Suppose $n$ is an odd perfect number. 

How to show that it exists a prime number $p$ such that $\frac{n}{p}$ is a square number?

My idea was:

$n$ is perfect if $\sigma(n)=2n$. 

Let $n=2k+1, \ k \in \mathbb{N\_0}$. 

Then it's $\sigma(n)=\sigma(2k+1)=\sum \limits\_{d \vert n}d=1+...+d\_{n-1}+d\_n=2n=2(2k+1)$ with $d\_1,...,d\_n$ odd.

So $\frac{1+...+d\_{n-1}}{p}
\Leftrightarrow\frac{2n}{2p}$.

But I don't know how to continue to show that such a $p$ exists, such that $\frac{n}{p}$ is a square number.

How can it be shown?