# proof that 3 is the only prime perfect totient number

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

$$p=\sum _{i=1}^{c+1}{\varphi}^{i}(n),$$ |

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

The first four primes are most easily examined empirically. Since $\varphi (2)=1$, 2 is deficient totient number. $\varphi (3)=2$, 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^{} $\varphi (n)>\sqrt{n}$ (true for all $n>6$). It is obvious that $\varphi (p)=p-1$, and by the foregoing, ${\varphi}^{2}(p)>3.162278$ (that is, it is sure to be more than the square root of 10), so it follows that $\varphi (p)+{\varphi}^{2}(p)>p+2.162278$ 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 |
---|---|

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 |