proof that 3 is the only prime perfect totient number

Given a prime numberMathworldPlanetmath p, only p=3 satisfies the equation


where ϕi(x) is the iterated totient function and c is the integer such that ϕc(n)=2. That is, 3 is the only perfect totient number that is prime.

The first four primes are most easily examined empirically. Since ϕ(2)=1, 2 is deficient totient number. ϕ(3)=2, 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 p>7 that we can avail ourselves of the inequalityMathworldPlanetmath ϕ(n)>n (true for all n>6). It is obvious that ϕ(p)=p-1, and by the foregoing, ϕ2(p)>3.162278 (that is, it is sure to be more than the square root of 10), so it follows that ϕ(p)+ϕ2(p)>p+2.162278 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
Canonical name ProofThat3IsTheOnlyPrimePerfectTotientNumber
Date of creation 2013-03-22 16:34:29
Last modified on 2013-03-22 16:34:29
Owner PrimeFan (13766)
Last modified by PrimeFan (13766)
Numerical id 5
Author PrimeFan (13766)
Entry type Proof
Classification msc 11A25