# Euler's totient function

• Jul 12th 2009, 04:41 PM
streethot
Euler's totient function
Find all positive integers n, such that $\displaystyle \phi(n)$ is a prime.
• Jul 12th 2009, 05:47 PM
PaulRS
Quote:

Originally Posted by o_O
We must have that $\displaystyle n$ is prime, since otherwise $\displaystyle \phi (n)$ would have more than one prime factor.

Careful, remember that $\displaystyle \phi(2)=1$. For example $\displaystyle \phi(6)=2$. (Wink)

To streethot: You should not have problems if you use the formula for the phi function (the one that uses the prime descomposition ), along with the observation o_O made that $\displaystyle \phi(n)$ is either an even number, or the number 1.
• Jul 12th 2009, 06:11 PM
streethot
If $\displaystyle \phi(n)$ is prime, then $\displaystyle \varphi(n)=2$, because if we have $\displaystyle \varphi(n)= n\prod_{p|n} \left( 1-\frac{1}{p} \right)$, its easy to see that $\displaystyle \prod_{p|n} \left( 1-\frac{1}{p} \right)\neq 1$ for $\displaystyle n\ge 2$

if $\displaystyle \varphi(n)=(p_1 - 1) p_1^{k_1-1}...(p_n -1)p_n^{k_n-1}$ we have this inequality:

$\displaystyle (p_1 - 1) p_1^{k_1-1}...(p_n -1)p_n^{k_n-1}\leq(p_1 - 1) p_1^{k_1}...(p_n -1)p_n^{k_n}$ for $\displaystyle \varphi(n)$

now, its possible to say that the n is in the form $\displaystyle 2^k.3^m$

$\displaystyle n=\{3,4,6\}$

Proceed?