Saudi Arabian Mathematical Competitions

Observe that
$$\begin{gathered} F_{2p} = \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^{2p} - \left( \frac{1-\sqrt{5}}{2} \right)^{2p} \right] \\ = \frac{1}{\sqrt{5}} \left[ \left( \frac{1+\sqrt{5}}{2} \right)^p - \left( \frac{1-\sqrt{5}}{2} \right)^p \right] \left[ \left( \frac{1+\sqrt{5}}{2} \right)^p + \left( \frac{1-\sqrt{5}}{2} \right)^p \right] \\ = F_p L_p \end{gathered}$$
where $L_n$ is the Lucas number, that is
$$L_n = \left( \frac{1+\sqrt{5}}{2} \right)^n + \left( \frac{1-\sqrt{5}}{2} \right)^n, \quad n=0,1, \ldots$$
it follows $F_{2p} - F_p = F_p L_p - F_p = F_p (L_p - 1)$. Now
$$\begin{aligned} & L_p - 1 = \left( \frac{1+\sqrt{5}}{2} \right)^p + \left( \frac{1-\sqrt{5}}{2} \right)^p - 1 \\ = & \frac{1}{2^{p-1}} \left[ \binom{p}{0} + \binom{p}{2} 5 + \ldots + \binom{p}{p-1} 5^{\frac{p-1}{2}} \right] - 1 \end{aligned}$$
$$= \frac{1}{2^{p-1}} \left[ \binom{p}{2} 5 + \binom{p}{4} 5^{2} + \ldots + \binom{p}{p-1} 5^{\frac{p-1}{2}} - \left(2^{p-1} - 1\right) \right],$$
hence $p \mid L_p - 1$, since $p$ divides $\binom{p}{2}, \binom{p}{4}, \ldots, \binom{p}{p-1}$, and $2^{p-1} - 1$, and $\gcd(p, 2) = 1$.