# proof that all powers of 3 are perfect totient numbers

Given an integer $x>0$, it is always the case that

$${3}^{x}=\sum _{i=1}^{c+1}{\varphi}^{i}({3}^{x}),$$ |

where ${\varphi}^{i}(x)$ is the iterated totient function and $c$ is the integer such that ${\varphi}^{c}(n)=2$. That is, all integer powers of three are perfect totient numbers.

The proof of this is easy and even considered trivial. Here it goes anyway:

Accepting as proven that $\varphi ({p}^{x})=(p-1){p}^{x-1}$, we can plug in $p=3$ and see that $\varphi ({3}^{x})=2({3}^{x-1})$, which falls short of ${3}^{x}$ by ${3}^{x-1}$. Given the proof that Euler $\varphi $ is a multiplicative function^{} (http://planetmath.org/ProofThatEulerPhiIsAMultiplicativeFunction) and the fact that $\varphi (2)=1$, it is obvious that $\varphi (2({3}^{x-1}))=\varphi ({3}^{x-1})$. Therefore, each iterate will be twice a power of three, with the exponent^{} gradually decreasing as the iterator nears $c$. To put it algebraically, ${\varphi}^{i}({3}^{x})=2({3}^{c-i})$ for $i\le c$. Adding up in ascending order starting at the $(c+1)$th iterate, we obtain $1+2+6+18+\mathrm{\cdots}+2({3}^{x-2})+2({3}^{x-1})={3}^{x}$.

## References

- 1 D. E. Ianucci, D. Moujie & G. L. Cohen, “On Perfect Totient Numbers” Journal of Integer Sequences, 6, 2003: 03.4.5

Title | proof that all powers of 3 are perfect totient numbers |
---|---|

Canonical name | ProofThatAllPowersOf3ArePerfectTotientNumbers |

Date of creation | 2013-03-22 16:34:32 |

Last modified on | 2013-03-22 16:34:32 |

Owner | PrimeFan (13766) |

Last modified by | PrimeFan (13766) |

Numerical id | 11 |

Author | PrimeFan (13766) |

Entry type | Proof |

Classification | msc 11A25 |