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 \[f(m)\]

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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 \[f(m)\]

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\[\frac{f(m) - f(n)}{f(m-n)}=k\]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 \[f(m) | f(n)\]
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?
thanks

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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!
I don't see a problem
though I would add more detail.
like? what cases are not covered that should be?
all the cases are covered...just more words to explain what is going on
well he still deserves a medal then :)
I don't like this sub forum...I'm only level 2 here :)
there ya go! on the way to guru again.
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.
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?
Actually I missed one step, which is setting m = -n and n equal to zero.\[\frac{f(-n)-f(0)}{f(-n)} \implies f(-n)| f(0)\]
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?
This means that f(0) is divisible by f(-n). w When m = 0, \[\frac{f(0)-f(n))}{f(-n)} \implies f(-n) | f(n)\]
you already had f(m)|f(0)...m was arbitrary...thus...
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?

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