I want to show that for each $n \geq 1$ it holds that:
$$2^{n-1} F_n \equiv n \pmod{5}$$
where $F_n$ is the $n$-th Fibonacci number.
Could you give a hint how we can show this?
The sequence $F_n$ of Fibonacci numbers is defined by the recurrence relation:
$$F_n=F_{n-1}+F_{n-2} \\ F_1=1, F_2=1.$$
Right? Do we use the definion in order to get to the desired result?