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Thread: Euler's phi function: multiplicity

  1. #1
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    Euler's phi function: multiplicity

    I am supposed to show that, for Euler's totient function:

    a.) If n is an odd integer, then $\displaystyle \phi(2n)=\phi(n)$.
    and
    b.) If n is an even integer, then $\displaystyle \phi(2n)=2\phi(n)$.

    I thought I would use the multiplicity of phi to establish this, i.e.
    $\displaystyle \phi(2n)=\phi(2)\cdot\phi(n)=1\cdot\phi(n)=\phi(n)$.

    Obviously this doesn't work, because while it works for (a), it contradicts (b). Not sure why.
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  2. #2
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    Sorry, title should be "multiplicativity", not "multiplicity".
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  3. #3
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    Quote Originally Posted by glowplug19 View Post
    I am supposed to show that, for Euler's totient function:

    a.) If n is an odd integer, then $\displaystyle \phi(2n)=\phi(n)$.
    and
    b.) If n is an even integer, then $\displaystyle \phi(2n)=2\phi(n)$.

    I thought I would use the multiplicity of phi to establish this, i.e.
    $\displaystyle \phi(2n)=\phi(2)\cdot\phi(n)=1\cdot\phi(n)=\phi(n)$.

    Obviously this doesn't work, because while it works for (a), it contradicts (b). Not sure why.
    Hi glowplug19.

    Be careful here. The formula $\displaystyle \phi(mn)=\phi(m)\phi(n)$ only works when $\displaystyle \gcd(m,n)=1.$ This is why it works for (a) but not for (b), as $\displaystyle \gcd(2,n)=1$ if and only if $\displaystyle n$ is odd.

    For (b) consider $\displaystyle S_1=\{1,2\ldots,n\}$ and $\displaystyle S_2=\{n+1,n+2\ldots,n+n=2n\}.$ If $\displaystyle n$ is even, the integers in $\displaystyle S_1$ which are coprime with $\displaystyle n$ excludes the even integers; hence an integer in $\displaystyle S_1$ is coprime with $\displaystyle 2n$ if and only if it is coprime with $\displaystyle n.$ For $\displaystyle S_2$ note that $\displaystyle n+r$ is coprime with $\displaystyle n$ if and only if $\displaystyle r$ is coprime with $\displaystyle n;$ hence the same argument can be applied to $\displaystyle S_2$ to show that there are $\displaystyle \phi(n)$ integers in $\displaystyle S_2$ coprime with $\displaystyle 2n.$ Thus $\displaystyle \phi(2n)=\phi(n)+\phi(n)=2\phi(n).$
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