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I have asked this question in MSE a few weeks back, but did not receive any responses. I have cross-posted it to MO, hoping that it is appropriate for this site.

Let $\sigma(x)$ be the (classical) sum of the divisors of $x$.

A number $N \in \mathbb{N}$ is called perfect if $\sigma(N)=2N$.

An even perfect number $U$ is said to be given in Euclidean form if $U=(2^p - 1){2^{p-1}}$ (where $2^p - 1$ is called the Mersenne prime). On the other hand, an odd perfect number $L$ is said to be given in Eulerian form if $L = {q^k}{n^2}$ (where $q$ is called the Euler prime).

Notice that for even perfect numbers, trivially we have $1 < 2^p - 1$ (where $1$ is the exponent of the Mersenne prime $2^p - 1$).

Does a similar statement hold for odd perfect numbers? That is, does $k < q$ always hold?

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  • $\begingroup$ Note that the statement is true when $n < q$, as then we have $1 = k < n < q$. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Apr 25, 2015 at 4:11
  • $\begingroup$ You left out the essential condition that $k$ is odd. $\endgroup$
    – Richard Stanley
    Commented Apr 25, 2015 at 13:34
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    $\begingroup$ Since squares are not perfect, I think that it is implicit that $k$ must be odd. (In fact, it is known that $k \equiv 1 \pmod 4$ holds.) $\endgroup$
    – Jose Arnaldo Bebita
    Commented May 15, 2015 at 6:21
  • $\begingroup$ Trivially, $k < q$ always holds when $k = 1$. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Jul 19, 2016 at 14:17
  • $\begingroup$ However, Brown has recently announced a proof for the inequality $q < n$. $\endgroup$
    – Jose Arnaldo Bebita
    Commented Jul 19, 2016 at 14:22

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