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Re: [PrimeNumbers] Lehmer Mack 2

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  • Phil Carmody
    ... Let n=2 Phi(2^2) = 2 = 2.(2-1) = k=2 = k/n = 2 Now, as someone with limited intelligence , I think that 2 actually _is_ an integer, but am prepared to
    Message 1 of 3 , Apr 2, 2002
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      --- Jon Perry <perry@...> wrote:
      > Consider phi(n^2)=k.(n-1)
      >
      > Obviously k contains some factor of n-1 in it's denominator.
      >
      > Therefore, as phi(n^2)=n.phi(n):
      >
      > phi(n)=(k/n).(n-1)
      >
      > And therefore k/n is never an integer.

      Let n=2
      Phi(2^2) = 2 = 2.(2-1)
      => k=2
      => k/n = 2

      Now, as someone with "limited intelligence", I think that 2 actually
      _is_ an integer, but am prepared to be proved wrong using your Most
      Valued mathematical techniques. After you've read Matthew 7 verse 5,
      that is.


      Phil

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    • Jon Perry
      Well spotted. I had confused the numerator/denominator. ... and also a prime! ... Don t read the Bible. This verse states? Jon Perry perry@globalnet.co.uk
      Message 2 of 3 , Apr 3, 2002
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        Well spotted. I had confused the numerator/denominator.

        > I think that 2 actually
        >_is_ an integer

        and also a prime!

        >After you've read Matthew 7 verse 5,that is.

        Don't read the Bible. This verse states?

        Jon Perry
        perry@...
        http://www.users.globalnet.co.uk/~perry/maths
        BrainBench MVP for HTML and JavaScript
        http://www.brainbench.com


        -----Original Message-----
        From: Phil Carmody [mailto:thefatphil@...]
        Sent: 03 April 2002 00:03
        To: primenumbers
        Subject: Re: [PrimeNumbers] Lehmer Mack 2


        --- Jon Perry <perry@...> wrote:
        > Consider phi(n^2)=k.(n-1)
        >
        > Obviously k contains some factor of n-1 in it's denominator.
        >
        > Therefore, as phi(n^2)=n.phi(n):
        >
        > phi(n)=(k/n).(n-1)
        >
        > And therefore k/n is never an integer.

        Let n=2
        Phi(2^2) = 2 = 2.(2-1)
        => k=2
        => k/n = 2

        Now, as someone with "limited intelligence", I think that 2 actually
        _is_ an integer, but am prepared to be proved wrong using your Most
        Valued mathematical techniques. After you've read Matthew 7 verse 5,
        that is.


        Phil

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