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I get $$\gcd(n^2,\sigma(n^2)) = \frac{n^2}{\sigma(q^k)/2} = \frac{\sigma(n^2)}{q^k} = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$ when $N = q^k n^2$ is an odd perfect number with special/Euler prime $q$ (satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$), which follows from $\sigma(N) = 2N$ and $\gcd(q^k, \sigma(q^k))=1$.

I also obtain $$\varphi(N) = \varphi(q^k n^2) = \varphi(q^k)\varphi(n^2) = q^{k-1} (q - 1)(n\varphi(n)).$$ Comparing with $\sigma(N)=2N$, I have $$\sigma(N) = \sigma(q^k n^2) = \sigma(q^k)\sigma(n^2) = 2 q^k n^2.$$ Dividing, we obtain $$\frac{\sigma(N)}{\varphi(N)} = \frac{2q^k n^2}{q^{k-1} (q - 1)(n\varphi(n))} = 2\cdot\frac{q}{q-1}\cdot\frac{n}{\varphi(n)}.$$

Therefore, we have the equation: $$(q-1)\varphi(n)\cdot\sigma(N) = 2qn\varphi(N).$$

Note that, since $q$ is the special prime, then $q \equiv 1 \pmod 4$, which implies that $q \geq 5$, from which we get $$\frac{q}{q - 1} \leq \frac{5}{4}.$$ Hence, we know that $$\frac{\varphi(n)}{n}\cdot\frac{\sigma(N)}{\varphi(N)} = 2\cdot\frac{q}{q-1} \leq \frac{5}{2} < 3.$$

In particular, $$\frac{\sigma(N)}{\varphi(N)}$$ is not an integer, as $(q - 1) \nmid 2qn$ (since $\gcd(q-1,q)=\gcd(q,n)=1$).

Is it possible to obtain a tight (numerical) upper bound for $\sigma(N)/\varphi(N)$ when $N=q^k n^2$ is an odd perfect number with special/Euler prime $q$?

Jose Arnaldo Bebita Dris
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  • Furthermore, $$\frac{\varphi(n)}{n}\cdot\frac{\sigma(N)/N}{\varphi(N)/N} = 2\cdot\frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = 2\cdot\frac{q}{q-1} \implies \frac{\varphi(n)}{n}\cdot\frac{N}{\varphi(N)} = \frac{q}{q-1} \leq \frac{5}{4}.$$ – Jose Arnaldo Bebita Dris Jul 05 '19 at 20:29
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    I'm sure it is clear in your own thinking, but it would be helpful to specify (perhaps parenthetically) in your lead sentence that $q$ is a prime. Moreover "special prime" has further requirements as in your earlier post. – hardmath Jul 05 '19 at 20:32
  • @JoseArnaldoBebita-Dris Your question title asks about an equation relating the values, but your text only looks at bounds. Please be clear in the question text itself what you're asking about, i.e., a specific equation (which you already have, of a sort) and/or upper bounds on the ratio. Thanks. – John Omielan Jul 05 '19 at 20:44
  • Sorry about that. Meant to include further details. Doing so now. – Jose Arnaldo Bebita Dris Jul 05 '19 at 20:52

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