5

Hereinafter, we will denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.

The topic of odd perfect numbers likely needs no introduction.

Let $q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. We wish to show that $q < n^{2/5}$.

MY ATTEMPT

Suppose to the contrary that $n^{2/5} \leq q$. Note that $n^{2/5} \neq q$ because otherwise we obtain $n^2 = q^5$ which contradicts $\gcd(q,n)=1$. Hence, by assumption, we have $$n < q^{5/2}.$$

Since $q^k < n^2$ (JIS - Dris, 2012), then note that this implies that $$q^k < n^2 < q^5$$ $$k < 5,$$ which forces $k=1$ since $k \equiv 1 \pmod 4$.

However, from this answer to a closely related question, we have the inequality $$1 = k < K < \frac{\log 2}{\log q} + 2\log_q {n} - \frac{\log \rho}{\log q}, \tag{1}$$ where the quantity $\rho$ satisfies $$\rho < \frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \gcd(n^2, \sigma(n^2))$$ (Note that we may take $\rho > 3374$ by this comment.)

We then get the estimate $$1 < K < \frac{\log 2}{\log q} + 2\log_q {n} - \frac{\log \rho}{\log q} < \frac{\log 2}{\log 5} + 2\log_q {q^{5/2}} - \frac{\log 3374}{\log 5}$$ $$= 5 + \frac{\log 2}{\log 5} - \frac{\log 3374}{\log 5} = \log_5\bigg(\frac{3125}{1687}\bigg) \approx 0.3830421, \tag{2}$$ which is a contradiction. Hence, we conclude that $q < n^{2/5}$.

Here are my:

QUESTIONS: Is this argument logically sound? (In particular, are the estimates in Steps (1) and (2) correct?) If not, how can it be mended so as to produce a valid proof?

Added from this recent comment: Just in case it is not clear - Another way to arrive at the contradiction in Estimate $(2)$ is as follows: $$K \log q < \log 2 + 2 \log n - \log \rho < \log 2 + 2 \log q^{5/2} - \log 3374$$ $$= \log 2 + 5 \log q - \log 3374 \implies (K - 5) \log q < \log 2 - \log 3374$$ $$\implies (K - 5) \log 5 \leq (K - 5) \log q < \log 2 - \log 3374 \implies K - 5 < \frac{\log 2 - \log 3374}{\log 5}$$ $$\implies K < 5 + \log_{5} 2 - \log_{5} 3374 = \log_{5} \left(\frac{3125}{1687}\right) \approx 0.3830421.$$

(Note that I am using the fact that the logarithm is a monotonically increasing function.)

Jose Arnaldo Bebita Dris
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  • Just in case it is not clear - Another way to arrive at the contradiction in Estimate $(2)$ is as follows: $$K \log q < \log 2 + 2 \log n - \log \rho < \log 2 + 2 \log q^{5/2} - \log 3374 = \log 2 + 5 \log q - \log 3374$$ $$\iff (K - 5)\log q < \log 2 - \log 3374 \implies (K - 5)\log 5 \leq (K - 5)\log q < \log 2 - \log 3374$$ $$\implies K - 5 < \frac{\log 2 - \log 3374}{\log 5} \implies K < 5 + \log_5 {2} - \log_5 {3374} = \log_5\left(\frac{3125}{1687}\right) \approx 0.3830421.$$ QED – Jose Arnaldo Bebita Dris Feb 23 '23 at 23:49
  • In the inequality $\dfrac{\log 2}{\log q} + 2\log_q {n} - \dfrac{\log \rho}{\log q} < \dfrac{\log 2}{\log 5} + 2\log_q {q^{5/2}} - \dfrac{\log 3374}{\log 5}$, it seems that you used $- \dfrac{\log \rho}{\log q} < - \dfrac{\log 3374}{\log 5}$, but this does not hold for $\rho=3375$ and $q=13$. – mathlove Feb 24 '23 at 11:42
  • Thank you for your time and attention, @mathlove! Please refer to my comment above for an alternative point of view. (Note that I am using the fact that $\log x$ is a monotonically increasing function of $x$.) – Jose Arnaldo Bebita Dris Feb 24 '23 at 12:40
  • In the inequality $(K−5)\log 5\leqslant (K−5)\log q$, you are assuming that $K-5\geqslant 0$. So, I think that what you've done is "Supposing that $n^{2/5}\leqslant q$ and $K\geqslant 5$ gives a contradiction". – mathlove Feb 24 '23 at 14:00
  • I do not need to make the assumption $K \geq 5$, @mathlove. – Jose Arnaldo Bebita Dris Feb 24 '23 at 14:09
  • If you don't assume $K\geqslant 5$, then the inequality $(K−5)\log 5\leqslant (K−5)\log q$ does not hold for $K=2$ and $q=13$. – mathlove Feb 24 '23 at 14:14
  • Hmm. I know that $1 = k < K$ (under the given assumption), and that $K$ "behaves" like $k \equiv 1 \pmod 4$. This may be taken to "imply" that $K \geq 5$ holds. – Jose Arnaldo Bebita Dris Feb 24 '23 at 14:23
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    Let $F(q):=\dfrac{\log 2 + 5 \log q - \log 3374}{\log q}$. Since $F'(q)>0$, we have $K <F(q) \lt \displaystyle\lim_{q\to\infty}F(q)=5$. So, $(K - 5) \log 5 \leqslant (K - 5) \log q$ does not hold. – mathlove Feb 25 '23 at 03:30
  • Thank you for the clarification, @mathlove! Please write out your comments into a full answer, so that I may be able to upvote and then accept. =) – Jose Arnaldo Bebita Dris Feb 25 '23 at 04:04

1 Answers1

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$$1 < K < \frac{\log 2}{\log q} + 2\log_q {n} - \frac{\log \rho}{\log q} < \frac{\log 2}{\log 5} + 2\log_q {q^{5/2}} - \frac{\log 3374}{\log 5}$$

It seems that you used $- \dfrac{\log \rho}{\log q} < - \dfrac{\log 3374}{\log 5}$, but this does not hold for $\rho=3375$ and $q=13$.

$$K \log q < \log 2 + 2 \log n - \log \rho < \log 2 + 2 \log q^{5/2} - \log 3374$$ $$= \log 2 + 5 \log q - \log 3374 \implies (K - 5) \log q < \log 2 - \log 3374$$ $$\implies (K - 5) \log 5 \leq (K - 5) \log q < \log 2 - \log 3374 $$

Let $F(q):=\dfrac{\log 2+5\log q−\log{3374}}{\log q}$.

Since $F′(q)>0$, we have $K<F(q)<\displaystyle\lim_{q\to\infty}F(q)=5$.

So, $(K−5)\log 5\leqslant (K−5)\log q$ does not hold.

mathlove
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