Find all functions $f(m)[(f(n))^2-1]=f(n)[f(m+n)-f(m-n)]$ Find all functions $f : \mathbb{N} \to \mathbb{N}$  satisfying
$$f(m)[(f(n))^2-1]=f(n)[f(m+n)-f(m-n)]$$ $\forall m>n$.

Attempted work :
Let $P(m,n)$ denote $f(m)[(f(n))^2-1]=f(n)[f(m+n)-f(m-n)], \forall m>n$.
Case 1 : $f(n)=1, \forall n \in \mathbb{N}$.
Case 2 : $\exists k \in \mathbb{N}, f(k)>1$
$P(m,k) : f(m)[(f(k))^2-1]=f(k)[f(m+k)-f(m-k)]$ 
Since $ \text{gcd} (f(k), f(k)^2-1) = 1$ so $f(k) \mid f(m), \forall m \in \mathbb{N}, m>k$
I don't know how to proceed.
 A: When $f(n)=a^n$ then $f$ verifies the functional equation.
$f(n)(f(m+n)-f(m-n))=a^n(a^{m+n}-a^{m-n})=a^m(a^{2n}-1)=f(m)(f(n)^2-1)$.

So let's set $f(1)=a\in\mathbb N^*$ and see if we can bring something by induction on $f(n)=a^n$.
$f(1)(f(m+1)-f(m-1))=f(m)(f(1)^2-1)\\\implies a(f(m+1)-a^{m-1})=a^m(a^2-1)\\\implies af(m+1)=a^{m+2}-a^m+a^m=a^{m+2}\\\implies
f(m+1)=a^{m+1}$
Unfortunately for the induction to work we need both $f(m)$ and $f(m-1)$ to be verified.
Thus we need $f(1)=a$ and $f(2)=a^2$ at initial step.
But we know nothing about $f(2)$

From now on, I'm using a CAS, because it is tedious to calculate everything


*

*$F_1=f(1)=a$

*$F_n=F_{n-2}+F_{n-1}(a^2-1)/a$
The values $F_n=f(n)$ obtained through formula $$f(1)(f(m+1)-f(m-1))=f(m)(f(1)^2-1)$$


*

*$B_n=F_{n-2}(F_2^2-1)/F_2+F_{n-4}$


The values $B_n=f(n)$ obtained through formula $$f(2)(f(m+2)-f(m-2))=f(m)(f(2)^2-1)$$

Let's set $F_2=b$
$F_3=a+ab-\frac ba$
Since $F_3$ integer then $a\mid b$

Let's set $F_2=ab$
$F_3=a+a^2b-b$
$F_4=-ab+a^2-1+a^3b+\frac ba$
Since $F_4$ integer then $a\mid b$

Let's set $F_2=a^2b$
$F_3=a+a^3b-ab$
$F_4=-a^2b+a^2-1+a^4b+b$
$F_5=-a-a^3b+ab+a^3+a^5b+\frac 1a-\frac ba$
Since $F_5$ integer then $a\mid (b-1)$

Let's set $F_2=a^2(1+ka)\quad$ with $k\ge 0$ 
patience, we are getting to it...
$F_3=a^3+ka^4-ka^2$
$F_4=-ka^3+a^4+a^5k+ka$
$F_5=-ka^4+ka^2+a^5+ka^6-k$
$F_6=ka^3-a^5k-ka+a^6+a^7k+\frac ka$
$B_5=\dfrac{(a^5+3ka^6+3a^7k^2+a^8k^3-ka^4-2a^5k^2-a^6k^3+k)}{(1+ka)}$
$B_6=\dfrac{-ka^6-2a^7k^2-a^8k^3+ka^2+a^7+3a^8k+3k^2a^9+a^{10}k^3+3a^5k^2+a^6k^3-k+2ka^4}{a(1+ka)}$
Now by definition $F_5=B_5$ and $F_6=B_6$
Unfortunately, the expressions are complicated, but I found that 
$(F_5-B_5)+a(F_6-B_6)=\dfrac{-ka^2\overbrace{(ka^3+a^2+1)}^{>0}\overbrace{(a^5k+a^4+1)}^{>0}}{1+ka}=0$
Thus $k=0$ is forced.
Consequently, $f(2)=a^2$ and the induction base hypothesis is verified.

Finally $f(n)=a^n$ for $a\in\mathbb N^*$ are the unique solutions of the equation.

