Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 4 / 7C / iii / Solution

The same recurrence matrix hasTherefore ${\mathbf{v}}_{n+5}\equiv 3{\mathbf{v}}_n(\mathrm{mod}5)$, and in particular $F_{n+5}\equiv 3F_n(\mathrm{mod}5)$. Starting from $F_0=0$ and iterating,Consequently every nonnegative index divisible by five has a Fibonacci number divisible by five, including index zero. This directly proves the requested divisibility without assuming a general Fibonacci divisibility theorem.