find functions that $f(m+k)=f(mk-n)$

Question

Find all functions from naturals excluding $0$ to reals that for a particular amount of $n$ we have:

$f(m+k)=f(mk-n):(mk>n , m,k,n \in \Bbb Z^+)$

by putting $k=1$ and $m>n$ we can see that the function is periodic with cycle of $n+1$.But I can't continue.

Answer 1

$k=n+1$, using the (n+1)-periodicity then for all $m$, \begin{align} f(m) &= f(m+n+1)\\ & =f((m-1)(n+1) + 1) = f(1) \end{align}