# On the nearest-square function - Part 2 and the quantity $m^2 - p^k$ where $p^k m^2$ is an odd perfect number

(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.

• 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. Nov 11, 2020 at 11:08
• 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 $\gcd(p,m)=1$. Nov 11, 2020 at 11:11
• 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 $m$. Nov 11, 2020 at 11:14
• @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. Nov 11, 2020 at 11:24
• @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. Nov 11, 2020 at 11:42

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.

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 $$N$$.

• 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.) Nov 11, 2020 at 14:49
• @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. Nov 11, 2020 at 14:56
• @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. Nov 11, 2020 at 14:58
• 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 Nov 11, 2020 at 15:01
• 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). Nov 11, 2020 at 15:12

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.