If $2 \mid F_n$, then $4 \mid F_{n+1}^2-F_{n-1}^2$, where $F_n$ is $n$-th Fibonacci number 
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?

Modulo two the Fibonacci sequence cyclically repeats the pattern of length three $0,1,1,0,1,1,0,1,1,\ldots$. So if $F_n$ is even then both $F_{n-1}$ and $F_{n+1}$ are odd. Therefore both $F_{n-1}^2$ and $F_{n+1}^2$ are congruent to $1\pmod 4$ (mod eight actually!). Therefore their difference is a multiple of eight.
Modulo three the cyclically repeating pattern has length eight: $$0,1,1,2,0,2,2,1,0,1,1,2,0,2,2,1,\ldots.$$ 
Again, we see that if $F_n$ is divisible by three, then $F_{n-1}$ and $F_{n+1}$ are either both $\equiv1\pmod 3$ or both $\equiv-1\pmod3$. 
In the former case their cubes are both $\equiv 1\pmod 9$ (check that $1^3\equiv4^3\equiv7^3\pmod9$. In the latter case both cubes are $\equiv -1\pmod9$ ($2^3\equiv5^3\equiv8^3$). Both these facts actually also follow from the fact that the group $\Bbb{Z}_9^*$ is cyclic of order six.
