Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for all positive integers $m$ and $n$ the number $f(m)+n-m$ is divisible by $f(n)$.
Now I wanted to verify my solution, which I feel is flawed somewhere, but I can't find the flaw. Also please refrain from providing any solution to this as I want to figure out myself. Please just point out where my method is flawed.
My Approach:
$f:\mathbb N \rightarrow \mathbb N$ such that $f(n)|f(m)+n-m$
$\Rightarrow f(n)\leq f(m)+n-m$ or $m-n\leq f(m)-f(n)$
So now when $m\geq n$ we have $f(m)\geq f(n)$
Thus $f$ is an increasing function
Now the inequality can be restated as $f(n)-f(m)\leq n-m$
Putting $n=m+1$ gives $f(m+1)-f(m)\leq 1$
Since $f:\mathbb N \rightarrow \mathbb N$, two cases are possible:
Case I: $f(m+1)=f(m)$. This means that $f(m)=k\forall\ m\in \mathbb N$, where $k$ is a constant
Case II: $f(m+1)=f(m)+1=f(m-1)+2=\cdots=f(1)+m$. This means that $f(m)=m+k-1\forall\ m\in \mathbb N$ where $k=f(1)$
Thus only two functions are possible.
Please help in finding any flaw in it, if present.
THANKS