# If $p^k m^2$ is an odd perfect number with special prime $p$, then $p^k < 2am$ for some positive integer $a < m$ [closed]

(Preamble: Andy Putman asserts, in the comments, that MO policy prohibits "requests to check completeness of proofs". I have therefore trimmed down my original question to the bare essentials. I hope this would already be OK.)

The following is a proof that $$m^2 - p^k$$ is not a square, if $$p^k m^2$$ is an odd perfect number with special prime $$p$$.

Assume that the estimate $$p < m$$ holds. We want to show that the quantity $$m^2 - p^k$$ is not a square. Notation-wise, we will denote this conclusion by the shorthand $$m^2 - p^k \neq \square$$. Suppose to the contrary that $$m^2 - p^k = s^2$$. This is true if and only if $$2s + 1 = p^k$$ and $$2m - 1 = p^k.$$ This implies that $$p < m < p^k$$, from which we obtain $$k > 1$$. Since $$k \equiv 1 \pmod 4$$, then we know that $$k \geq 5$$. We can now use a proof by anonymous MSE user FredH to show that $$m^2 - p^k \neq \square$$ (under the assumption $$p < m$$), as follows:

Since $$N = p^k m^2$$ is (odd) perfect, then we have the defining equation $$\sigma(N) = 2N,$$ from which it follows that $$\sigma(p^k)\sigma(m^2) = 2p^k m^2.$$

We know that $$\sigma(p^k) = (p^{k+1} - 1)/(p - 1)$$. Since we have shown that $$m = (p^k + 1)/2$$, then we have the equation $$2(p^{k+1} - 1)\sigma(m^2) = p^k (p - 1)(p^k + 1)^2. \hspace{0.76in} (*)$$

FredH considered the $$GCD$$ of $$p^{k+1} - 1$$ with the right-hand side of Equation $$(*)$$: $$p^{k+1} - 1 = \gcd\left(p^{k+1} - 1, p^k (p - 1)(p^k + 1)^2\right) \leq (p - 1)\left(\gcd(p^{k+1} - 1, p^k + 1)\right)^2$$ where FredH used the fact that $$\left(p^{k+1} - 1\right) \mid RHS$$ and the property that $$\gcd(x,yz) \leq \gcd(x,y)\gcd(x,z).$$ But FredH also noticed that $$p^{k+1} - 1 = p(p^k + 1) - (p + 1)$$, whence FredH did also find $$\gcd(p^{k+1} - 1, p^k + 1) = \gcd(p + 1, p^k + 1),$$ which is $$p + 1$$ because $$k$$ is odd. Thus, $$(p - 1)\left(\gcd(p^{k+1} - 1, p^k + 1)\right)^2 = (p - 1)(p + 1)^2.$$

Hence, the inequality $$p^{k+1} - 1 \leq (p - 1)(p + 1)^2$$ holds.

Since $$k \geq 5$$, we obtain $$p^5 < p^{k+1} - 1 \leq (p - 1)(p + 1)^2 < p^4,$$ which is a contradiction.

Hence, we now have the implication $$p < m \Rightarrow m^2 - p^k \neq \square.$$

In other words, we have the contrapositive $$m^2 - p^k = \square \Rightarrow m < p.$$
Now, suppose to the contrary that $$m^2 - p^k = \square$$. This implies that $$m < p$$. Since $$p^k < m^2$$, we then have the implication $$m < p \Rightarrow k = 1$$. Therefore, $$k = 1$$. But we know (from the considerations above) that $$m^2 - p^k = \square \iff m = (p^k + 1)/2.$$

Since $$k = 1$$, we infer that $$m = (p + 1)/2$$, or in other words, $$p = 2m - 1$$. From Acquaah and Konyagin's results, we have the unconditional estimate $$p < m \sqrt{3}$$. This implies that $$2m - 1 = p < m \sqrt{3}$$, from which we infer that $$m(2 - \sqrt{3}) < 1$$ which contradicts the fact that $$\omega(m) > 4$$. (In fact, we do know that $$m > {10}^{375}$$, by using Ochem and Rao's lower bound $$N > {10}^{1500}$$ for the magnitude of an odd perfect number $$N$$, together with $$p^k < m^2$$.)

We conclude that $$m^2 - p^k \neq \square$$.

Now, here goes the part where I am a bit unsure about its logical tightness, and is also my main question in this post:

Does the following statement necessarily hold? "Since $$m^2 - p^k$$ is not a square, then it is between two (consecutive) squares."

If so, WLOG we may assume that $$(m - a)^2 < m^2 - p^k < (m - a + 1)^2$$ for some positive integer $$a$$. We may likewise assume that $$m > a$$.

If I am not mistaken, these assumptions will then yield a proof for the inequality $$m < p^k < 2am$$ except for the (problematic) case $$a=1$$, where we can only derive $$p^k < 2m - 1.$$

Either way, I think the inequalities can be summarized as $$p^k < 2am$$ for some positive integer $$a < m$$.

• This question is off-topic. Requests to check completeness of proofs are not allowed. The bounty prevents me from voting to close at the moment, but this question absolutely should be closed. Commented Jun 10 at 15:35
• It's not a matter of how it is worded: as far as I can tell, this question is fundamentally a request for someone to either find the flaw in your proof or fill in some missing details, and no amount of rephrasing will make that on-topic. Commented Jun 10 at 15:54
• I have trouble following the reasoning pretty much by the first display line. However, the answer to the question "Does the following statement necessarily hold? 'Since $m^2 - p^k$ is not a square, then it is between two (consecutive) squares.' " is yes, and the proof is very easy. Since the natural numbers are well-ordered, there is a least square $M^2$ greater than this $N := m^2 - p^k$. Then $(M-1)^2 \leq N$; otherwise, $M^2$ wouldn't have been the least square. $(M-1)^2 < N$ since $N$ is assumed not to be a square. So $(M-1)^2 < N < M^2$. Commented Jun 10 at 18:52
• I do not understand why you devoted so many efforts to show that $m^2-p^k$ is not a square. Indeed, if $m>2$ and $m^2-p^k\ge 0$ then $p^k<2(m-1)m$ and so we can put $a=m-1$. On the other hand, if $m^2-p^k<0$ then it cannot be between two squares, so your finishing arguments fail. Commented Jun 15 at 7:09
• If $p^k<m^2$ always holds then $p^k<2(m-1)m$ and so we can put $a=m-1$, which provides an affirmative answer to your question, right? Commented Jun 15 at 8:43

Since $$p^k, we have $$p^k<2(m−1)m$$, and so we can put $$a=m−1$$.

(This is not a direct answer to the original question, as posed, but rather are some thoughts that recently occurred to me, which would be too long to fit in the Comments section.)

Let $$p^k m^2$$ be an odd perfect number with special prime $$p$$.

Since $$m^2 - p^k \neq \square$$, then $$p^k \neq 2m - 1$$.

If $$p^k < 2m - 1$$ holds, then $$p < 2m - 1$$ is true. (In particular, note that we get $$p \leq p^k < 2m - 1 < 2m$$.)

Now assume that $$2m - 1 < p^k$$. We get $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)}.$$

Claim #1: $$\sigma(p^k) > 2m$$

Proof: $$\sigma(p^k) \geq p^k + 1 > 2m$$. QED

Claim #2: $$m \neq \frac{2pm - p - 1}{2(p - 1)}$$

Proof: $$2pm - 2m = 2(p - 1)m = 2pm - p - 1$$, which is equivalent to $$p = 2m - 1 > m$$. This implies that $$k = 1$$. But then $$m^2 - p^k = m^2 - p = m^2 - 2m + 1 = (m - 1)^2 = \square$$, contradicting our result.

It thus remains to rule out the case $$\sigma(p^k)/2 > m > \frac{2pm - p - 1}{2(p - 1)}.$$ The RHS inequality yields $$p > 2m - 1 > m$$, which is equivalent to $$k = 1$$, since the assumption $$2m - 1 < p^k$$ implies $$m < p^k$$, which in turn implies that the biconditional $$m < p \iff k = 1$$ holds. Substituting $$k = 1$$ into the LHS inequality yields $$(p + 1)/2 > m.$$ From Acquaah and Konyagin's results, we have $$p < m\sqrt{3}.$$ This implies that $$m < (p + 1)/2 < \frac{m\sqrt{3} + 1}{2}.$$ This is a contradiction.

The only way out of the contradictions is to have $$\sigma(p^k)/2 > \frac{2pm - p - 1}{2(p - 1)} > m,$$ which implies, under the assumption $$2m - 1 < p^k$$, that $$p < 2m - 1.$$ In particular, we obtain $$k \neq 1$$. Since the assumption $$2m - 1 < p^k$$ implies $$m < p^k$$, and because $$m < p^k$$ implies the biconditional $$m < p \iff k = 1$$ holds, then $$k \neq 1$$ is equivalent to $$p < m$$.

Either way, we conclude that $$p < 2m$$. (Note that Acquaah and Konyagin has already proved that $$p < m\sqrt{3}$$, so all of this is but an academic exercise.)