## You are here

Homeproof that 3 is the only prime perfect totient number

## Primary tabs

# 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.

## Mathematics Subject Classification

11A25*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections