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?
 A: (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
$$q > \dfrac{3125}{781} \approx 4.0012804.$$
