# 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 $p$.

• For starters, WolframAlpha says that $$\dfrac{(x+1)+\sqrt{x^2 - 6x + 1}}{2x}$$ has no global maxima in the domain $0 < x < 3/20$. – Jose Arnaldo Bebita-Dris May 23 '18 at 14:43
• Additionally, WolframAlpha says that $$\dfrac{(x+1) - \sqrt{x^2 - 6x + 1}}{2x}$$ has no global minima in the domain $0 < x < 3/20$. – Jose Arnaldo Bebita-Dris May 23 '18 at 14:46

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 $x$.
To get an approximation to $\dfrac{(x+1) + \sqrt{x^2 - 6x + 1}}{2x}$ you can write $$\dfrac{(x+1) + \sqrt{x^2 - 6x + 1}}{2x}=\frac 12+\frac 1{2x}+\frac {\sqrt{(x-3)^2-8}}{2x}\\=\frac 12+\frac 1{2x}+\frac 12\sqrt{\left(1-\frac 3x\right)^2-\frac 8{x^2}}$$ The last term forces $0 \lt x \lt 3-2\sqrt 2$ or $x \gt 3+2\sqrt 2$ When $x$ gets small this blows up like $\frac 1x$. When $x$ gets large the square root is less than and close to $1$, so we can bound this by $1+\frac 1{2x}$
• Just a minor comment, @RossMillikan: Shouldn't that be $$\dfrac{(x+1)+\sqrt{x^2 - 6x + 1}}{2x} = \dfrac{1}{2}\cdot\bigg(1+\dfrac{1}{x}\bigg) + \dfrac{\sqrt{(x-3)^2 - 8}}{2x}?$$ – Jose Arnaldo Bebita-Dris May 24 '18 at 11:39
• @JoseArnaldoBebitaDris: Yes, I dropped a factor $2$ from the first two terms. Fixed. – Ross Millikan May 24 '18 at 13:01