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Modulo and prime numbers

  1. Nov 4, 2012 #1
    [tex]f(x)[/tex] will give us the smallest integer [tex]m[/tex] such that [tex]y^m \equiv 1 \bmod{x}[/tex] given that x and y are coprime

    how would one go about showing that this function is multiplicative? I'm trying to do some Number Theory self study, and the problems I'm encountering are quite difficult to figure out from text alone.

    I'm guessing that the Chinese remainder theorem is applicable here.

    Also if we let p be prime then I know that f(p) will give the result of (p-1). This is basically proved with Euler's theorem.

    Is this true or is Euler's theorem not so easily applicable here? I guess the fact that here m is the smallest integer, where's in Euler's it's not the smallest, but it's easy to prove that this doesn't matter using groups.
     
    Last edited: Nov 4, 2012
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  3. Nov 9, 2012 #2

    haruspex

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    How does y come in? Is it another given? If so, isn't it f(x, y)?
     
  4. Nov 10, 2012 #3
    I don't think so because y is dependent on x just as m is. I.E. y is any number coprime with x.
     
    Last edited: Nov 10, 2012
  5. Nov 10, 2012 #4

    haruspex

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    Rereading it, I think we're both wrong. It should say "for all y coprime to x". (Otherwise the later part about f(p) = p-1 is wrong.) In short, this f(x) is Euler's totient function.
     
  6. Nov 10, 2012 #5
    i don't think it's Euler's totient function.

    for example, consider x = 37; y = 3.
    both prime, but the smallest m value is not 36, its 18.
     
  7. Nov 10, 2012 #6

    haruspex

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    As I said, I think the question is intended as "for all y". I.e. what is the smallest m that s.t. for every y prime to x the equation is true.
     
  8. Nov 11, 2012 #7
    The totient function gives you an upper bound for f(x), but it is not necessarily the same as f(x). A counterexample occurs with x=12.

    The coprimes to x are 1, 5, 7 and 11. An exponent of 1 won't work, but 2 will:
    1^2 == 1 (mod 12)
    5^2 = 25 == 1 (mod 12)
    7^2 = 49 == 1 (mod 12)
    11^2 = 121 == 1 (mod 12)

    so f(12)=2, while phi(12)=4.

    (I I understood the problem correctly, f(x) would be the order of the multiplicative group modulo x. As such, f(x) divides phi(x), but it's not necessarily equal.)

    (By the way, f(3)=2 and f(4)=2, but f(12)=2 and not 4, so f(x) cannot be multiplicative.)
     
    Last edited: Nov 11, 2012
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