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 $$f(m)$$

Question

anonymous · Answer

$$\frac{f(m) - f(n)}{f(m-n)}=k$$Setting n = 0, $$f(m) | f(0)$$Setting m = 0, $$f(-n)| f(n)$$Setting m = 0, n = -n, $$f(n)| f(-n)$$$$f(n) = f(-n)$$Setting m = 2n, $$f(n) |f(2n)$$$$f(2n) \ge f(n)$$Setting n as -n $$f(m+n)|(f(m) - f(n))$$$$f(m) - f(n) \ge f(m+n)$$ Assume $$f(m+n) > f(m)$$, then$$f(m+n) > f(m) - f(n)$$However, this contradicts to $$f(m) - f(n) \ge f(m+n)$$, which means $$f(m+n) $$ is either smaller or equal to f(m). Assume $$f(m+n) < f(m)$$then setting m = n, $$f(2n) < f(n)$$, which contradicts to $$f(2n) \ge f(n)$$, Thus $$f(m+n) = f(m)$$ Substitute this back to the original equation$$\frac{f(m+n)-f(n)}{f(m)}$$Thus, we proved $$f(m) | f(n)$$

turingtest · Answer

Wow moneybird this looks really great!
I'll have to give it special attention on a non-holiday though, I don't think I'm smart enough to really evaluate it. Certainly not right now at least. I'll be sure to get others to look at this, this was an IMO problem, yeah?

anonymous · Answer

thanks

turingtest · Answer

Having had a bit more time to digest this I can say I follow the logic and see no mistakes. That said I don't think I'm qualified to give you the all clear myself. I'll be curious to hear other opinions.
Again, excellent job!

zarkon · Answer

I don't see a problem

zarkon · Answer

though I would add more detail.

turingtest · Answer

like? what cases are not covered  that should be?

zarkon · Answer

all the cases are covered...just more words to explain what is going on

turingtest · Answer

well he still deserves a medal then :)

zarkon · Answer

I don't like this sub forum...I'm only level 2 here :)

turingtest · Answer

there ya go! on the way to guru again.

jamesj · Answer

Interesting proof, because it seems you've proven something stronger.  When you get to the statement
$$ f(m+n) = f(m) $$
m and n are arbitrary, and hence f is constant.

asnaseer · Answer

I'm not clear on how setting m=0 leads to:$$f(-n)| f(n)$$on the 5th line of the proof.
same goes for the 7th line of the proof where setting m=0 and n=-n seems to lead to:$$f(n)| f(-n)$$how do we reach these conclusions?

anonymous · Answer

Actually I missed one step, which is  setting m = -n and n equal to zero.$$\frac{f(-n)-f(0)}{f(-n)} \implies f(-n)| f(0)$$

asnaseer · Answer

is that a legal step?
you are (in the same step) setting m=-n AND n=0. isn't that equivalent to setting m=0?

anonymous · Answer

This means that f(0) is divisible by f(-n). w
When m = 0, $$\frac{f(0)-f(n))}{f(-n)} \implies f(-n) | f(n)$$

zarkon · Answer

you already had f(m)|f(0)...m was arbitrary...thus...

asnaseer · Answer

I understand the 1st step where you showed $f(m)|f(0)$, but I am afraid I don't follow the rest of the arguments.
this is probably beyond my expertise as I canot see how the next steps are deduced?