On a conjectured relationship between the least prime factor and the Euler prime of an odd perfect number (Note: This question has been cross-posted to MO.)
Let $\sigma$ be the classical sum-of-divisors function.  For example,
$$\sigma(6)=1+2+3+6=12={2}\cdot{6}.$$
If $\sigma(N)=2N$ and $N$ is odd, then $N$ is called an odd perfect number.  Currently, the problem of the existence of odd perfect numbers is still open.
Euler proved that an odd perfect number $N$ must have the form
$$N=q^k n^2,$$
where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Dris [2008] showed that the inequality 
$$\sigma(q^k) \leq \dfrac{2n^2}{3}$$
holds, and subsequent improvements were given by other authors (Dris, Luca (2016), Chen, Chen (2012), Broughan, Delbourgo, Zhou (2013), and Chen, Chen (2014)).
Let $p_1$ be the least prime factor of the odd perfect number $N$ with Euler prime $q$.  It is relatively easy to show that $p_1 \neq q$. Suppose to the contrary that the Euler prime $q$ is also the least prime factor of the odd perfect number $N$.  Since $\sigma(q) \mid \sigma(q^k) \mid \sigma(N) = 2N$, we get that
$$\dfrac{q+1}{2} = \dfrac{\sigma(q)}{2} \mid N,$$
so that $(q+1)/2$ divides $N$.  If $(q+1)/2$ is prime, then this will contradict the assumption that $q$ is the least prime factor of $N$ (as $(q+1)/2 < q$).  If $(q+1)/2$ is composite, then $(q+1)/2$ will have a prime factor (necessarily smaller) that likewise divides $N$. Again, this is a contradiction.  QED
Therefore, $p_1 < q$.
Here is my question:

Will it be possible to either prove or rule out the following equality, using the current state of knowledge that we have on odd perfect numbers?
  $$\dfrac{q+1}{2} = {p_1}$$

 A: This is a partial answer, and takes off from the hint in user243301's comment here.

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Let
$$n = \prod_{i=1}^{\omega(n)}{{p_i}^{\alpha_i}}$$
be the canonical prime factorization of $n$.  (That is, all of the $\alpha_i$'s are positive, and $p_1 < p_2 < \ldots < p_{\omega(n)}$ are prime numbers.)
Suppose to the contrary that
$$\frac{q+1}{2}=p_1.$$
user243301 obtained the following equation:
$$\sigma(q^k) + \frac{q}{q-1}\sum_{j=0}^{k-1}\binom{k}{j}q^j = \frac{q(2^k {p_1}^k) - 1}{q - 1},$$
where $\sigma(x)=\sigma_1(x)$ is the classical sum of divisors of the positive integer $x$.
However, we know that we have the upper bound
$$\frac{\sigma(q^k)}{q^k} < \frac{q}{q-1}$$
since $q$ is prime.  This means that
$$\frac{q(2^k {p_1}^k) - 1}{q - 1} = \sigma(q^k) + \frac{q}{q-1}\sum_{j=0}^{k-1}\binom{k}{j}q^j > \sigma(q^k)\left(1 + \sum_{j=0}^{k-1}\binom{k}{j}q^{j-k}\right)$$
$$> \sigma(q^k)\cdot\left(\sum_{j=0}^{k}q^{j-k}\right) = \sigma(q^k)\cdot{\frac{\sigma(q^k)}{q^k}}.$$
This implies that
$$\frac{q(2^k {p_1}^k) - 1}{q^k (q - 1)} > \left(\frac{\sigma(q^k)}{q^k}\right)^2 \geq \left(\frac{q+1}{q}\right)^2.$$
But the fraction $q/(q - 1)$ is bounded from above by
$$\frac{q}{q - 1} \leq \frac{5}{4}$$.
Hence, it follows that
$$\left(\frac{q+1}{q}\right)^2 < \frac{q(2^k {p_1}^k) - 1}{q^k (q - 1)} < \frac{q(2^k {p_1}^k)}{q^k (q - 1)} \leq \frac{5}{4}\cdot\frac{2^k {p_1}^k}{q^k} = \frac{5}{4}\cdot\left(\frac{2p_1}{q}\right)^k = \frac{5}{4}\cdot\left(\frac{q+1}{q}\right)^k,$$
whence the inequality
$$\left(\frac{q+1}{q}\right)^{2-k} < \frac{5}{4},$$
or equivalently the inequality
$$\left(\frac{q}{q+1}\right)^{k-2} < \frac{5}{4}, \tag{1}$$
holds.
WolframAlpha then gives the following inequality plot:

and the following real solutions (???) to Inequality $(1)$:
$$k > 2, q = 0$$
$$k = n_1 + 2, -1 < q < -\frac{\sqrt[n_1]{5}}{2^{2/{n_1}} + \sqrt[n_1]{5}}, n_1 \in \mathbb{Z}, n_1 \leq -1$$
$$k = n_1 + 2, -\frac{\sqrt[n_1]{5}}{2^{2/{n_1}} + \sqrt[n_1]{5}} < q < 0, n_1 \in \mathbb{Z}, n_1 \geq 1$$
$$k = n_2 + 2, -1 < q < 0, n_2 \in \mathbb{Z}$$
$$k < 2, q < -1.$$
(Note that these real solutions violate $q \geq 5$.)
But the minimum value for $q/(q+1)$ occurs at $5/6$ (for $q=5$), hence
$$\left(\frac{5}{6}\right)^{k-2} \leq \left(\frac{q}{q+1}\right)^{k-2} < \frac{5}{4}. \tag{2}$$
Solving Inequality $(2)$ for $k$ yields
$$k > \dfrac{\log\left(\dfrac{144}{125}\right)}{\log\left(\dfrac{6}{5}\right)} \approx 0.776099,$$
which does not rule out $k=1$.

(This answer is currently under construction.)
