I want to show that
- If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$
- If $3 \mid F_n$, then $9 \mid F_{n+1}^3-F_{n-1}^3$
where $F_n$ is the $n$-th Fibonacci number.
I have tried the following so far:
Since $F_1=F_2=1$, we suppose that $n \geq 3$.
$$\begin{align} F_{n+1}&=F_n+F_{n-1} \\ F_{n+1}^2&=(F_n+F_{n-1})^2=F_n^2+2F_n F_{n-1}+F_{n-1}^2 \end{align}$$
$$\begin{align} F_{n-1} &=F_{n-2}+F_{n-3} \\ F_{n-1}^2&=(F_{n-2}+F_{n-3})^2=F_{n-2}^2+2F_{n-2}F_{n-3}+F_{n-3}^2 \end{align}$$ so that $$ F_{n+1}^2-F_{n-1}^2=F_n^2+2F_n F_{n-1}+F_{n-1}^2-F_{n-2}^2-2 F_{n-2} F_{n-3}-F_{n-3}^2 $$
How can we deduce that the latter is divisible by $4$?
Or do we show it somehow else, for example by induction?