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Thread: Euler Phi primes with just definition of Phi

  1. #1
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    Question Euler Phi primes with just definition of Phi

    So I was given a pretty easy problem
    Calculate $\displaystyle \phi(2p)$ where $\displaystyle p$ is prime.
    Which I can solve with what I have read online:
    * Given $\displaystyle \phi(ab)$ if $\displaystyle (a,b)=1$, then $\displaystyle \phi(a) \phi(b)$

    Then it's pretty easy:
    $\displaystyle \phi(2p)$ and since$\displaystyle (2,p)=1$, then$\displaystyle \phi(2) \phi(p) = 1*(p-1) = p-1$

    But the problem is, in class we haven't proved * ,so we can't use it. Pretty much all we know about$\displaystyle \phi(m)$ is that it is the number of positive integers wich are less than m and relatively prime to m. AND $\displaystyle \phi(p) = p-1 $.

    I'm just wondering how I would go about doing this?
    Thanks

    EDIT:
    After some thought and more reading I came to this:

    CASE 1 : if p != 2
    $\displaystyle \phi(2p) = \phi(2^1p^1)=$
    $\displaystyle 2p(1-1/2)(1-1/p) = p-1$

    Case 2:$\displaystyle p = 2$
    $\displaystyle \phi(2*2) = \phi(4) =\phi(2^2)=$
    $\displaystyle 4(1-1/2) = 2$

    Am I correct in my reasoning?
    Last edited by Th3sandm4n; Mar 31st 2009 at 08:56 PM. Reason: lightbulb!
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  2. #2
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    I think they are expecting you to calculate the value from the definition of $\displaystyle \phi(n)$.

    How many numbers below 2p are coprime to 2?
    How many numbers below 2p are coprime to p?
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  3. #3
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    Quote Originally Posted by SimonM View Post
    I think they are expecting you to calculate the value from the definition of $\displaystyle \phi(n)$.

    How many numbers below 2p are coprime to 2?
    How many numbers below 2p are coprime to p?
    Hmm,
    I get something like
    when p != 2
    How many numbers below 2p are coprime to 2? $\displaystyle 2p-2$
    How many numbers below 2p are coprime to p? $\displaystyle (1/2) 2p$
    with overlapping = $\displaystyle p-1$, but I don't know how to show this other than example values of p?
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