Is it possible to improve on the bounds for $\varphi(N)/N$, if $N = q^k n^2$ is an odd perfect number with special prime $q$?

Question

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 $\gcd(q,n)=1$.

From a comment underneath this earlier question, we have the equation (and corresponding inequalities) $$1 < \frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1} \leq \frac{5}{4}$$ since $q$ is prime with $q \equiv 1 \pmod 4$ implies that $q \geq 5$. This implies that $$\frac{4}{5} \leq \dfrac{\dfrac{\varphi(N)}{N}}{\dfrac{\varphi(n)}{n}} = \frac{q-1}{q} < 1.$$

But 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 $$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{1}{2}.$$

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 $$\frac{120}{217\zeta(3)} \approx 0.4600409433626.$$

Here is my question:

Is it possible to improve on the bounds for $\varphi(N)/N$, if $N = q^k n^2$ is an odd perfect number with special prime $q$?

MOTIVATION FOR THE INQUIRY

It can be shown that the equation $$\frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1}$$ together with the bounds $$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} < \frac{1}{2}$$ and $$\frac{120}{217\zeta(3)} < \frac{\varphi(n)}{n} < \frac{5}{8}$$ imply $$0.92 \approx \dfrac{\dfrac{120}{217\zeta(3)}}{\dfrac{1}{2}} < \frac{q}{q-1} < \dfrac{\dfrac{5}{8}}{\dfrac{120}{217\zeta(3)}} \approx 1.358574729,$$ from which we obtain trivial bounds.

Nonetheless, it can be shown that the equation $$\frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1}$$ together with the upper bound $\varphi(N)/N < 1/2$ implies that $$q < \frac{x}{x-1}$$ where $$x = \frac{2\varphi(n)}{n}.$$ Thus, if we can improve the upper bound for $\varphi(N)/N$ to something smaller than $1/2$ (say $1/2 - \varepsilon$ for some tiny $\varepsilon > 0$), then we can improve the coefficient of $\frac{\varphi(n)}{n}$ in $x$ to some number bigger than $2$. Likewise, if we can get a better lower bound for $\varphi(N)/N$, then we will be able to get an improved lower bound for $\varphi(n)/n$. Together, they would translate (hopefully!) to a numerical upper bound for the special/Euler prime $q$!

Answer 1

Eureka!!!

Let $N = q^k n^2$ be an odd perfect number with special/Euler prime $q$.

From the equation and lower bound for $\varphi(N)/N$ $$\frac{120}{217\zeta(3)} < \frac{\varphi(N)}{N} = \frac{\varphi(q^k)}{q^k}\cdot\frac{\varphi(n)}{n}$$ and the equation $$\frac{\varphi(q^k)}{q^k} = \frac{q - 1}{q},$$ we get the lower bound $$2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1} < \frac{2\varphi(n)}{n} = x.$$ This implies that we have the upper bound $$q < \frac{x}{x-1} < \frac{2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}}{\bigg(2\cdot\frac{120}{217\zeta(3)}\cdot\frac{q}{q - 1}\bigg) - 1}$$ which can be solved using WolframAlpha, yielding the upper bound $$q < \frac{217\zeta(3)}{217\zeta(3) - 240} \approx 12.5128,$$ from which it follows that $q=5$, since $q$ is a prime satisfying $q \equiv 1 \pmod 4$. QED