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?

$$F_{n+1}^2-F_{n-1}^2=(F_{n+1}-F_{n-1})(F_{n+1}+F_{n-1})
=F_n(F_{n+1}+F_{n-1}).$$
Also $F_{n+1}=F_n+F_{n-1}\equiv F_{n-1}\pmod{F_n}$, so
$F_{n+1}+F_{n-1}\equiv 2F_{n-1}\pmod{F_n}$. If $F_n$ is even,
then so is $F_{n+1}+F_{n-1}$.
Similarly
$$F_{n+1}^3-F_{n-1}^3=F_n(F_{n+1}^2+F_{n+1}F_{n-1}+F_{n-1}^2)$$
and
$$F_{n+1}^2+F_{n+1}F_{n-1}+F_{n-1}^2\equiv 3F_{n-1}^2\pmod{F_n}$$
etc.
