Exercise 2.8.16 ($(2^m-1, 2^n+1) = 1$ if $m$ is odd)

Proof. Assume for contradiction that $d=(2^m-1)\wedge (2^n+1)>1$. Since $d$ is odd, $2\wedge d=1$, so the order of $2$ modulo $d>1$ exists. Let $h$ be this order.

Since $2^m\equiv 1(\mathit{mod}d)$, $h\mid m$, where $m$ is odd, therefore $h$ is odd.

From $2^n\equiv -1$, we deduce $2^{2n}\equiv 1$. Hence

But $h$ is odd, and $h\mid 2n$, thus $h\mid n$. This is a contradiction. Therefore

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