final answer

$\boxed{16}$

The derived answer matches the stored correct answer.