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TuringTest

  • 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 \[f(m)\]

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  1. moneybird
    • 2 years ago
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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 \[f(m) | f(n)\]

  2. TuringTest
    • 2 years ago
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    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?

  3. moneybird
    • 2 years ago
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    thanks

  4. TuringTest
    • 2 years ago
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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!

  5. Zarkon
    • 2 years ago
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    I don't see a problem

  6. Zarkon
    • 2 years ago
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    though I would add more detail.

  7. TuringTest
    • 2 years ago
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    like? what cases are not covered that should be?

  8. Zarkon
    • 2 years ago
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    all the cases are covered...just more words to explain what is going on

  9. TuringTest
    • 2 years ago
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    well he still deserves a medal then :)

  10. Zarkon
    • 2 years ago
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    I don't like this sub forum...I'm only level 2 here :)

  11. TuringTest
    • 2 years ago
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    there ya go! on the way to guru again.

  12. JamesJ
    • 2 years ago
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    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.

  13. asnaseer
    • 2 years ago
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    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?

  14. moneybird
    • 2 years ago
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    Actually I missed one step, which is setting m = -n and n equal to zero.\[\frac{f(-n)-f(0)}{f(-n)} \implies f(-n)| f(0)\]

  15. asnaseer
    • 2 years ago
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    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?

  16. moneybird
    • 2 years ago
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    This means that f(0) is divisible by f(-n). w When m = 0, \[\frac{f(0)-f(n))}{f(-n)} \implies f(-n) | f(n)\]

  17. Zarkon
    • 2 years ago
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    you already had f(m)|f(0)...m was arbitrary...thus...

  18. asnaseer
    • 2 years ago
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    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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