One of my favourite easy group theory problems:
Any prime number $p$ divides $f_{2p(p^2-1)}$, where $f_n$ is the Fibonacci sequence.
Proof: Let
$$G:= \{ A \in M_2 (Z_p) | \det(a)= \pm1 \}$$
and let
$$F=\left( \begin{array}{c c} 1 & 1 \\ 1 & 0 \\ \end{array}\right) $$$$F=\left( \begin{array}{c c} 1 & 1 \\\\ 1 & 0 \\\\ \end{array}\right) $$
Then $F \in G$ and $G$ is a group of order $2p(p^2-1)$.
Thus
$$\left( \begin{array}{c c} f_{n+1+2p(p^2-1)} & f_{n+2p(p^2-1)} \\ f_{n+2p(p^2-1)} & f_{n+2p(p^2-1)-1} \end{array}\right)$$$$\left( \begin{array}{c c} f_{n+1+2p(p^2-1)} & f_{n+2p(p^2-1)} \\\\ f_{n+2p(p^2-1)} & f_{n+2p(p^2-1)-1} \end{array}\right)$$
$$= F^{n+2p(p^2-1)}= F^n = \left( \begin{array}{c c} f_{n+1} & f_{n} \\ f_{n} & f_{n-1} \end{array}\right) \mod p \,.$$$$= F^{n+2p(p^2-1)}= F^n = \left( \begin{array}{c c} f_{n+1} & f_{n} \\\\ f_{n} & f_{n-1} \end{array}\right) \mod p \,.$$
P.S. Better periods can be obtained by solving the linear reccurence in $Z_p$ if $p =\pm1 \mod 5$ or in an algebraic extension if $p =\pm 2 \mod 5$, but that's exactly the same thing as calculating the order of the matrix $F$ by diagonalizing it.
The same idea can be used for any linear recurrence, but one has to replace $G$ by $GL_2(Z_p)$, and discuss the cases when $p$ divides or doesn't divide the free term of the polynomial associated to teh recurrence.
PPS: Can anyone please fix my matrices.
