(MO link: https://mathoverflow.net/questions/470088)

(

**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. 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 $m^2 - p^k$ is not a 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

$$p < m \Rightarrow m^2 - p^k \text{ is not a square}.$$

In other words, we have the contrapositive

$$m^2 - p^k \text{ is a square } \Rightarrow m < p.$$

Now, suppose to the contrary that $m^2 - p^k$ is a 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

$$m^2 - p^k \text{ is a square } \iff m = (p^k + 1)/2.$$

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 $m^2 - p^k$ is not a 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 $a < m$.

----------------------------------------------------------------------------------

__Posted Answer__

(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 $m^2 - p^k \text{ is not a square}$, 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 $m^2 - p^k = m^2 - p = m^2 - 2m + 1 = (m - 1)^2$ is a square, contradicting our result.

**QED**

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.)