I was working through some basic Number Theory Problems when I came across :
Given an integer $m$ $≥ 2$ such that $(2^{m} -1)$ is a prime, and $n$ $=$ $(2^{m-1})$$(2^{m} -1)$, then show that $\sigma(n)$ $=$ $2n$ where $\sigma(n)$ is the sum of the divisors of an integer $x$
I am not able to make progress ,can someone help me out ; even a hint will suffice