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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 $\gcd(k,n)>1$?

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Hint: Find some proper divisors of $kn$ that add up to exactly $kn$. Are any divisors missing?

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sigma(kn)= kn + k +n+.. ? – user1932595 Apr 4 '13 at 6:39
@user1932595 Look at an example, like how $6 = 1 + 2 + 3$ is perfect and $12 = 2 + 4 + 6$ is not – Cocopuffs Apr 4 '13 at 7:13

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 $kn$.

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Hint: If $I(x) = \sigma(x)/x$ is the abundancy index of $x$, then $$y \mid z \implies I(y) \leq I(z),$$ with equality if $y = z$.

Can you take it from here?

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