 Solution

Put ${\mathbf{v}}_n=(F_{n+1},F_n{)}^T$. The Fibonacci recurrence matrix modulo an integer is $A=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$, so ${\mathbf{v}}_{n+r}=A^r{\mathbf{v}}_n$. Direct matrix multiplication givesTaking the second coordinate provesFor explicit arithmetic behind the second power, squaring $A^2=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$ gives $A^4=\left(\begin{array}{cc}5& 3\\ 3& 2\end{array}\right)$, whose square is the displayed $A^8$. Thus the modular period assertions follow from calculations and the recurrence, not from quoting a period theorem.