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 $1+2+6+18+\mathrm{\cdots }+2(3^{x-2})+2(3^{x-1})=3^x$.

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