## I created new phi function identites.

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• Hi again,... ... take a look at the algebra for Euler s /*phi*/ function ... if and only if... phi(n^2 -1) = 2*phi(n +1)*phi(n -1) can be proved, then
Message 1 of 1 , Jun 1, 2013
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Hi again,...
...
take a look at the "algebra" for Euler's /*phi*/ function
...
if and only if... phi(n^2 -1) = 2*phi(n +1)*phi(n -1) can be
proved, then because phi(n^k) = n^(k -1)*phi(n), we can surely
understand the following two equations:
...
phi(n+1) = [(n -1)*phi(n^2 -1)] / [2*phi((n -1)^2)] and...
...
phi(n -1) = [phi((n-1)^2) / (n -1)].
...
everyone can understand them! they must be written in the
subtractive format, and it must be noted that 'n' is prime.
Also, I can see how the 1's could be replaced by 'k's such
that we'd come to know that phi(n +k) & phi(n -k) can be
meaningful, iff gcd(n, k) =1 is closely monitored.
...
Bill
www.oddperfectnumbers.com; Lehmer's Totient Problem Solved!
...
"Numerical precision is the very soul of mathematics."
(Sir D'Arcy Wentworth Thompson)
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