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

This question has been cross-posted from this MSE question and is an offshoot of this other MSE question.

(Note that MSE user mathlove has posted an answer in MSE, which I could not completely understand. I have therefore cross-posted this question in MO, hoping the sages here would be able to give some enlightenment. I hope this is okay, and that the question is research-level.)

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 arguments:

Theorem 1 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

Theorem 2 If $$n = p^k m^2$$ is an odd perfect number satisfying $$m^2 - p^k = 40$$, then $$m < p^k$$.

Proof

Let $$p^k m^2$$ be an odd perfect number satisfying $$m^2 - p^k = 40$$.

Then $$(m+7)(m-7)=m^2 - 49=p^k - 9,$$ from which it follows that $$(m+7) \mid (p^k - 9)$$ which implies that $$m < m+7 \leq p^k - 9 < p^k.$$

QED

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.

(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.)

So now consider the equation $$m^2 - p^k = 4z$$. Following our proof strategy, we have:

Subtracting the smallest square that is larger than $$m^2 - p^k$$, we obtain

$$m^2 - \bigg(\lceil{\sqrt{m^2 - p^k}}\rceil\bigg)^2 = p^k + \Bigg(4z - \bigg(\lceil{\sqrt{m^2 - p^k}}\rceil\bigg)^2\Bigg).$$

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

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

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

• "We now endeavor to disprove the Dris Conjecture" I'm not sure what you mean by this. Wouldn't a disproof by nature need to mean one had an actual odd perfect number as a counterexample? – JoshuaZ Nov 14 '20 at 12:12
• Thank you for your comment, @JoshuaZ! Yes, of course, if you want then you could go "Assume that $p^k m^2$ is an odd perfect number with special prime $p$, satisfying $m^2 - p^k = 4z$ and $p^k < m$." You will still get the same conclusion that $m < p^k$. It is in this sense that "we endeavor(ed) to disprove the Dris Conjecture (that $p^k < m$)". – Arnie Bebita-Dris Nov 14 '20 at 20:07
• If indeed $m-a=\sqrt{m^2 - p^k}$, then $m^2 - 2am + a^2 = m^2 - p^k$. This implies that $p^k = 2am - a^2 = a(2m - a)$, from which it follows that $0 < a < 1$. We conclude that $p^k < 2m$. – Arnie Bebita-Dris Nov 21 '20 at 15:03
• In particular, this means that, under the problematic case, $m^2 - p^k > m^2 - 2m$, so that the smallest square larger than $m^2 - p^k$ is $m^2 - 2m + 1 = (m - 1)^2$. Does this proof suffice, @mathlove? – Arnie Bebita-Dris Nov 21 '20 at 15:19
• I don't know why you can say that the smallest square larger than $m^2-p^k$ is $(m-1)^2$. It is possible that $m^2-2m\lt (m-1)^2\lt m^2-p^k$. – mathlove Nov 21 '20 at 15:39

## 2 Answers

Middle of page 6 of https://arxiv.org/pdf/1312.6001v10.pdf

" we always have $$0 < n−\lceil\sqrt{n^2−q^k}\rceil$$ "

No, this requires that $$q^k\ge 2n-1$$, an helpful assumption when the goal is to prove $$q^k > n$$.

You are asking if $$m\lt p^k$$ can be proved in the following way :

We have $$\Bigg(m + \left\lceil{\sqrt{m^2 - p^k}}\right\rceil\Bigg)\Bigg(m - \left\lceil{\sqrt{m^2 - p^k} }\right\rceil\Bigg) = p^k +4z - \left\lceil{\sqrt{m^2 - p^k}}\right\rceil^2$$ which implies $$\bigg(m + \left\lceil{\sqrt{m^2 - p^k}}\right\rceil\bigg) \mid \bigg(p^k +4z - \left\lceil{\sqrt{m^2 - p^k}}\right\rceil^2\bigg)$$ from which it follows that $$m < m + \left\lceil{\sqrt{m^2 - p^k}}\right\rceil \leq p^k +4z - \left\lceil{\sqrt{m^2 - p^k}}\right\rceil^2 < p^k.\quad\square$$

This is not correct since this does not work when $$m =\left\lceil{\sqrt{m^2 - p^k} }\right\rceil$$.

(If it is true that $$m \not=\left\lceil{\sqrt{m^2 - p^k} }\right\rceil$$, then your method works.)

In the comments, you are trying to prove $$m \not=\left\lceil{\sqrt{m^2 - p^k} }\right\rceil$$ in the following way :

Suppose that $$m=\left\lceil{\sqrt{m^2 - p^k} }\right\rceil$$. Then, there is an $$a\in[0,1)$$ such that $$m-a=\sqrt{m^2-p^k}$$. Squaring the both sides, we get $$p^k=2am-a^2$$ which implies $$p^k\lt 2m$$ to have $$m^2-p^k\gt (m-1)^2-1$$. So, we see that the smallest square larger than $$m^2-p^k$$ is $$(m-1)^2$$, which is a contradiction.$$\quad\square$$

This is not correct since it is possible that $$(m-1)^2-1\lt (m-1)^2\lt m^2-p^k$$.