Saudi Arabia Mathematical Competitions

Observe that $$\begin{array}{c}{F}_{2p}=\frac{1}{\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^{2p}-{\left(\frac{1-\sqrt{5}}{2}\right)}^{2p}\right]\\ =\frac{1}{\sqrt{5}}\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^p-{\left(\frac{1-\sqrt{5}}{2}\right)}^p\right]\left[{\left(\frac{1+\sqrt{5}}{2}\right)}^p+{\left(\frac{1-\sqrt{5}}{2}\right)}^p\right]\\ =F_p{L}_p\end{array}$$ where $L_n$ is the Lucas number, that is $$L_n={\left(\frac{1+\sqrt{5}}{2}\right)}^n+{\left(\frac{1-\sqrt{5}}{2}\right)}^n,n=0,1,\dots$$ it follows $F_{2p}-F_p=F_p{L}_p-F_p=F_p(L_p-1)$. Now $$\begin{array}{rl}& L_p-1={\left(\frac{1+\sqrt{5}}{2}\right)}^p+{\left(\frac{1-\sqrt{5}}{2}\right)}^p-1\\ =& \frac{1}{2^{p-1}}\left[\left(\genfrac{}{}{0ex}{}{p}{0}\right)+\left(\genfrac{}{}{0ex}{}{p}{2}\right)5+\dots +\left(\genfrac{}{}{0ex}{}{p}{p-1}\right)5^{\frac{p-1}{2}}\right]-1\end{array}$$ $$=\frac{1}{2^{p-1}}\left[\left(\genfrac{}{}{0ex}{}{p}{2}\right)5+\left(\genfrac{}{}{0ex}{}{p}{4}\right)5^2+\dots +\left(\genfrac{}{}{0ex}{}{p}{p-1}\right)5^{\frac{p-1}{2}}-\left(2^{p-1}-1\right)\right],$$ hence $p\mid L_p-1$, since $p$ divides $\left(\genfrac{}{}{0ex}{}{p}{2}\right),\left(\genfrac{}{}{0ex}{}{p}{4}\right),\dots ,\left(\genfrac{}{}{0ex}{}{p}{p-1}\right)$, and $2^{p-1}-1$, and $\gcd (p,2)=1$.