The title says it all.
If $N = q^k n^2$ is an odd perfect number, is it possible to have $I(n^2) = I(q^k) + c$, for some constant $c > 0$?
Here $I(x)$ is defined to be the ratio $$I(x) = \dfrac{\sigma(x)}{x} = \dfrac{\sum_{d \mid x}{d}}{x}.$$
Note that we have $$I(q^k)I(n^2) = I(q^k n^2) = 2$$ since $\sigma(x)$ is weakly multiplicative and $N = q^k n^2$ is perfect.
My Attempt
Since the Euler prime $q$ satisfies $q \equiv 1 \pmod 4$, we have $q \geq 5$.
In particular, this means that $$1 < \dfrac{q + 1}{q} = I(q) \leq I(q^k) < \dfrac{q}{q - 1} \leq \dfrac{5}{4}$$ from which it follows that $$\dfrac{8}{5} < I(n^2) = \dfrac{2}{I(q^k)} < 2.$$
Suppose that we have the equation $$I(n^2) = I(q^k) + c.$$ Then we have $$\dfrac{7}{20} = \dfrac{8}{5} - \dfrac{5}{4} < c = I(n^2) - I(q^k) < 2 - 1 = 1.$$ Also, from the equation $$I(n^2) = I(q^k) + c,$$ we get $$I(q^k)I(n^2) = \left(I(q^k)\right)^2 + c\cdot{I(q^k)}$$ by multiplying both sides by $I(q^k)$. We also get $$\left(I(n^2)\right)^2 = I(q^k)I(n^2) + c\cdot{I(n^2)}$$ by multiplying both sides by $I(n^2)$. We get the equations $$\left(I(q^k)\right)^2 + c\cdot{I(q^k)} - 2 = 0$$ and $$\left(I(n^2)\right)^2 - c\cdot{I(n^2)} - 2 = 0.$$ We now use the quadratic formula to solve these two equations. From the first equation, we get that $$I(q^k) = \dfrac{-c \pm \sqrt{c^2 + 8}}{2}$$ while from the second equation, we obtain $$I(n^2) = \dfrac{c \pm \sqrt{c^2 + 8}}{2}.$$
Since $7/20 < c < 1$, we have that
(1) $$-2 = \dfrac{-1 - 3}{2} = \dfrac{-1 - \sqrt{(1)^2 + 8}}{2} < I({q_1}^{k_1}) = \dfrac{-c - \sqrt{c^2 + 8}}{2} < \dfrac{-(7/20) - \sqrt{(7/20)^2 + 8}}{2} = \dfrac{-(7/20) -(57/20)}{2} = \dfrac{-64}{40} = -\dfrac{8}{5},$$ which contradicts $1 < I({q_1}^{k_1}) < 5/4$. Hence, $$I(q^k) \neq \dfrac{-c - \sqrt{c^2 + 8}}{2}.$$
(2) $$\dfrac{37}{40} = \dfrac{-1 + (57/20)}{2} = \dfrac{-1 + \sqrt{(7/20)^2 + 8}}{2} < I({q_2}^{k_2}) = \dfrac{-c + \sqrt{c^2 + 8}}{2} < \dfrac{-(7/20) + \sqrt{(1)^2 + 8}}{2} = \dfrac{(-7/20) + 3}{2} = \dfrac{53}{40}$$ which does not contradict $1 < I({q_2}^{k_2}) < 5/4$. Hence, $$I(q^k) = \dfrac{-c + \sqrt{c^2 + 8}}{2}.$$
(3) $$\dfrac{-53}{40} = \dfrac{(7/20) - 3}{2} = \dfrac{(7/20) - \sqrt{(1)^2 + 8}}{2} < I({n_1}^2) = \dfrac{c - \sqrt{c^2 + 8}}{2} < \dfrac{1 - \sqrt{(7/20)^2 + 8}}{2} = \dfrac{1 - (57/20)}{2} = -\dfrac{37}{40}$$ which contradicts $8/5 < I({n_1}^2) < 2$. Hence, $$I(n^2) \neq \dfrac{c - \sqrt{c^2 + 8}}{2}.$$
(4) $$\dfrac{8}{5} = \dfrac{64}{40} = \dfrac{(7/20) + (57/20)}{2} = \dfrac{(7/20) + \sqrt{(7/20)^2 + 8}}{2}< I({n_1}^2) = \dfrac{c + \sqrt{c^2 + 8}}{2} < \dfrac{1 + \sqrt{(1)^2 + 8}}{2} = \dfrac{1 + 3}{2} = 2$$ which does not contradict $8/5 < I({n_2}^2) < 2$. Hence, $$I(n^2) = \dfrac{c + \sqrt{c^2 + 8}}{2}.$$
We now check: $$2 = I(q^k n^2) = I(q^k)I(n^2) = \left(\dfrac{-c + \sqrt{c^2 + 8}}{2}\right)\cdot\left(\dfrac{c + \sqrt{c^2 + 8}}{2}\right) = \dfrac{\left(\sqrt{c^2 + 8}\right)^2 - c^2}{4} = \dfrac{|c^2 + 8| - c^2}{4} = \dfrac{(c^2 + 8) - c^2}{4} = \dfrac{8}{4} = 2,$$ whence there is no contradiction.
Here is my question:
Will it be possible to derive a contradiction from assuming $$I(q^k) = \dfrac{-c + \sqrt{c^2 + 8}}{2}$$ and $$I(n^2) = \dfrac{c + \sqrt{c^2 + 8}}{2},$$ where $7/20 < c = I(n^2) - I(q^k) < 1$ and $c$ is a constant?
In general, we have $$c = I(n^2) - I(q^k) = \dfrac{2}{I(q^k)} - I(q^k) = \dfrac{2 - \left(I(q^k)\right)^2}{I(q^k)} = \dfrac{2 - \left(\dfrac{q^{k+1} - 1}{{q^k}(q-1)}\right)^2}{\dfrac{q^{k+1} - 1}{{q^k}(q-1)}}$$ which is not independent of $k$ and $q$. Hence, the difference $I(n^2) - I(q^k)$ is not constant.
In particular, even if $k=1$, we have $$c' = I(n^2) - I(q) = \dfrac{2 - \left(\dfrac{q^{2} - 1}{{q}(q-1)}\right)^2}{\dfrac{q^{2} - 1}{{q}(q-1)}} = \dfrac{q^2 - 2q - 1}{q(q+1)},$$ which is still dependent on the Euler prime $q$.
In the same manner, for the case of even perfect numbers ${2^{p-1}}(2^p - 1)$, we have $$d = I(2^{p-1}) - I(2^p - 1) = 2 - \dfrac{1}{2^{p-1}} - \left(1 + \dfrac{1}{2^p - 1}\right) = 1 - \dfrac{1}{2^{p-1}} - \dfrac{1}{2^p - 1},$$ which is dependent on $p$.