Find all functions $f:\Bbb N\rightarrow \Bbb R $ such that for a given value $n\in \Bbb N$ , the following identity holds: $$f(m+k)=f(mk-n) ,m,k \in \Bbb N , mk>n$$
This problem has already been asked here . I tried to get more deep solutions using my reputation but failed,so I repeat it here.
Firstly we show that the function must be periodic with period $n+1$. We do this by picking appropriate values of $m$ and $k$.
For any $l \in \mathbb{N}$ we can set $m = n + l$ and $k = 1$, and observe from the relation that $$f(n + l + 1) = f(m + k) = f(mk - n) = f(l)$$ which indeed shows that $f$ must be periodic.
The second observation is that the function $f$ must be constant.
We see this by setting $m = l$ and $k = n+1$, so that $$f(l + n + 1) = f(m + k) = f(mk- n) = f\big(l(n+1) - n\big)$$ But since $f$ is ($n+1$)-periodic, we have $$f(l + n + 1) = f(l)$$ and $$f\big(l(n+1) - n\big) = f\big(1\cdot(n+1) - n\big) = f(1)$$
So putting it together we find that for every $l \in \mathbb{N}$, $$f(l) = f(1)$$.
Thus $f$ must be constant. Conversely, it's immediately clear that any constant function $f$ does satisfy the property in question.