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, 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.

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 $j=3$.

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 $(*)$.

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, 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.

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 $j=3$.

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 $(*)$.