# proof that 3 is the only prime perfect totient number

Given a prime number  $p$, only $p=3$ satisfies the equation

 $p=\sum_{i=1}^{c+1}\phi^{i}(n),$

where $\phi^{i}(x)$ is the iterated totient function and $c$ is the integer such that $\phi^{c}(n)=2$. That is, 3 is the only perfect totient number that is prime.

The first four primes are most easily examined empirically. Since $\phi(2)=1$, 2 is deficient totient number. $\phi(3)=2$, 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 inequality  $\phi(n)>\sqrt{n}$ (true for all $n>6$). It is obvious that $\phi(p)=p-1$, and by the foregoing, $\phi^{2}(p)>3.162278$ (that is, it is sure to be more than the square root of 10), so it follows that $\phi(p)+\phi^{2}(p)>p+2.162278$ 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 ProofThat3IsTheOnlyPrimePerfectTotientNumber 2013-03-22 16:34:29 2013-03-22 16:34:29 PrimeFan (13766) PrimeFan (13766) 5 PrimeFan (13766) Proof msc 11A25