A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 3 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 \[f(m)\]
 3 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 \[f(m)\]

This Question is Closed

moneybird
 3 years ago
Best 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 \[f(m)  f(n)\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0Wow 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?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.0Having 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
 3 years ago
Best ResponseYou've already chosen the best response.2though I would add more detail.

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

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

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

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

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

JamesJ
 3 years ago
Best ResponseYou've already chosen the best response.0Interesting 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
 3 years ago
Best ResponseYou've already chosen the best response.0I'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?

moneybird
 3 years ago
Best ResponseYou've already chosen the best response.2Actually 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
 3 years ago
Best ResponseYou've already chosen the best response.0is that a legal step? you are (in the same step) setting m=n AND n=0. isn't that equivalent to setting m=0?

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

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0I 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?
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.