# Thread: a question on square-free numbers

1. ## a question on square-free numbers

Suppose that $2^m-1$ and $2^m+1$ are both square-free numbers. Denote by $\phi$ the Euler function, and $\pi(n)$ the set of prime divisors of $n$. If $\pi(\phi(2^m-1))\subseteq\pi(2^m+1)\cup\{2\}$, and $\pi(\phi(2^m+1))\subseteq\pi(2^m-1)\cup\{2\}$, then whether or not $m$ is 2,4?

2. Originally Posted by rulin
Let $2^m-1$ and $2^m+1$ be both square-free numbers. Denote by $\phi$ Euler function, and $\pi(n)$ the set of prime divisors of $n$. If $\pi(\phi(2^m-1))\subseteq\pi(2^m+1)\cup\{2\}$, and $\pi(\phi(2^m+1))\subseteq\pi(2^m-1)\cup\{2\}$, whether or mot $m=2,4$?