Q:

Would like to get numerical (lower [and upper?]) bounds for $p$

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
$$3 - \dfrac{p - 2}{p(p-1)} = \dfrac{3p^2 - 4p + 2}{p(p-1)} < I(p^k) + I(m^2) \leq \dfrac{3p^2 + 2p + 1}{p(p+1)} = 3 - \frac{p - 1}{p(p+1)}.$$
Now, setting $x := 3 - \bigg(I(p^k) + I(m^2)\bigg)$, we have the simultaneous inequalities
$$\dfrac{p-1}{p(p+1)} \leq x < \dfrac{p - 2}{p(p - 1)}$$
resulting in the inequalities
$$\begin{cases}
{
(p - 2) > xp(p-1) \\
(p - 1) \leq xp(p+1).
}
\end{cases}$$
Notice that it is known that
$$\dfrac{57}{20} < I(p^k)+I(m^2) < 3$$
so that we know that
$$0 < x < \dfrac{3}{20}.$$
We now solve the inequalities one by one.
WolframAlpha computation for $(p - 2) > xp(p - 1)$

Solution is
  $$\dfrac{(x+1) - \sqrt{x^2 - 6x + 1}}{2x} < p < \dfrac{(x+1) + \sqrt{x^2 - 6x + 1}}{2x}$$

WolframAlpha computation for $(p - 1) \leq xp(p + 1)$

Solution is
  $$p \in \bigg(-\infty, \dfrac{1 - x - \sqrt{x^2 - 6x + 1}}{2x}\bigg] \bigcup \bigg[\dfrac{1 - x + \sqrt{x^2 - 6x + 1}}{2x},\infty\bigg)$$

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 $p$.

A:

It is easily seen that$$\dfrac{(x+1) - \sqrt{x^2 - 6x + 1}}{2x}<2, \dfrac{1 - x - \sqrt{x^2 - 6x + 1}}{2x}<2$$so your conditions reduces to$$\dfrac{1 - x + \sqrt{x^2 - 6x + 1}}{2x}\le p<\dfrac{1+x + \sqrt{x^2 - 6x + 1}}{2x}$$
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 $x$.

