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?