Odd Perfect Number An odd perfect number n is of the form $n=q^{k}m^{2}$ where $q$ is prime and both $q,k \equiv 1 \mod 4$. Also, n satisfies $\sigma (n)=2n$ so that $\sigma (q^{k}m^{2})=2q^{k}m^{2}$. 
My questions are about $q,k$. 
1) Is it known whether it is possible that $k>q$ ? and also if so
2) Can $k=a*q$ where is a positive integer?
 A: We can't answer these questions at present. We can't rule out almost any specific choice of this sort, such as $k=1$ and $q=5$. That said, we can rule out a few choices of $q$ and $k$. For example, we can rule out $q=1049$ using that $q+1|2n$, and so if $1049=q$ then $105|n$ but it isn't hard to show that any number of the form in OP that is divisible by 105 must be abundant (since 9(5)(49) is abundant and any multiple of an abundant number is abundant). But other than examples of $k$ and $q$ where they force an abundant divisor of $n$, we can't say anything much. 
A: Let $n = q^k m^2$ be an odd perfect number with special / Euler prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,m)=1$.
Descartes, Frenicle, and subsequently Sorli conjectured that $k=1$ always holds for all odd perfect numbers.
If proven true, then the Descartes-Frenicle-Sorli Conjecture (on odd perfect numbers) would imply that:

(1) It is not possible to have $k > q$, as we would then have the contradiction
  $$1=k>q \geq 5.$$
(2) It cannot be the case that $k = aq$ (where $q \geq 5$ and $a$ is a positive integer).

As commented by Ed Pegg: "This is one of the famous unsolved math questions."
