Prove that $F^3_n \mid F_{mn} - F^m_{n + 1} + F^m_{n - 1} (m, n \in \mathbb N, m, n \ge 1)$. 
Let $F_n$ be the $n^{th}$ number of the Fibonacci sesquence $(n \in \mathbb N, n \ge 1)$. Prove that $$\large F^3_n \mid F_{mn} - F^m_{n + 1} + F^m_{n - 1}$$ $(m \in \mathbb N, m \ge 1)$

I have proven that $$\large F^2_n \mid F_{mn + 1} - F_{mn - 1} - F^m_{n + 1} + F^m_{n - 1}$$
But I am not sure if this contributes anything to the solution.
 A: First, we recall two well-known lemmas.
Lemma $1$: $F_{m+n}=F_m F_{n+1}+F_{m-1}F_n$.
Lemma $2$: $F_{m} \mid F_{mn}$.
Second, we prove another lemma.
Lemma $3$: $F_n^2 \mid F_{n-1}^m-F_{mn-1}$.
Proof: Use induction. Imagine the statement is true for $m$. We are going to prove that the relation holds for $m+1$ (Notice that obviously, $F_n^2 \mid F_{n-1}-F_{n-1}$). So, we will have:
$$A=F_{n-1}^{m+1}-F_{(m+1)n-1}=(F_{n-1}^{m})F_{n-1}-F_{(mn)+(n-1)}=(F_{n-1}^{m})F_{n-1}-F_{mn}F_n+F_{mn-1}F_{n-1}$$
One can easily put $F_{mn-1}$ instead of $F_{n-1}^m$ by the induction assumption. Therefore: $$A=F_{mn-1}F_{n-1}-F_{mn}F_n+F_{mn-1}F_{n-1}=F_{mn}F_{n}$$
Thus, the lemma $3$ is proved.
Now, we straight go to prove the main statement. Again, we use induction. Assume the statement is true for $m$. We are going to prove that the relation holds for $m+1$ (Notice that obviously, $F_n^3 \mid F_{n}-F_{n+1}+F_{n-1}$).
Again we re-write the terms as below:
$$B=F_{(m+1)n}-F_{n+1}^{m+1}+F_{n-1}^{m+1}=F_{mn+n}-F_{n+1}^{m+1}+F_{n-1}^{m+1}$$
$$=F_{mn}F_{n+1}+F_{mn-1}F_n-F_{n+1}^{m+1}+F_{n-1}^{m+1}$$
Again, we can put $F_{n+1}^{m}-F_{n-1}^{m}$ instead of $F_{mn}$ by the induction assumption. So, we will have: $$B=(F_{n+1}^{m}-F_{n-1}^{m})F_{n+1}+F_{mn-1}F_n-F_{n+1}^{m+1}+F_{n-1}^{m+1}=F_{mn-1}F_n+F_{n-1}^m(F_{n-1}-F_{n+1})$$ Thus, $$B=F_{mn-1}F_n+F_{n-1}^m(-F_{n})$$
To be done, we just need to prove that $F_n^2 \mid F_{mn-1}-F_{n-1}^m$, which is clear by the lemma $3$.
