What is the 5th solution of ?
This is of the form where $\displaystyle \phi$ x = x (1 - 1/p1 ).... (1 - 1/pr)
where p1....pr refer to the prime factorization of x
ok maybe i am missing something fundamental here because thats what i assumed you had meant originally.... but if any 3^k is a solution, then take 3^2 = 9 for example....
we get p1=3 and p2=3
so (1) = (3-1)(3-1)/9 = 4/9 which does not equal 2/3 as required?
The 'Fundamental Theorem of Arithmetic'm extablishes that any n>1 can be written as...
(1)
... so that...
(2)
... depends only from the 'prime factors' and not from the 'exponents' . In Your example is...
... because 9 and 3 have both the only prime factor 3...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$