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I am trying to show that if $n=2^{m-1}(2^{m}-1)$, where $m$ is a positive integer such that $2^{m}-1$ is composite, then $n$ is abundant. This is my proof thus far:

Proof. Let $n=2^{m-1}(2^{m}-1)$, and let $m$ be a positive integer such that $2^{m}-1$ is composite. It follows that $2^{m-1}$ is even and $2^{m}-1$ is odd. Therefore, $(2^{m-1},2^{m}-1)=1$, and $\sigma(n)=\sigma(2^{m-1}(2^{m}-1))=\sigma(2^{m-1})\sigma(2^{m}-1)=(2^{m}-1)\sigma(2^{m}-1)$.

However, I get stuck here. I know that all I am left to do is show that $\sigma(2^{m}-1)>2^{m}$, but I do not see how I could accomplish that (since $2^{m}-1$ is composite). What do you guys think?

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I just had an idea: Could I perhaps point out that $\sigma(2^{m}-1)=1+\cdots +2^{m}-1=2^{m}+\cdots$, which implies that $\sigma(2^{m}-1)\geq2^{m}$, but since $n$ is not a perfect number, it must then be that $\sigma(2^{m}-1)>2^{m}$? – EsX_Raptor Feb 17 '12 at 15:38
up vote 2 down vote accepted

Recall that $\sigma(n)$ is the sum of the divisors of $n$, and that $n$ is abundant if $\sigma(n)>2n$.

There is a related function $s(n)$, which is the sum of the divisors of $n$ other than $n$ itself. Then $n$ is abundant if $s(n)>n$. We work with $\sigma(n)$, because it is more nicely behaved than $s(n)$.

Suppose that $n=2^{m-1}(2^m-1)$, where $2^m-1$ is composite. Then because $2$ is the only possible prime divisor of $2^{m-1}$, and $2^m-1$ is odd, we conclude that the two numbers are relatively prime. (You sort of said it is because $2^{m-1}$ is even, but that's not true, we can have two numbers, one even and one odd, which are not relatively prime, like $6$ and $9$.)

By the multiplicativity of $\sigma$, we have $$\sigma(n)=\sigma(2^{m-1})\sigma(2^m-1)=(2^m-1)\sigma(2^m-1).\qquad(\ast)$$ Since $2^m-1$ is composite, $2^m-1$ has the divisors $1$, $2^m-1$, and at least one additional divisor. So $\sigma(2^m-1)>2^m$. It follows from $(\ast)$ that $$\sigma(n)>(2^m-1)2^m=2n,$$ and thus $n$ is abundant.

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