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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 \[f(m)\]
 2 years ago
 2 years ago
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 \[f(m)\]
 2 years ago
 2 years ago

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moneybirdBest ResponseYou've already chosen the best response.2
\[\frac{f(m)  f(n)}{f(mn)}=k\]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 \[f(m)  f(n)\]
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.0
Wow moneybird this looks really great! I'll have to give it special attention on a nonholiday 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?
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.0
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!
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.2
though I would add more detail.
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.0
like? what cases are not covered that should be?
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.2
all the cases are covered...just more words to explain what is going on
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.0
well he still deserves a medal then :)
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.2
I don't like this sub forum...I'm only level 2 here :)
 2 years ago

TuringTestBest ResponseYou've already chosen the best response.0
there ya go! on the way to guru again.
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
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.
 2 years ago

asnaseerBest ResponseYou've already chosen the best response.0
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?
 2 years ago

moneybirdBest ResponseYou've already chosen the best response.2
Actually I missed one step, which is setting m = n and n equal to zero.\[\frac{f(n)f(0)}{f(n)} \implies f(n) f(0)\]
 2 years ago

asnaseerBest ResponseYou've already chosen the best response.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?
 2 years ago

moneybirdBest ResponseYou've already chosen the best response.2
This means that f(0) is divisible by f(n). w When m = 0, \[\frac{f(0)f(n))}{f(n)} \implies f(n)  f(n)\]
 2 years ago

ZarkonBest ResponseYou've already chosen the best response.2
you already had f(m)f(0)...m was arbitrary...thus...
 2 years ago

asnaseerBest ResponseYou've already chosen the best response.0
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?
 2 years ago
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