2

is this correct?

1) Show that if $\sigma (n) > 2n$ it follows $ \sigma (kn) > 2(kn)$.

Proof: $\sigma (kn) \ge \sum_{d|n} kd = k\cdot \sigma(n) > k2n = 2kn$.

How can I show $\sigma (kn) \ge \sum_{d|n} kd$?

2) Show that if $\sigma (n) = 2n$ it follows $ \sigma (kn) = 2kn$.

???

user91500
  • 5,606
Vazrael
  • 2,281

1 Answers1

1

Statement 2 is not true, indeed

If $\sigma(n)=2n$, it follows that $\sigma(kn)>2kn$

Proof: $$\sigma(kn)=\sum_{x|kn}x\ge\color{red}1+\sum_{y|n}ky=1+k\sigma(n)=1+2kn>2kn $$ (Since $\color{red}1$ is not of the form $ky$ for a divisor $y$ of $n$, but $\color{\red}1$ certainly divided $kn$.)

This proof also can prove statement 1.

user91500
  • 5,606
  • The first inequality in your proof is reversed. The double-sum overcounts divisors of $kn$ whenever $k$ and $n$ are not coprime, since one $x$ can be represented in multiple ways as $yz$. While it's true that $\sigma(kn) > 2kn$, this is not a valid way to show it. – Erick Wong Apr 07 '14 at 07:59
  • @ErickWong: Thanks, You are right and I've edited my answer. – user91500 Apr 08 '14 at 04:58