TITLE: How to prove that the Fibonacci sequence is periodic mod 5 without using induction?
QUESTION [42 upvotes]: The sequence $(F_{n})$ of Fibonacci numbers is defined by the recurrence relation
$$F_{n}=F_{n-1}+F_{n-2}$$ 
for all $n \geq 2$
with $F_{0} := 0$
and 
$F_{1} :=1$.
Without mathematical induction,
how can I show that
$$F_{n}\equiv F_{n+20}\pmod 5$$
for all $n \geq 2$?

REPLY [4 votes]: Since you asked for non-inductional ways, i'll use an expression attributable to François Édouard Lucas:
$$
f_n =\frac{\left(\phi \right)^n - \left(1- \phi \right)^{-n}}{\sqrt5}
$$
$$
We´re \space looking\space for \space a \space difference(D) \space such\space that \qquad{D \equiv0 \mod 5}\qquad \qquad
Note\space that \qquad{ D= f_{n+20}- f_n=\frac{\left(\phi \right)^{n+20} - \left(1- \phi \right)^{-(n+20)}}{\sqrt5}-\frac{\left(\phi \right)^n - \left(1- \phi \right)^{-n}}{\sqrt5}=\left( \frac{\left(\phi \right)^{10} - \left(1- \phi \right)^{-10}}{\sqrt5} \right) \left({\left(\phi \right)^{n+10} - \left(1- \phi \right)^{-(n+10)}}\right)=(f_{10})\left({\left(\phi \right)^{n+10} - \left(1- \phi \right)^{-(n+10)}}\right)}
$$
$$
but\qquad 5\mid (f_{10}=55) \Longrightarrow 5\mid D\Longrightarrow{f_n≡f_{n+20}(mod5)} 
$$