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  • leavemsg1
    Dear Group, ... I made an observation of a conjecture that was posted on Carlos Rivera s website. It was named The Goldbach Temptation. Can someone prove
    Message 1 of 6 , May 31 7:31 PM
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      Dear Group,
      ...
      I made an observation of a conjecture that was posted
      on Carlos Rivera's website. It was named The Goldbach
      Temptation. Can someone prove whether phi(n^2 -k^2) =
      (n-1)^2 -k^2, working for ONLY some n's and some k's,
      is equivalent to Goldbach's conjecture? I proposed it
      just recently as ... phi(n^2-k^2)=2*phi(n+k)*phi(n-k)
      such that 'n' is prime, gcd(n,k) = 1 w/the hopes that
      someone might find a way to prove the latter equation.
      Are both representations equivalent??? I believe that
      they are... but people were too busy to question it.
      ...
      Rewards, Bill
      ...
      P.S. I remember an old quote... "If you want something
      done, give it to a busy person."
    • 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 2 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 3 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 4 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 5 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 6 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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