= Solution
Put $\mathbf v_n=(F_{n+1},F_n)^T$. The <Fibonacci recurrence matrix modulo an integer> is $A=\begin{pmatrix}1&1\\1&0\end{pmatrix}$, so $\mathbf v_{n+r}=A^r\mathbf v_n$. Direct <matrix multiplication> gives
$$
A^3=\begin{pmatrix}3&2\\2&1\end{pmatrix}\equiv I\pmod2,\qquad A^8=\begin{pmatrix}34&21\\21&13\end{pmatrix}\equiv I\pmod3.
$$
Taking the second coordinate proves
$$
\boxed{F_{n+3}\equiv F_n\pmod2,\qquad F_{n+8}\equiv F_n\pmod3},\qquad n\geq0.
$$
For explicit arithmetic behind the second power, squaring $A^2=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ gives $A^4=\begin{pmatrix}5&3\\3&2\end{pmatrix}$, whose square is the displayed $A^8$. Thus the modular period assertions follow from calculations and the recurrence, not from quoting a period theorem.
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