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]

Question

(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.)

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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 $p$.

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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 $m^2 - p^k \neq \square$.

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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 $a < m$.

Answer 1

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

Answer 2

(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 $p^k \neq 2m - 1$.

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

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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 $p < m$.

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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.)