The title says it all.
What is wrong with this proof that $q < n$, if $N = qn^2$ is an odd perfect number with Euler prime $q$ and $\gcd(q, n) = 1$?
Acquaah and Konyagin showed that $q < (3N)^{1/3}$. The following proof (communicated to me by Dr. Patrick Brown) is a modification of theirs to strengthen the result to show $q < n$. (Update (Feb 6 2016): Patrick's preprint containing a proof for $q < n$ (for all odd perfect numbers) as well as for $q^k < n$ (under some mild conditions) has appeared in the arXiv.)
For the three cases of the proof we write
$$N = q{p^{2b}}{{r_1}^{2\beta_1}}{{r_2}^{2\beta_2}}\cdots{{r_k}^{2\beta_k}}$$
where $p$ is the unique prime whereby $q \mid \sigma(p^{2b})$. When convenient we will let $w^2 = {{r_2}^{2\beta_2}}\cdots{{r_k}^{2\beta_k}}$, and write $N = q{p^{2b}}{{r_1}^{2\beta_1}}{w^2}$.
Case 1
$$q = \sigma(p^{2b})$$
Note that the assumption $q = \sigma(p^{2b})$ means $p \not{\mid} \sigma(q)$ since $q + 1 \equiv 2 \pmod p$. So we let $p^{c_i} || \sigma({r_i}^{2\beta_i})$ for $1 \leq i \leq k$. It is possible that $p^{c_i} = \sigma({r_i}^{2\beta_i})$ for any particular $i$, but since we know $N$ has at least ten components, at least one of the $\sigma({r_i}^{2\beta_i})$ has to have factors other than $p$. Thus we may rewrite subscripts and assume:
$$\sigma({r_1}^{2\beta_1}) = p^{c_1}{r_2}v$$
for $c_1 \geq 0$, where $v$ is equal to any other primes dividing $\sigma({r_1}^{2\beta_1})$, including multiplicities of $r_2$ should they appear.
We now have what we need to prove this case. Observe,
$$2N = \sigma(N) = \sigma(q)\sigma(p^{2b})\sigma({r_1}^{2\beta_1})\sigma(w^2).$$
Since $p \not{\mid} \sigma(q)$, then $p^{2b - c_1} \mid \sigma(w^2)$, thus
$$2N > (q + 1)q({p^{c_1}}{r_2})(p^{2b - c_1})$$
$$2N > {q^2}{r_2}{p^{2b}}$$
Now, $r_2$ being an odd prime means $r_2 \geq 3$. We also note that $p^{2b} > (2/3)\sigma(p^{2b})$.
Consequently,
$$2N > {q^2}(3)\frac{2}{3}\sigma(p^{2b})$$
$$N > q^3$$
from which it easily follows that $q < n$.
Case 2
$$q{r_1} \mid \sigma(p^{2b}), p \not{\mid} \sigma(q)$$
In this case, since $q < \sigma(p^{2b})$, then there is another prime dividing $\sigma(p^{2b})$. We assume without loss of generality that $r_1 \mid \sigma(p^{2b})$. This time, we set $w^2 = {{r_1}^{2\beta_1}}\cdots{{r_k}^{2\beta_k}}$. Observe that $p \not{\mid} \sigma(q)$ implies $p^{2b} || \sigma(w^2)$. This is all the machinery we need to prove $q < n$ for this case.
$$2N = \sigma(N) = \sigma(q)\sigma(p^{2b})\sigma(w^2)$$
$$2N > (q + 1)q(r_1)(p^{2b})$$
$$2N > {q^2}(r_1)\frac{2}{3}\sigma(p^{2b})$$
$$2N > {q^2}(r_1)\frac{2}{3}{r_1}q$$
$$2N > \frac{2}{3}{r_1}^2{q^3}$$
As above, $r_1 \geq 3$. Therefore,
$$N > 3q^3$$
and again, $q < n$ easily follows.
Case 3
$$q{r_1} \mid \sigma(p^{2b}), p \mid \sigma(q)$$
We borrow the same proof method Acquaah and Konyagin borrowed from Luca and Pomerance. We also would not really utilize the hypothesis that $r_1 \mid \sigma(p^{2b})$ as it would not buy us the extra factor we need. For that, we look back to case 1, and let $p^{c_i} || \sigma({r_1}^{2\beta_i})$, for $1 \leq i \leq k$ and $p^{c_q} || \sigma(q)$. Again, we assume without loss of generality that $\sigma({r_1}^{2\beta_1}) = p^{c_1}{r_2}v$ for $c_1 \geq 0$ as we did in case 1.
Let $u = \sigma(p^{2b})/q$. Since
$$\sigma(p^{2b}) \equiv 1 \pmod p, q \equiv -1 \pmod p$$
we know $u \equiv -1 \pmod p$. Since $u$ is odd we know $u \neq p - 1$ and thus $u \geq 2p - 1$.
By assumption, $c_q \geq 1$. For $w^2 = {{r_2}^{2\beta_2}}\cdots{{r_k}^{2\beta_k}}$, we have $p^{2b - c_q - c_1} || \sigma(w^2)$, which implies
$$\sigma(w^2) \geq p^{2b - c_q - c_1}.$$
Observe now,
$$p^{2b+1} - 1 = (p - 1)\sigma(p^{2b}) = (p - 1)uq = (p - 1)u\sigma(q) - (p - 1)u.$$
Therefore, $(p - 1)u \equiv 1 \pmod{p^{c_q}}$, which implies that $(p - 1)u > p^{c_q}$.
Combining inequalities yields,
$$\sigma(w^2)(p - 1)u > p^{2b - c_1} \Longrightarrow \sigma(w^2)u > \frac{p^{2b - c_1}}{p - 1}.$$
This should be all we need (for $w$ as defined in case 1):
$$2N = \sigma(N) = \sigma(q)\sigma(p^{2b})\sigma({r_1}^{2\beta_1})\sigma(w^2)$$
$$2N = (q + 1)uq(p^{c_1}{r_2}v)\sigma(w^2)$$
$$2N > {q^2}\frac{p^{2b - c_1}}{p - 1}{p^{c_1}{r_2}}$$
$$2N > {q^2}{r_2}\frac{p^{2b}}{p - 1}$$
$$2N > {q^2}{r_2}\frac{2\sigma(p^{2b})}{3(p - 1)}$$
$$2N > {q^2}{r_2}\frac{2uq}{3(p - 1)}$$
Recall that $u \geq 2p - 1$ and again $r_2$ being an odd prime means $r_2 \geq 3$.
$$2N > {q^3}(3)\frac{2}{3}\frac{2p - 1}{p - 1}$$
$$2N > {q^3}(3)\frac{2}{3}(2)$$
$$N > 2q^3$$
And again, we get $q < n$.
