# A question on (odd) perfect numbers

(Note: This has been cross-posted to MO.)

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 $k < q$ is true when $n < q$, as then we have $1 = k < n < q$. However, Brown has recently announced a proof for the inequality $q < n$. – Jose Arnaldo Bebita-Dris Jul 19 '16 at 14:20
• Trivially, $k < q$ always holds when $k=1$. – Jose Arnaldo Bebita-Dris Jul 19 '16 at 14:20

(The following answer has also been posted to MO.)

This is only a partial answer. I just wanted to collect some of my recent thoughts on this problem.

Suppose to the contrary that $q \leq k$. Since $q \equiv k \equiv 1 \pmod 4$ and $q$ is prime, we have $q \geq 5$, which then implies that $k \geq 5$. (That is, $k=1$ does not hold when $q \leq k$.)

We compute bounds for the abundancy index $$I(q^k) = \dfrac{q^{k+1} - 1}{{q^k}(q - 1)} = \dfrac{q}{q - 1} - \dfrac{1}{q - 1}\cdot\bigg(\dfrac{1}{q}\bigg)^k$$ which is known to lie in the range $(1, 5/4)$ when $k \neq 1$. (Details are in this MSE question.) The abundancy index of $q^k$ has no global maxima nor global minima for $q \leq k$. Hence we proceed as follows:

We have the simultaneous inequalities (1) $\dfrac{k}{k - 1} - \dfrac{1}{q - 1}\cdot\bigg(\dfrac{1}{q}\bigg)^k \leq I(q^k) < \dfrac{5}{4}$ and (2) $1 < I(q^k) \leq \dfrac{q}{q - 1} - \dfrac{1}{k - 1}\cdot\bigg(\dfrac{1}{k}\bigg)^k,$ which can then be solved for $q$ in terms of $k$.

I tried using WolframAlpha, but it is returning a "Standard computation time exceeded..." message on both inequalities (1) and (2).

On the other hand, if we just assume $q = k$, then we still have that $k=1$ does not hold, as well as the stronger bound $$I(q^q) = \dfrac{q^{q+1} - 1}{{q^q}(q - 1)} \leq \dfrac{3906}{3125} = 1.24992.$$ (Computation of this upper bound in WolframAlpha here.)

Added March 04 2017 (Manila time)

Following the methodology in this paper, we get the lower bound $$I(q^q) + I(n^2) \geq 2\cdot\dfrac{3125}{3906} + \dfrac{3906}{3125} = \dfrac{17394043}{6103125} \approx 2.850022406554.$$

Since $k \neq 1$, we have the upper bound $$\dfrac{3q^2 + 2q + 1}{q(q+1)} = 3 - \dfrac{q - 1}{q(q+1)} > I(q^q) + I(n^2).$$

WolframAlpha then gives the trivial lower bound $$q > \dfrac{3125}{781} \approx 4.0012804.$$