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

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  • 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 1 of 6 , Jun 1, 2013
      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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