Prove that if $n$ is a perfect number, $kn$ is not.
If $\gcd(k,n)=1$ then this is clear.
(assume $\sigma(n)=2n$ , $\sigma(nk)=2kn$ , then $k=\sigma(k)$ but $\sigma(k)>k)$.
But what about $\gcd(k,n)>1$?
Hint: Find some proper divisors of $kn$ that add up to exactly $kn$. Are any divisors missing?
Let $a_1,...a_r$ be the divisors of $n$, we know that $\sum a_i = n$. So $kn$ is divisible by $ka_1,...,ka_n$ which sum to $kn$ but also by $a_1$ so the sum of the divisors of $kn$ is greater than $kn$.