Prove that if $\displaystyle \phi(n)|(n-1)$ then $\displaystyle n$ is prime, where $\displaystyle \phi(n)$. is the Euler's totient function.

I am able to show that $\displaystyle n$ is square-free under the above condition.

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- Mar 6th 2012, 03:55 AMabhishekkgpphi(n) divides n-1 implies n is prime
Prove that if $\displaystyle \phi(n)|(n-1)$ then $\displaystyle n$ is prime, where $\displaystyle \phi(n)$. is the Euler's totient function.

I am able to show that $\displaystyle n$ is square-free under the above condition. - Mar 6th 2012, 04:40 AMprincepsRe: phi(n) divides n-1 implies n is prime
- Mar 6th 2012, 05:09 AMabhishekkgpRe: phi(n) divides n-1 implies n is prime