On the equation $\frac{1}{q}=1-\frac{\varphi(N)/N}{\varphi(n)/n}$, where $N=q^k n^2$ is an odd perfect number with special prime $q$

Question

Preamble: This question is an offshoot of this earlier MSE post.

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$, and the Euler totient function of $x$ by $\varphi(x)$.

Let $N = q^k n^2$ be an odd perfect number given in Eulerian form. That is, $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. (Note that this implies that $q \geq 5$.)

From the following source: Advanced Problem H-661, On Odd Perfect Numbers, Proposed by J. L´opez Gonz´alez, Madrid, Spain and F. Luca, Mexico (Vol. 45, No. 4, November 2007), Fibonacci Quarterly, we have the bounds $$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{N}{\sigma(N)} = \frac{1}{2}.$$

However, we also have $$\frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}.$$ Notice that $$\frac{4}{5} \leq \frac{\varphi(q^k)}{q^k} = \frac{q^k \bigg(1 - \frac{1}{q}\bigg)}{q^k} = \frac{q - 1}{q} < 1.$$ Therefore, we have the bounds $$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{\varphi(n)}{n},$$ and $$\frac{4}{5}\cdot\frac{\varphi(n)}{n} \leq \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n} < \frac{1}{2},$$ which implies that $$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{5}{8}.$$

WolframAlpha gives the rational approximation $$\frac{120}{217\zeta(3)} \approx 0.4600409433626.$$

Note that from the equation $$\frac{\varphi(N)}{N}=\frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n},$$ we obtain $$\frac{\varphi(q^k)}{q^k}=\frac{q-1}{q}=1-\frac{1}{q}=\frac{\varphi(N)/N}{\varphi(n)/n},$$ from which we finally get $$q = \frac{1}{1 - \Bigg(\frac{\varphi(N)/N}{\varphi(n)/n}\Bigg)}.$$ But since we have obtained $$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{1}{2},$$ and $$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{5}{8},$$ then we get $$\frac{192}{217\zeta(3)} < \frac{\varphi(N)/N}{\varphi(n)/n} < \frac{217\zeta(3)}{240},$$ which implies that $$1 - \frac{217\zeta(3)}{240} < 1 - \frac{\varphi(N)/N}{\varphi(n)/n} < 1 - \frac{192}{217\zeta(3)},$$ where we have the rational approximations $$1 - \frac{217\zeta(3)}{240} \approx -0.086859783273466499715596587699936$$ and $$1 - \frac{192}{217\zeta(3)} \approx 0.26393449061983486641861638.$$

Note that we are sure that $$\frac{\varphi(n)}{n} > \frac{\varphi(N)}{N},$$ since otherwise we would get $$1 - \frac{\varphi(N)/N}{\varphi(n)/n} = \frac{1}{q} \leq 0,$$ which is a contradiction.

Consequently, $$0 < \frac{1}{q} = 1 - \frac{\varphi(N)/N}{\varphi(n)/n} < 1 - \frac{192}{217\zeta(3)},$$ so that $$q > \frac{217\zeta(3)}{217\zeta(3) - 192} \approx 3.788818951443435495481230888.$$

Here is my:

QUESTION: Do you see a way on how to improve the lower bound for $q$, given my approach in this post? If so, how could this be done?

MY ATTEMPT

Since $$\frac{1}{q} = 1 - \frac{\varphi(N)/N}{\varphi(n)/n}$$ and $q \geq 5 > 0$, then $$0 < \frac{1}{q} = 1 - \frac{\varphi(N)/N}{\varphi(n)/n}.$$ It follows that $$\frac{\varphi(N)/N}{\varphi(n)/n} < 1.$$

Alas, this is where I get stuck.

Answer 1

This is too long for a comment: I merely wanted to record some thoughts that recently occurred to me, which are related to the problem at hand.

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Since $$\dfrac{1}{q} = \dfrac{\dfrac{\varphi(n)}{n}-\dfrac{\varphi(N)}{N}}{\dfrac{\varphi(n)}{n}}$$ and $q$ is (the special) prime satisfying $q \equiv 1 \pmod 4$, then $q \geq 5$, so that $$\dfrac{1}{q} \leq \dfrac{1}{5},$$ which implies that $$\dfrac{\varphi(n)}{n}-\dfrac{\varphi(N)}{N} \leq \dfrac{1}{5}\cdot\dfrac{\varphi(n)}{n} < \dfrac{1}{5}\cdot\dfrac{5}{8} = \frac{1}{8},$$ since $\varphi(n)/n < 5/8$, from which it follows that $$\dfrac{\varphi(n)}{n} < \dfrac{\varphi(N)}{N} + \dfrac{1}{8} < \dfrac{1}{2}+\dfrac{1}{8} = \dfrac{5}{8},$$ which does not improve on the known upper bound $\varphi(n)/n < 5/8$.

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We also obtain $$\dfrac{\varphi(N)}{N} < \dfrac{\varphi(n)}{n} < \dfrac{5}{8},$$ which does not improve on the known upper bound $\varphi(N)/N < 1/2$.