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Thread: a question on square-free numbers

  1. #1
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    a question on square-free numbers

    Suppose that $\displaystyle 2^m-1$ and $\displaystyle 2^m+1$ are both square-free numbers. Denote by $\displaystyle \phi$ the Euler function, and $\displaystyle \pi(n)$ the set of prime divisors of $\displaystyle n$. If $\displaystyle \pi(\phi(2^m-1))\subseteq\pi(2^m+1)\cup\{2\}$, and $\displaystyle \pi(\phi(2^m+1))\subseteq\pi(2^m-1)\cup\{2\}$, then whether or not $\displaystyle m$ is 2,4?
    Last edited by rulin; Apr 28th 2010 at 10:06 PM.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by rulin View Post
    Let $\displaystyle 2^m-1$ and $\displaystyle 2^m+1$ be both square-free numbers. Denote by $\displaystyle \phi$ Euler function, and $\displaystyle \pi(n)$ the set of prime divisors of $\displaystyle n$. If $\displaystyle \pi(\phi(2^m-1))\subseteq\pi(2^m+1)\cup\{2\}$, and $\displaystyle \pi(\phi(2^m+1))\subseteq\pi(2^m-1)\cup\{2\}$, whether or mot $\displaystyle m=2,4$?
    You should rephrase your question, your grammar is a bit off.
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