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?

We have more general answer: 

If $k\mid a-b$ then $k^2\mid a^k-b^k$

Proof: Write $a-b=km$ for some integer $m$. Then we have
$$a^k-b^k=(a-b)(\underbrace{a^{k-1}+a^{k-2}b+a^{k-3}b^2+...+b^{k-1}}_E)$$
\begin{eqnarray}E&\equiv_k& b^{k-1}+b^{k-2}b+b^{k-3}b^2+...+b^{k-1}\\
&\equiv_k&k\cdot b^{k-1} \\
&\equiv_k&0 \\
\end{eqnarray}
so $E = k\cdot p$ for some interger $p$, so we have $k^2\mid a^k-b^k$
