## Re: I find it very interesting...

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• ... You may find this useful: phi(a*b)=phi(a)*phi(b)*g/phi(g), where g=gcd(a,b). David
Message 1 of 6 , Jun 1, 2013
"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
• 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 2 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 :-)

http://math.stackexchange.com/questions/249982/how-to-calculate-euler-totient-from-nearby-values

> **
>
>
> --- 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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