# Help with “A Simpler Dense Proof regarding the Abundancy Index.”

arithmetic-functionsdivisor-sumnumber theoryperfect numbers

I'm reading Richard Ryan's article "A Simpler Dense Proof regarding the Abundancy Index" and got stuck in his proof for Theorem 2. The Theorem is stated as follows:

Suppose we have a fraction of the form $$\frac{2n-1}{n}$$, where $$2n-1$$ is prime.

(i) …

(ii) If $$n$$ is odd and $$I(b)=\frac{2n-1}{n}$$ for some $$b$$, then $$b$$ is odd; moreover, if $$2n-1$$ does not divide $$b$$, then $$b(2n-1)$$ is a perfect number.

Ryan's proof: Suppose that $$n$$ is odd and $$b$$ is even. Let $$m$$ be the greatest integer such that $$2^m$$ divides $$b$$. Once again, there is a prime factor, $$q$$, of $$\sigma(2^m)$$ that divides $$b$$. Thus $$I(b) > I(2^mq) > 2$$ and we have a contradiction. Finally, if the prime number $$2n – 1$$ does not divide $$b$$ then, since $$I$$ is multiplicative, $$I(b(2n – 1)) = 2$$.

The 'once again' comes from (i): … let $$I(b)=\frac{2n-1}{n}$$ for even $$b$$ and $$m$$ be the greatest integer such that $$2^m$$ divides $$b$$. There is a prime factor, $$q$$, of $$\sigma(2^m)$$ that
also divides $$b$$ since $$\sigma(2^m) = 2^{m+1} – 1 \neq 2n – 1$$. Thus

$$I(b) > I(2^mq) \geq \frac{2^{m+1}-1}{2^m}\cdot\frac{2^{m+1}}{2^{m+1}-1}=2$$,

which contradicts $$I(b)=\frac{2n-1}{n}$$.

I see that $$\sigma(2^m)=2^{m+1}-1\neq 2n-1$$, because $$n$$ is odd and thus $$2n$$ is not a power of two. It is also possible to easily calculate, that $$b\neq2^m$$ so $$b$$ must have some other factor in addition to $$2^m$$ and that factor has a prime factor, $$p$$, for which $$2^mp$$ divides $$b$$. But Ryan writes that $$\sigma(2^m)$$ must have the prime factor $$q$$ and arrives at a contradiction through that. Could someone explain this to me? Also, I don't understand how $$I(q)\geq\frac{2^{m+1}}{2^{m+1}-1}$$

EDIT: As Jose pointed out, the answer below is incorrect. I think I made some progress in the right direction below.

$$I(b)=\frac{\sigma(b)}{b}=\frac{2n-1}{n}\implies n\sigma(b)=b(2n-1)\implies n\sigma(2^m)\sigma(a)=b(2n-1)$$, because $$b$$ is even but not a power of two. Now $$\sigma(2^m)=2^{m+1}-1\neq2n-1$$ from Ryan's proof makes sense: $$\sigma(2^m)$$ divides b so it must have the prime factor $$q$$, which divides $$b$$. Thus

$$I(b)>I(2^mq)=\frac{2^{m+1}-1}{2^m}\cdot\frac{q+1}{q}$$. However, I still do not understand how $$I(q)=\frac{\sigma(q)}{q}=\frac{q+1}{q}\geq\frac{2^{m+1}}{2^{m+1}-1}$$.

Abundancy Index $$I$$ is defined as $$I(n)=\frac{\sigma(n)}{n}$$, where $$\sigma$$ counts the sum of divisors. Both $$I$$ and $$\sigma$$ are multiplicative.

EDIT: There's no tag for Abundancy Index.

Hint: Since $$q \mid \sigma(2^m)$$, then $$q \leq 2^{m+1} - 1,$$ which implies that $$\frac{1}{q} \geq \frac{1}{2^{m+1} - 1}.$$