Prove Corollary (4.11).
Corollary (4.11) says that if $n$ is composite then $2^n− 1$ is also composite.
We start with Proposition (4.9) which says that, for $m,n$ positive integers
$$ m \mid n \implies (2^m - 1) \mid (2^n-1) $$
This gives us the desired conclusion, because it states that if $n$ is composite, with a non-trivial factor $m$, where $1<m<n$, then $2^n-1$ is composite, with a non-trivial factor $2^m-1$, where $1 < 2^m - 1< 2^n-1 $.