This question is an offshoot of this earlier MSE question.
Let \sigma(z) denote the sum of divisors of z \in \mathbb{N}, the set of positive integers. Denote the abundancy index of z by I(z) := \sigma(z)/z.
If N={p^k}{m^2} is an odd perfect number with Euler prime p satisfying p \equiv k \equiv 1 \pmod 4 and \gcd(p,m)=1, then it is somewhat trivial to prove that
3 - \frac{p - 2}{p(p-1)} = \frac{3p^2 - 4p + 2}{p(p-1)} < I(p^k) + I(m^2)
and
I(p^k) + I(m^2) \leq \frac{3p^2 + 2p + 1}{p(p+1)} = 3 - \frac{p - 1}{p(p+1)}.
Now, setting x := 3 - \bigg(I(p^k) + I(m^2)\bigg), we have the simultaneous inequalities
\frac{p-1}{p(p+1)} \leq x < \frac{p - 2}{p(p - 1)}
resulting in the inequalities
\bigg((p - 2) > xp(p-1)\bigg) \land \bigg((p - 1) \leq xp(p+1)\bigg).
Notice that it is known that
\frac{57}{20} < I(p^k)+I(m^2) < 3
so that we know that
0 < x < \frac{3}{20}.
We now solve the inequalities one by one.
Solution is
\frac{(x+1) - \sqrt{x^2 - 6x + 1}}{2x} < p < \frac{(x+1) + \sqrt{x^2 - 6x + 1}}{2x}
Solution is
p \in \bigg(-\infty, \frac{1 - x - \sqrt{x^2 - 6x + 1}}{2x}\bigg] \bigcup \bigg[\frac{1 - x + \sqrt{x^2 - 6x + 1}}{2x},\infty\bigg)
Now, this is where the computations start to get messy. Can I ask for some help?
Basically, I would like to get numerical (lower [and upper?]) bounds for p.
It is easily seen that
\frac{(x+1) - \sqrt{x^2 - 6x + 1}}{2x} < 2, \frac{1 - x - \sqrt{x^2 - 6x + 1}}{2x} < 2
so your conditions reduces to
\frac{1 - x + \sqrt{x^2 - 6x + 1}}{2x} \leq p < \frac{1+x + \sqrt{x^2 - 6x + 1}}{2x}
then, one can see that the lower bound and upper bound differs by 1 and both are unbounded in the given range of x. That is, there can be only one prime p for any given value x.
Related Paper: The abundancy index of divisors of odd perfect numbers - Part III, http://nntdm.net/volume-23-2017/number-3/53-59/
