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Re: [PrimeNumbers] Re: I find it very interesting...

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  • Sebastian Martin Ruiz
    See conjecture 33 at primepuzzles Enviado desde Yahoo! Mail con Android [Non-text portions of this message have been removed]
    Message 1 of 6 , Jun 1 5:03 AM
      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 2 of 6 , Jun 1 10:33 AM
        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 3 of 6 , Jun 1 1:34 PM
          --- 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 4 of 6 , Jun 1 10:44 PM
            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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