In what follows, we let $N = r^s u^2$ be an odd perfect number given in *Eulerian form*, i.e. $r$ is the *special prime* satisfying $r \equiv s \equiv 1 \pmod 4$ and $\gcd(r,u)=1$.  In this [preprint](https://arxiv.org/abs/1602.01591), Brown claimed a proof for the estimate $r < u$.  Starni, on the other hand, proved the inequality $r < u$ using a different method in this [paper](https://nntdm.net/volume-24-2018/number-1/5-9/).

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Say a perfect number $M$ is *Euclidean* if $M = m_1 m_2 \cdots m_j$, where the factors are pairwise coprime, $j > 1$, $\sigma(m_i) = m_{i+1}$ for $i < j$, and $\sigma(m_j) = 2 m_1$.

We give them this name since Euclid gave a formula for even perfect numbers of this form (with $j = 2$).  As we know, Euler showed that all even perfect numbers have this form with $j = 2$.

Consider now *Euclidean odd perfect numbers*.  It is not hard to prove the conjecture that $r < u$ for *Euclidean odd perfect numbers* in case $j=2$ or in case $j > 3$.  However,
the case $j=3$ seems hard.  Say $M = q m^2 p^{2a}$, where the $3$ factors are pairwise coprime, and $p, q$ are primes.  We might have 
$$\sigma(m^2) = p^{2a},$$ 
$$\sigma(p^{2a}) = q,$$
and
$$\sigma(q) = 2m^2.$$
For such an odd perfect number $M$ we would have $q > p^{2a} > m^2$, so that $q^2 > m^2 p^{2a}$, which gives $q > m p^a$.

Hence, to prove the conjecture $r < u$, it would seem that one has to rule out this kind of *Euclidean odd perfect number* with $j=3$.

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Here is our:
> **QUESTION:** Do you see a way of ruling out the following system of equations?
$$\begin{cases}
{
\sigma(m^2) = p^{2a} \\ 
\sigma(p^{2a}) = q \\
\sigma(q) = 2m^2
}
\end{cases}$$

If my hunch is correct, one needs to concentrate on $\sigma(p^{2a}) = q$, to get
$$\sigma(p^{2a})=\frac{p^{2a+1} - 1}{p - 1}=q. \tag{*}$$

However, I am currently unfamiliar with methods on how to solve Equation $(*)$.