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I was reading the following theorem.

Theorem. There are infinitely-many even abundant numbers.

Proof. For a positive integer $a$, let $n=2^a\cdot3$, and compute$$\sigma(2^a\cdot3)=\sigma(2^a)\sigma(3)=(2^{a+1}-1)(3+1)=2(2^a\cdot3)+2^{a+1}-4=2n+2^{a+1}-4,$$ which is greater than $2n$ whenever $a\geq2$. This demonstrates infinitely-many even abundant numbers. $\square$

I know that this proof assumes knowledge of the fact that the smallest even abundant number is $12=2^2\cdot3$, and that any multiple of an abundant number is also abundant. However, It is not immediately clear to me that $2^{a+1}-4$ is a multiple of $2n$ for $n\geq2$. What am I missing?

Thanks in advance.

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1 Answer 1

up vote 2 down vote accepted

All that the proof is demonstarting is that each number of the form $2^a \cdot 3$ is abundant when $a \geq 2$.

Recall a number $n$ is abundant when $\sigma(n) > 2n$. This follows by the above calculation since:

$$\begin{align} \sigma(2^a \cdot 3) = 2(2^{a}\cdot 3) + 2^{a+1} - 4 > 2 (2^a \cdot 3) &\iff 2^{a+1} - 4 > 0 \\ &\iff 2^{a+1} > 2^2 \\ &\iff a > 1. \end{align}$$

This argument does not rely on $12$ being the smallest abundant number, or that a multiple of an abundant number is again abundant. It just states that every number of the form $n = 2^a \cdot 3$ is abundant when $a \geq 2$. This set of integers forms in infinite set of even abundant numbers.

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