# A question on (odd) perfect numbers

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?

• Note that the statement is true when $n < q$, as then we have $1 = k < n < q$. – Arnie Bebita-Dris Apr 25 '15 at 4:11
• You left out the essential condition that $k$ is odd. – Richard Stanley Apr 25 '15 at 13:34
• 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.) – Arnie Bebita-Dris May 15 '15 at 6:21
• Trivially, $k < q$ always holds when $k = 1$. – Arnie Bebita-Dris Jul 19 '16 at 14:17
• However, Brown has recently announced a proof for the inequality $q < n$. – Arnie Bebita-Dris Jul 19 '16 at 14:22