Re: [PrimeNumbers] Re: I find it very interesting...
- 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
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
On Sat, Jun 1, 2013 at 4:11 PM, djbroadhurst <d.broadhurst@...>wrote:
> --- In email@example.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.
"Mathematics is the queen of the sciences and number theory is the queen of
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