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Thread: Euler Totient problem

  1. #1
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    Euler Totient problem

    Find all n such that:

    $\displaystyle \varphi(n)=12$
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  2. #2
    MHF Contributor chisigma's Avatar
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    The $\displaystyle \varphi(*)$ is a multiplicative function so that if m and n are coprime, then...

    $\displaystyle \varphi(m\cdot n)= \varphi(m)\cdot \varphi(n)$ (1)

    For $\displaystyle n=1$ and $\displaystyle n=2$ is $\displaystyle \varphi(n)=1$ and $\displaystyle \forall n>2$ $\displaystyle \varphi(n)$ is an even number, so that the solutions of the equation...

    $\displaystyle \varphi (m\cdot n)=12$ (2)

    ... are obtained by systematic search of couples of coprime numbers m and n for which is...

    $\displaystyle \varphi(m)=12, \varphi(n)=1$

    $\displaystyle \varphi(m)=6, \varphi(n)=2$ (3)

    Since $\displaystyle \varphi(13)=12$ solutions are...

    $\displaystyle 13 \cdot 1 =13$

    $\displaystyle 13\cdot 2 = 26$

    Since $\displaystyle \varphi(7)=6$ also solutions are...

    $\displaystyle 7 \cdot 3 = 21$

    $\displaystyle 7 \cdot 4= 28$

    $\displaystyle 7 \cdot 6 = 42$

    Finally is $\displaystyle \varphi (9)= 6$ so that solution is...

    $\displaystyle 9\cdot 4 = 36$

    The numbers k for which is $\displaystyle \varphi(k)=12$ are...

    $\displaystyle k=13,21,26,28,36,42$

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    Quote Originally Posted by chisigma View Post
    The $\displaystyle \varphi(*)$ is a multiplicative function so that if m and n are coprime, then...

    $\displaystyle \varphi(m\cdot n)= \varphi(m)\cdot \varphi(n)$ (1)

    For $\displaystyle n=1$ and $\displaystyle n=2$ is $\displaystyle \varphi(n)=1$ and $\displaystyle \forall n>2$ $\displaystyle \varphi(n)$ is an even number, so that the solutions of the equation...

    $\displaystyle \varphi (m\cdot n)=12$ (2)

    ... are obtained by systematic search of couples of coprime numbers m and n for which is...

    $\displaystyle \varphi(m)=12, \varphi(n)=1$

    $\displaystyle \varphi(m)=6, \varphi(n)=2$ (3)

    Since $\displaystyle \varphi(13)=12$ solutions are...

    $\displaystyle 13 \cdot 1 =13$

    $\displaystyle 13\cdot 2 = 26$

    Since $\displaystyle \varphi(7)=6$ also solutions are...

    $\displaystyle 7 \cdot 3 = 21$

    $\displaystyle 7 \cdot 4= 28$

    $\displaystyle 7 \cdot 6 = 42$

    Finally is $\displaystyle \varphi (9)= 6$ so that solution is...

    $\displaystyle 9\cdot 4 = 36$

    The numbers k for which is $\displaystyle \varphi(k)=12$ are...

    $\displaystyle k=13,21,26,28,36,42$


    $\displaystyle \chi$ $\displaystyle \sigma$
    Hi, thank you for directing me here. Its good to see an example, but i dont understand this method.

    Why do you multiply 7 by 3, 4 and 6? And how do you get $\displaystyle \phi(7) = 6$ and $\displaystyle \phi(9) = 6$. I know you can easily work these out but aren't both of these just an instance of the same sort of question as $\displaystyle \phi(m) = 12$? Aren't there more $\displaystyle \phi(m) = 6$?

    Sorry if i'm being a bit slow, all this is very confusing for me.
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  4. #4
    MHF Contributor chisigma's Avatar
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    The basic assumptions are...

    a) is $\displaystyle \varphi(n)=1$ only for $\displaystyle n=1$ and $\displaystyle n=2$. For all $\displaystyle n>2$ , $\displaystyle \varphi (n)$ is an even number...

    b) is $\displaystyle \varphi(m\cdot k)=\varphi(m)\cdot \varphi(k)$ if and only if m and k are coprime...

    On the basis of b) the identification of an n for which $\displaystyle \varphi(n)=12$ is for the fact that must be $\displaystyle n=m\cdot k$ where m and k are coprime and $\displaystyle \varphi(m)\cdot \varphi(k) = 12$. There are the following possibilities...

    $\displaystyle \varphi(m)=12$ , $\displaystyle \varphi(k)=1$

    $\displaystyle \varphi(m)=6$ , $\displaystyle \varphi(k)=2$

    The situation...

    $\displaystyle \varphi(m)=4$ , $\displaystyle \varphi(k)=3$

    ... as proposed by tonio in the original post, is impossible on the basis of a)...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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