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(Note: This question has been cross-posted to MO.)

This question is an offshoot of this earlier one and this other question.

Let $n = p^k m^2$ be an odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.

It was conjectured in Dris (2008) and Dris (2012) that the inequality $p^k < m$ holds.

Brown (2016) showed that the Dris Conjecture (that $p^k < m$) holds in many cases.

It is trivial to show that $m^2 - p^k \equiv 0 \pmod 4$. This means that $m^2 - p^k = 4z$, where it is known that $4z \geq {10}^{375}$. (See this MSE question and answer, where the case $m < p^k$ is considered.) Note that if $p^k < m$, then $$m^2 - p^k > m^2 - m = m(m - 1),$$ and that $${10}^{1500} < n = p^k m^2 < m^3$$ where the lower bound for the magnitude of the odd perfect number $n$ is due to Ochem and Rao (2012). This results in a larger lower bound for $m^2 - p^k$. Therefore, unconditionally, we have $$m^2 - p^k \geq {10}^{375}.$$ We now endeavor to disprove the Dris Conjecture.

Consider the following sample proof argument:

Theorem If $n = p^k m^2$ is an odd perfect number satisfying $m^2 - p^k = 8$, then $m < p^k$.

Proof

Let $p^k m^2$ be an odd perfect number satisfying $m^2 - p^k = 8$.

Then $$(m + 3)(m - 3) = m^2 - 9 = p^k - 1.$$

This implies that $(m + 3) \mid (p^k - 1)$, from which it follows that $$m < m + 3 \leq p^k - 1 < p^k.$$ We therefore conclude that $m < p^k$.

QED

So now consider the equation $m^2 - p^k = 4z$. Following our proof strategy, and the formula in the accepted answer to the first hyperlinked question, we have:

$$m^2 - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2 = p^k + \Bigg(4z - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg).$$

So the only remaining question now is whether it could be proved that $$\Bigg(4z - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg) = -y < 0$$ for some positive integer $y$?

In other words, is it possible to prove that it is always the case that $$\Bigg((m^2 - p^k) - \bigg(\lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\bigg)^2\Bigg) < 0,$$ if $n = p^k m^2$ is an odd perfect number with special prime $p$?

(Additionally, note that it is known that $m^2 - p^k$ is not a square, if $p^k m^2$ is an OPN with special prime $p$. See this MSE question and the answer contained therein.)

If so, it would follow that $$\Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg)\Bigg(m - \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) = p^k - y$$ which would imply that $$\Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) \mid (p^k - y)$$ from which it follows that $$m < \Bigg(m + \lfloor{\sqrt{m^2 - p^k} + \frac{1}{2}}\rfloor\Bigg) \leq p^k - y < p^k.$$

Update (November 11, 2020 - 10:21 PM Manila time) Please check out the recently posted answer for a minor adjustment to the logic that should make the general proof argument work.

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  • $\begingroup$ Are there any more restrictions on $p$ and $k$? Or can $p$ be any prime, and $k$ and $m$ any integers? I ask, because your sample proof doesn't use the fact that $p^km^2$ is perfect or odd. $\endgroup$
    – Servaes
    Commented Nov 11, 2020 at 11:08
  • $\begingroup$ Thank you for your comment, @Servaes! $p$ is the special prime of the odd perfect number $n = p^k m^2$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$. $\endgroup$ Commented Nov 11, 2020 at 11:11
  • $\begingroup$ You call it 'the' special prime; does every odd perfect number have precisely one such prime divisor? Do you have any helpful results on these special primes? I ask because I know absolutely nothing abouth such numbers, only that it is very easy to construct examples without the restriction that $p^km^2$ is perfect. So it would be nice to know what restriction implies for $p$ and $m$. $\endgroup$
    – Servaes
    Commented Nov 11, 2020 at 11:14
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    $\begingroup$ @Servaes: Yes, every odd perfect number (OPN) has exactly one (special) prime divisor $p$ occurring to an odd exponent $k \equiv 1 \pmod 4$. There are lots of results on the special prime of an OPN, the most notable of which, in my opinion, is due to Cohen and Sorli (2012). I just do not know whether Cohen and Sorli's results on the special prime of an OPN would be helpful for the purpose of resolving this question. $\endgroup$ Commented Nov 11, 2020 at 11:24
  • $\begingroup$ @Servaes: Additionally, note that it is known that $m^2 - p^k$ is not a square, if $p^k m^2$ is an OPN with special prime $p$. See this MSE question and the answer contained therein. $\endgroup$ Commented Nov 11, 2020 at 11:42

2 Answers 2

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If you don't have a proof that the smallest square larger than $m^2-p^k$ is not $m^2$, then your method does not work.

Otherwise, your method works.

Using your idea, one can prove that if $\lfloor\sqrt{4z}+1\rfloor\lt m$, then $m\lt p^k$.

Proof :

Subtracting $\lfloor\sqrt{4z}+1\rfloor^2$ which is the smallest square larger than $4z$ from the both sides of $$m^2=p^k+4z$$ gives $$m^2-\lfloor\sqrt{4z}+1\rfloor^2=p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z$$ which can be written as $$(m-\lfloor\sqrt{4z}+1\rfloor)(m+\lfloor\sqrt{4z}+1\rfloor)=p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z\tag1$$

So, we can say that $$m+\lfloor\sqrt{4z}+1\rfloor\mid p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z\tag2$$

If $\lfloor\sqrt{4z}+1\rfloor\lt m$, then LHS of $(1)$ is positive, so RHS of $(1)$ is positive. So, we can say that$$(2)\implies m+\lfloor\sqrt{4z}+1\rfloor\le p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z$$from which we can have$$m\lt m+\lfloor\sqrt{4z}+1\rfloor\le p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z\lt p^k.\quad\blacksquare$$


If $m=\lfloor\sqrt{4z}+1\rfloor$, then letting $\sqrt{4z}=N+a$ where $N\in\mathbb Z$ and $0\le a\lt 1$, we have $$p^k-m=(N+1)^2-(N+a)^2-N-1=(1-2a)N-a^2$$ whose sign depends on $a$ and $N$.

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  • $\begingroup$ Thank you for your answer, @mathlove! I think it would always be possible to subtract the larger square that is nearest to a given value of $4z = m^2 - p^k$, hence $m < p^k$ is always true? (Please see other answer.) $\endgroup$ Commented Nov 11, 2020 at 14:49
  • $\begingroup$ @Arnie Bebita-Dris : I don't think so because $$m+\lfloor\sqrt{4z}+1\rfloor \mid p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z\implies m+\lfloor\sqrt{4z}+1\rfloor\le p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z$$ is true only when $p^k-\lfloor\sqrt{4z}+1\rfloor^2+4z$ is positive. $\endgroup$
    – mathlove
    Commented Nov 11, 2020 at 14:56
  • $\begingroup$ @Arnie Bebita-Dris : In your answer, you are assuming that $p^k-9$ is positive. In other words, you are considering only $p,k$ satisfying $p^k-9$ is positive. $\endgroup$
    – mathlove
    Commented Nov 11, 2020 at 14:58
  • $\begingroup$ As an extreme example, we have $m^2 - p^k = 48$. Subtracting the smallest square that is larger than $48$, we get $(m+7)(m-7)=m^2 - 49=p^k - 1$, from which we obtain $(m+7) \mid (p^k - 1)$ and consequently, $m < m+7 \leq p^k - 1 < p^k$. QED $\endgroup$ Commented Nov 11, 2020 at 15:01
  • $\begingroup$ That $p^k - 9$ is indeed positive is automatic from $m^2 - p^k = 40$ since we have the lower bound $m^2 > {10}^{750}$ from the estimates $p^k < m^2$ (Dris, 2012) and ${10}^{1500} < n = p^k m^2$ (Ochem and Rao, 2012). $\endgroup$ Commented Nov 11, 2020 at 15:12
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Let me illustrate what I have in mind for a small value of $z$, say $z=10$.

Then we have $$m^2 - p^k = 4z = 40$$ $$m^2 - 49 = p^k - 9$$ $$(m+7)(m-7) = p^k - 9.$$ This implies that $$(m+7) \mid (p^k - 9)$$ from which it follows that $$m < m+7 \leq p^k - 9 < p^k.$$

Note that $49$ is not the nearest square to $40$ ($36$ is), but rather the nearest square larger than $40$.

With this minor adjustment in the logic, I would expect the general proof argument to work.

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