How to show that $\sigma(n)=k$ has a finite # of solutions $k\in\mathbb{N}$? I wish the question is clearly explained in the title. Thanks in advance.
Since $n$ divides $n$ then 
$
k=\sigma(n) \ge n.
$ 
Hence there are at most $k$ solutions.
Since $n$ divides $n$ then 
$
k=\sigma(n) \ge n.
$ 
Hence there are at most $k$ solutions.