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Re: I find it very interesting...

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  • djbroadhurst
    ... Let n and k be positive integers with n k+1. Then eulerphi(n^2-k^2) = (n-1)^2 - k^2 if and only if n+k and n-k are both prime. David
    Message 1 of 6 , Jun 1, 2013
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      --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...> wrote:

      > phi(n^2-k^2) = (n-1)^2 - k^2, working for ONLY some n's and some k's

      Let n and k be positive integers with n > k+1. Then
      eulerphi(n^2-k^2) = (n-1)^2 - k^2
      if and only if n+k and n-k are both prime.

      David
    • Sebastian Martin Ruiz
      See conjecture 33 at primepuzzles Enviado desde Yahoo! Mail con Android [Non-text portions of this message have been removed]
      Message 2 of 6 , Jun 1, 2013
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        See conjecture 33 at primepuzzles

        Enviado desde Yahoo! Mail con Android



        [Non-text portions of this message have been removed]
      • leavemsg1
        Also, I wondered about the algebra of Euler s phi function. Is it true that phi(n^3 -k^3) = y*phi(n -k)*phi(n^2 +nk +k^2)? and... Is it true that phi(n^3
        Message 3 of 6 , Jun 1, 2013
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          Also, I wondered about the "algebra" of Euler's phi function.
          Is it true that phi(n^3 -k^3) = y*phi(n -k)*phi(n^2 +nk +k^2)?
          and...
          Is it true that phi(n^3 +k^3) = z*phi(n +k)*phi(n^2 -nk +k^2)?
          ...
          of course, when 'n' is prime, and gcd(n, k) = 1; it appears
          that y, z = 1 in the smaller cases of 'n' and 'k', but when
          I tried n= 5 and k= 4, I noticed that z= 3/2; at what point
          do things tend to break down?? or Can 'y' and 'z' be computed
          in terms of n's and k's.
          Bill

          --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
          >
          > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
          >
          > > phi(n^2-k^2) = (n-1)^2 - k^2, working for ONLY some n's and some k's
          >
          > Let n and k be positive integers with n > k+1. Then
          > eulerphi(n^2-k^2) = (n-1)^2 - k^2
          > if and only if n+k and n-k are both prime.
          >
          > David
          >
        • djbroadhurst
          ... You may find this useful: phi(a*b)=phi(a)*phi(b)*g/phi(g), where g=gcd(a,b). David
          Message 4 of 6 , Jun 1, 2013
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            --- In primenumbers@yahoogroups.com,
            "leavemsg1" <leavemsg1@...> wrote:

            > I wondered about the "algebra" of Euler's phi function

            You may find this useful:
            phi(a*b)=phi(a)*phi(b)*g/phi(g), where g=gcd(a,b).

            David
          • Mohsen Afshin
            That surely derives from the fact that if both n-k and n+k are primes then phi(n^2-k^2) = phi((n-k)(n+k)) = phi(n-k) * phi (n+k) = (n-k-1) * (n+k-1) = n^2 - 2n
            Message 5 of 6 , Jun 1, 2013
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              That surely derives from the fact that if both n-k and n+k are primes then

              phi(n^2-k^2) = phi((n-k)(n+k)) = phi(n-k) * phi (n+k) = (n-k-1) * (n+k-1) =
              n^2 - 2n + 1 - k^2 = (n-1)^2 - k^2
              Example
              n = 105, k = 4

              phi(105^2-4^2) = (105-1)^2 - 4^2 = 10800 = (101-1)*(109-1)

              But my question is how we can approach to an additive feature of EulerPhi?

              If there was one such a feature then the primality tests would become so
              easy. As an example consider n!+1, we know phi(n!) without any computation
              and using that dreamed feature we could easily check if phi(n!+1) equals n!
              or not :-)

              I already tried some here
              http://math.stackexchange.com/questions/249982/how-to-calculate-euler-totient-from-nearby-values


              On Sat, Jun 1, 2013 at 4:11 PM, djbroadhurst <d.broadhurst@...>wrote:

              > **
              >
              >
              > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...> wrote:
              >
              > > phi(n^2-k^2) = (n-1)^2 - k^2, working for ONLY some n's and some k's
              >
              > Let n and k be positive integers with n > k+1. Then
              > eulerphi(n^2-k^2) = (n-1)^2 - k^2
              > if and only if n+k and n-k are both prime.
              >
              > David
              >
              >
              >



              --
              "Mathematics is the queen of the sciences and number theory is the queen of
              mathematics."
              --Gauss


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