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Homeproof that all powers of 3 are perfect totient numbers

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# proof that all powers of 3 are perfect totient numbers

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

$3^{x}=\sum_{{i=1}}^{{c+1}}\phi^{i}(3^{x}),$ |

where $\phi^{i}(x)$ is the iterated totient function and $c$ is the integer such that $\phi^{c}(n)=2$. 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 $\phi(p^{x})=(p-1)p^{{x-1}}$, we can plug in $p=3$ and see that $\phi(3^{x})=2(3^{{x-1}})$, which falls short of $3^{x}$ by $3^{{x-1}}$. Given the proof that Euler $\phi$ is a multiplicative function and the fact that $\phi(2)=1$, it is obvious that $\phi(2(3^{{x-1}}))=\phi(3^{{x-1}})$. Therefore, each iterate will be twice a power of three, with the exponent gradually decreasing as the iterator nears $c$. To put it algebraically, $\phi^{i}(3^{x})=2(3^{{c-i}})$ for $i\leq c$. Adding up in ascending order starting at the $(c+1)$th iterate, we obtain $1+2+6+18+\cdots+2(3^{{x-2}})+2(3^{{x-1}})=3^{x}$.

# References

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

## Mathematics Subject Classification

11A25*no label found*

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