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Lehmer Mack 2

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  • Jon Perry
    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
    Message 1 of 3 , Apr 2, 2002
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      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.

      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry/maths
      BrainBench MVP for HTML and JavaScript
      http://www.brainbench.com
    • 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 2 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 3 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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