<s>[INST] How to prove that the Fibonacci sequence is periodic mod 5 without using induction? 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 $n \geq 2$?
 [/INST] $$\begin{bmatrix}
F_{n+1} & F_{n} \\
F_{n} & F_{n - 1} 
\end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$$
So
$$\begin{align}
\begin{bmatrix}
F_{n+21} & F_{n+20} \\
F_{n+20} & F_{n + 19} 
\end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{20 + n} &\pmod 5 \\
%
&= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{20}
   \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n} &\pmod 5 \\
%
&= \begin{bmatrix} 10946 & 6765 \\ 6765 & 4181 \end{bmatrix}
   \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n} &\pmod 5 \\
%
&= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}
   \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{n} &\pmod 5 \\
%
&= \begin{bmatrix}
F_{n+1} & F_{n} \\
F_{n} & F_{n - 1} 
\end{bmatrix} &\pmod 5 \\
\end{align}$$
Other cycles can be found similarly, for example
$$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{24} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \pmod 6$$
Giving a cycle of length $24$ modulo $6$.
</s>