a question on square-free numbers

**rulin** · April 28th 2010, 06:15 PM

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?

**chiph588@** · April 28th 2010, 09:45 PM

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 $m=2,4$ ?

You should rephrase your question, your grammar is a bit off.

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