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