IMO 2011: Prove that, for all integers $m$ and $n$ with $f(m)
Problem: Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m-n)$. Prove that, for all integers $m$ and $n$ with $f(m)<f(n)$, the number $f(n)$ is divisible by $f(m).$ (Resource: IMO $2011$)
My method:
$$\frac {f(m)-f(n)}{f(m-n)}\in\mathbb{Z}$$
If $f(m)=f(n)$ , $\frac{f(n)}{f(m)}=1\in \mathbb {Z^{+}}$
I can accept $f(n)>f(m)$.
It is obvious, $f(n)-f(m)≥f(m-n)$

$$ \begin{cases} m \mapsto m & \\ n \mapsto m-n& \end{cases} \Rightarrow \begin{cases} f(m) \mapsto f(m) & \\ f(n) \mapsto f(m-n) & \end{cases} $$
Now,  I will prove that $f(m-n)=f(m)$ must be. 
It is obvious $$\frac {f(m)-f(n)}{f(m-n)}\in\mathbb{Z} \Rightarrow \frac {f(m)-f(m-n)}{f(n)}\in\mathbb{Z}$$
If $f(m)≠f(m-n)$, we can write $\mid  f(m)-f(m-n) \mid ≥f(n)$. Considering $f(m)>0 , f(m-n)>0$ and $f(n)>f(m)$ we get  $f(m-n)>f(m)$ must be.
Case $1.$ 
$$f(m-n)-f(m)≥f(n) $$
Case $2.$
$$f(m)=f(m-n)$$
Let $n=0$, for Case $1$, we can write $f(n)≤f(m-n)-f(m) \Rightarrow f(0)≤0$  But, this is a contradiction. Because, $E(f)>0$. So, we get,  if $f(n)>f(m)$ then $f(m)=f(m-n)$ must be.
Finally,
$$\frac {f(m)-f(n)}{f(m-n)}\in\mathbb{Z} \Rightarrow \frac {f(m)-f(n)}{f(m)}\in\mathbb{Z} \Rightarrow \frac {f(n)}{f(m)} \in \mathbb{Z^{+}} $$ Q.E.D.

Can You verify my solution? Because, I'm not so sure. I don't have a teacher to approve the solution.
 A: Your proof looks correct to me.

Now,  I will prove that $f(m-n)≥f(m)$ must be.

I think that you have a typo here. It should be $f(m-n)=f(m)$.

It is obvious $$\frac {f(m)-f(n)}{f(m-n)}\in\mathbb{Z} \Rightarrow \frac {f(m)-f(m-n)}{f(n)}\in\mathbb{Z}$$

Yes, $f(m)-f(m-n)$ is divisible by $f(m-(m-n))=f(n)$.

If $f(m)≠f(m-n)$, we can write $\mid  f(m)-f(m-n) \mid ≥f(n)$. Considering $f(m)>0$ and $f(m-n)>0$, we get  $f(m-n)>f(m)$ must be. Because, $f(n)>f(m).$
Case $1.$
$$f(m-n)-f(m)≥f(n) $$
Case $2.$
$$f(m)=f(m-n)$$
Let $n=0$, for Case $1$, we can write $f(n)≤f(m-n)-f(m) \Rightarrow f(0)≤0$  But, this is a contradiction. Because, $E(f)>0$. So, case $1$ is impossible.

I think there is no need to separate it into cases as follows :
"Suppose that $f(m)\not =f(m-n)$. Then, we can write $\mid  f(m)-f(m-n) \mid \ge f(n)$. Considering $f(m)>0$ and $f(m-n)>0$, we get  $f(m-n)>f(m)$ because $f(n)>f(m).$ It follows that $f(m-n)-f(m)\ge f(n)$. Let $n=0$. Then, we can write $f(n)≤f(m-n)-f(m) \implies f(0)≤0$  which contradicts $f(0)\gt 0$. So, we have $f(m)=f(m-n)$."

Another way to prove $f(m-n)=f(m)$.
We have $$-f(n)\lt f(m)-f(n)\le -f(m-n)\lt 0$$ from which
$$0\lt f(m-n)\lt f(n)$$
follows.
From  $$f(m)-f(m-n)\lt f(m)+f(m-n)\le f(n)$$
and
$$f(m-n)\lt f(n)\lt f(n)+f(m)\implies -f(n)\lt f(m)-f(m-n)$$
we get
$$-f(n)\lt f(m)-f(m-n)\lt f(n)$$
Since $f(m)-f(m-n)$ is divisible by $f(m-(m-n))=f(n)$, we get $f(m)-f(m-n)=0$.
