# On odd perfect numbers $p^k m^2$ with special prime $p$ satisfying $m^2 - p^k = 2^r t$ - Part VI

(Note: This question has been cross-posted to MO.)

The topic of odd perfect numbers likely needs no introduction.

Denote the classical sum of divisors of the positive integer $$x$$ by $$\sigma(x)=\sigma_1(x)$$.

If $$n$$ is odd and $$\sigma(n)=2n$$, then we call $$n$$ an odd perfect number. Euler proved that a hypothetical odd perfect number must necessarily have the form $$n = p^k m^2$$ where $$p$$ is the special prime satisfying $$p \equiv k \equiv 1 \pmod 4$$ and $$\gcd(p,m)=1$$.

Descartes, Frenicle, and subsequently Sorli conjectured that $$k=1$$ always holds. Dris conjectured that the inequality $$p^k < m$$ is true in his M. Sc. thesis, and Brown (2016) eventually produced a proof for the weaker inequality $$p < m$$.

Now, recent evidence suggests that $$p^k < m$$ may in fact be false.

THE ARGUMENT

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

Since $$p \equiv k \equiv 1 \pmod 4$$ and $$m$$ is odd, then $$m^2 - p^k \equiv 0 \pmod 4$$. Moreover, $$m^2 - p^k$$ is not a square (Dris and San Diego (2020)).

This implies that we may write $$m^2 - p^k = 2^r t$$ where $$2^r \neq t$$, $$r \geq 2$$, and $$\gcd(2,t)=1$$.

It is trivial to prove that $$m \neq 2^r$$ and $$m \neq t$$, so that we consider the following cases:

$$\text{Case (1): } m > t > 2^r$$ $$\text{Case (2): } m > 2^r > t$$ $$\text{Case (3): } t > m > 2^r$$ $$\text{Case (4): } 2^r > m > t$$ $$\text{Case (5): } t > 2^r > m$$ $$\text{Case (6): } 2^r > t > m$$

We can easily rule out Case (5) and Case (6), as follows:

Under Case (5), we have $$m < t$$ and $$m < 2^r$$, which implies that $$m^2 < 2^r t$$. This gives $$5 \leq p^k = m^2 - 2^r t < 0,$$ which is a contradiction.

Under Case (6), we have $$m < 2^r$$ and $$m < t$$, which implies that $$m^2 < 2^r t$$. This gives $$5 \leq p^k = m^2 - 2^r t < 0,$$ which is a contradiction.

Under Case (1) and Case (2), we can prove that the inequality $$m < p^k$$ holds, as follows:

Under Case (1), we have: $$(m - t)(m + 2^r) > 0$$ $$p^k = m^2 - 2^r t > m(t - 2^r) = m\left|2^r - t\right|.$$

Under Case (2), we have: $$(m - 2^r)(m + t) > 0$$ $$p^k = m^2 - 2^r t > m(2^r - t) = m\left|2^r - t\right|.$$

So we are now left with Case (3) and Case (4).

Under Case (3), we have: $$(m + 2^r)(m - t) < 0$$ $$p^k = m^2 - 2^r t < m(t - 2^r) = m\left|2^r - t\right|.$$

Under Case (4), we have: $$(m - 2^r)(m + t) < 0$$ $$p^k = m^2 - 2^r t < m(2^r - t) = m\left|2^r - t\right|.$$

Note that, under Case (3) and Case (4), we actually have $$\min(2^r,t) < m < \max(2^r,t).$$

But the condition $$\left|2^r - t\right|=1$$ is sufficient for $$p^k < m$$ to hold.

Our inquiry is:

QUESTION: Is the condition $$\left|2^r - t\right|=1$$ also necessary for $$p^k < m$$ to hold, under Case (3) and Case (4)?

Note that the condition $$\left|2^r - t\right|=1$$ contradicts the inequality $$\min(2^r,t) < m < \max(2^r,t),$$ under the remaining Case (3) and Case (4), and the fact that $$m$$ is an integer.