proof that 3 is the only prime perfect totient number
Given a prime number , only satisfies the equation
The first four primes are most easily examined empirically. Since , 2 is deficient totient number. , so, per the previous remark, it is a perfect totient number. For 5, the iterates are 4, 2 and 1, adding up to 7, hence 5 is an abundant totient number. The same goes for 7, with its iterates being 7, 6, 2, 1.
It is for that we can avail ourselves of the inequality (true for all ). It is obvious that , and by the foregoing, (that is, it is sure to be more than the square root of 10), so it follows that and thus it is not necessary to examine any further iterates to see that all such primes are abundant totient numbers.
|Title||proof that 3 is the only prime perfect totient number|
|Date of creation||2013-03-22 16:34:29|
|Last modified on||2013-03-22 16:34:29|
|Last modified by||PrimeFan (13766)|