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4.3.22

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7.2.22

A new approach to odd perfect numbers via GCDs

Here is the Scribd link.  (This same document has likewise been uploaded to Academia.)

Update (February 17, 2022 - 9:39 AM [Manila time]):  This article has just been announced on math.NT in arXiv.

Here is the abstract:

Let $q^k n^2$ be an odd perfect number with special prime $q$.  Define the GCDs

$$G = \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$

$$H = \gcd\bigg(n^2,\sigma(n^2)\bigg)$$

and

$$I = \gcd\bigg(n,\sigma(n^2)\bigg).$$

We prove that $G \times H = I^2$.  (Note that it is trivial to show that $G \mid I$ and $I \mid H$ both hold.)  We then compute expressions for $G, H,$ and $I$ in terms of $\sigma(q^k)/2, n,$ and $\gcd\bigg(\sigma(q^k)/2,n\bigg)$.  Afterwards, we prove that if $G = H = I$, then $\sigma(q^k)/2$ is not squarefree.  Other natural and related results are derived further.  Lastly, we conjecture that the set

$$\mathscr{A} = \{m : \gcd(m,\sigma(m^2))=\gcd(m^2,\sigma(m^2))\}$$

has asymptotic density zero.

2.2.22

On factors of odd perfect numbers

Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

In a previous blog post, I showed that the following equation must hold:

$$GH = I^2$$

where $G, H,$ and $I$ are defined as

$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg)$$

$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg)$$

$$I := \gcd\bigg(n,\sigma(n^2)\bigg).$$

Note that

$$H = i(q)$$

is the index of the odd perfect number $N$ at the prime $q$, and that

$$i(q) := \frac{\sigma(n^2)}{q^k} = \frac{2n^2}{\sigma(q^k)} = \frac{D(n^2)}{s(q^k)} = \frac{2s(n^2)}{D(q^k)}. \tag{1}$$

Note further that

$$i(q) = q\sigma(n^2) - 2(q - 1)n^2$$

so that it follows from this blog post that

$$\frac{qn^2}{i(q)} = \frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)} = \frac{q\sigma(q^k)}{2}.$$

Since $q \equiv k \equiv 1 \pmod 4$ and $q$ is prime, then

$$\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4,$$

so that $\sigma(q^k)/2$ is an integer.

Hence, $H = i(q) \mid n^2 \mid N$, so that $H = \gcd\bigg(n^2,\sigma(n^2)\bigg)$ is a factor of the odd perfect number $N$.

Next, we show that $I \mid H$.

Since $a \mid b$ implies that $\gcd(a, c) \mid \gcd(b, c)$, then setting $a = n$, $b = n^2$, and $c = \sigma(n^2)$, we get that

$$n \mid n^2 \Rightarrow \gcd\bigg(n,\sigma(n^2)\bigg) = I \mid H = \gcd\bigg(n^2,\sigma(n^2)\bigg).$$

Since $n \mid n^2$ is true in general, we conclude that $I \mid H$.

Lastly, since $GH = I^2$ is true, then we obtain

$$\frac{H}{I} = \frac{I}{G}. \tag{2}$$

Because $I \mid H$, the LHS of Equation (2) is an integer.  It follows that the RHS of Equation (2) is also an integer.  Consequently, the divisibility constraint $G \mid I$ holds.


Hence, we have the following proposition:

THEOREM (02/02/2022): If $N = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, and $G, H,$ and $I$ defined as follows

$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg),$$

$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg),$$

and

$$I := \gcd\bigg(n,\sigma(n^2)\bigg),$$

then we have the equation

$$G \times H = I^2$$

and the divisibility chain

$$G \mid I \mid H.$$


In particular, we have the following corollary:

COROLLARY (02/02/2022):  If $N = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, and $G, H,$ and $I$ defined as follows

$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg),$$

$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg),$$

and

$$I := \gcd\bigg(n,\sigma(n^2)\bigg),$$

then $G, H,$ and $I$ are factors of $N$.


Remark:

(1)   Note that $G = H$ holds if and only if $H = I$.

(2)  Suppose that $\sigma(q^k)/2$ is squarefree.  We want to show that $G \neq H$.  Assume to the contrary that $G = H$.  The assumption that $\sigma(q^k)/2$ is squarefree implies that $\sigma(q^k)/2 \mid n$ (since $\sigma(q^k)/2 \mid n^2$ holds in general).  But the condition $\sigma(q^k)/2 \mid n$ is equivalent to $n \mid \sigma(n^2)$, because of the equation

$$\frac{\sigma(n^2)}{n}=\frac{q^k n}{\sigma(q^k)/2}.$$

Hence, since $n \mid \sigma(n^2)$, then it follows that

$$G = \gcd(\sigma(q^k),\sigma(n^2))=\frac{\sigma(q^k)}{2}$$

since we can compute

$$\gcd(\sigma(q^k),\sigma(n^2))=\frac{\bigg(\gcd(n,\sigma(n^2))\bigg)^2}{2n^2/\sigma(q^k)}=\frac{\sigma(q^k)}{2}\cdot\Bigg(\gcd\bigg(1,\frac{\sigma(n^2)}{n}\bigg)\Bigg)^2.$$

But we know from Equation (1) that

$$H = \frac{n^2}{\sigma(q^k)/2}.$$

It follows from the assumption $G = H$ that $\sigma(q^k)/2 = n$.  This contradicts Steuerwald (1937) who showed that $n$ must contain a square factor.

(Reference: R. Steuerwald, "Verschärfung einer notwendigen Bedingung für die Existenz einer ungeraden vollkommenen Zahl," S.-B. Math.-Nat. Abt. Bayer. Akad. Wiss., 1937, pp. 68-73. - See this MathOverflow answer for more information.)

Thus, we also have the following proposition:

THEOREM (02/02/2022): If $N = q^k n^2$ is an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$, $\sigma(q^k)/2$ is squarefree, and $G, H,$ and $I$ defined as follows

$$G := \gcd\bigg(\sigma(q^k),\sigma(n^2)\bigg),$$

$$H := \gcd\bigg(n^2,\sigma(n^2)\bigg),$$

and

$$I := \gcd\bigg(n,\sigma(n^2)\bigg),$$

then we have $G < I < H$.

24.12.21

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

(The first part of this blog post is taken from this MSE question, which was asked on June 12, 2021, and the answer contained therein, which was last updated on June 20, 2021.)


QUESTION

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 rule out Case (5) and Case (6), and under Case (1) and Case (2), we can prove that the inequality $m < p^k$ holds.


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


Under both cases left under consideration, we have

$$(m - 2^r)(m - t) < 0$$

$$m^2 + 2^r t < m(2^r + t)$$

$$m^2 + (m^2 - p^k) < m(2^r + t)$$

$$2m^2 < m(2^r + t) + p^k.$$


Since we want to prove $m < p^k$, assume to the contrary that $p^k < m$.  We get

$$2m^2 < m(2^r + t) + p^k < m(2^r + t) + m < m(2^r + t + 1)$$

which implies, since $m > 0$, that

$$2m < 2^r + t + 1.$$


Here then is our question:

Will it be possible to derive a contradiction from the inequality

$$2m < 2^r + t + 1,$$

under Case (3) and Case (4) above, considering that $2m$ is large?  (In fact, it is known that $m > {10}^{375}$.)


ANSWER BY mathlove


On OP's request, I am converting my comment into an answer.


$p^k\lt m$ is equivalent to $$m\lt\frac{1+\sqrt{1+2^{r+2}t}}{2}\tag7$$ since we have

$$p^k\lt m \iff m^2-2^rt\lt m \iff m^2-m-2^rt\lt 0 \iff m\lt\frac{1+\sqrt{1+2^{r+2}t}}{2}$$


$(7)$ is better than $2m\lt 2^r+t+1$ since

$$\frac{1+\sqrt{1+2^{r+2}t}}{2}\lt \frac{2^r+t+1}{2}\tag8$$

holds. To see that $(8)$ holds, note that

$$(8) \iff \sqrt{1+2^{r+2}t}\lt 2^r+t \iff 1+2^{r+2}t\lt 2^{2r}+2^{r+1}t+t^2 \iff (2^r-t)^2\gt 1 \iff |2^r-t|\gt 1$$

which does hold.


We can say that $$\bigg(\frac{1+\sqrt{1+2^{r+2}t}}{2}-t\bigg)\bigg(\frac{1+\sqrt{1+2^{r+2}t}}{2}-2^r\bigg)\lt 0\tag9$$

since

$$(9) \iff \bigg(\frac{1+\sqrt{1+2^{r+2}t}}{2}\bigg)^2-\frac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0 \iff \frac{1+\sqrt{1+2^{r+2}t}+2^{r+1}t}{2}-\frac{1+\sqrt{1+2^{r+2}t}}{2}(t+2^r)+2^rt\lt 0$$

$$\iff 1+\sqrt{1+2^{r+2}t}+2^{r+1}t-(1+\sqrt{1+2^{r+2}t})(t+2^r)+2^{r+1}t\lt 0 \iff 2^{r+2}t-2^r-t+1\lt (t+2^r-1)\sqrt{1+2^{r+2}t}$$ 

$$\iff (2^{r+2}t-2^r-t+1)^2\lt (t+2^r-1)^2(1+2^{r+2}t) \iff 2^{r + 2} t (2^r - t - 1) (2^r - t + 1)\gt 0 \iff (2^r-t)^2\gt 1 \iff |2^r-t|\gt 1$$

which does hold.


It follows from $(7)(9)$ that if $p^k\lt m$ with $(m-t)(m-2^r)\lt 0$, then $$\min(t,2^r)\lt m\lt\dfrac{1+\sqrt{1+2^{r+2}t}}{2}\lt \max(t,2^r).$$


(The second part of this blog post is taken from this MO question.)


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.

29.8.21

New preprint by Dagal and Dris - If $q^k n^2$ is an odd perfect number with special prime $q$, then $k$ is bounded.

Dagal and myself have a new preprint out there, uploaded to the math.GM subject area in arXiv.

Here is the abstract:

In this note, we continue an approach pursued in an earlier paper of the second author and thereby attempt to produce an improved lower bound for the sum $I(q^k) + I(n^2)$, where $q^k n^2$ is an odd perfect number with special prime $q$ and $I(x)$ is the abundancy index of the positive integer $x$.  In particular, this yields an upper bound for $k$.

This paper has already been submitted to a journal for possible publication.

31.7.21

A failed attempt at a proof for the nonexistence of odd perfect numbers

The following hyperlink points to a failed attempt of the author of this blog post at a proof for the nonexistence of odd perfect numbers.

The ideas contained therein may prove useful to somebody else in the future.

13.7.21

Interesting identity involving divisors of odd perfect numbers that I discovered today (July 13, 2021)

Denote the classical sum of divisors of the positive integer $x$ by $\sigma(x)=\sigma_1(x)$.  Denote the abundancy index of $x$ as $I(x)=\sigma(x)/x$.

I discovered an interesting identity involving divisors of odd perfect numbers given in the Eulerian form $N = q^k n^2$ today (July 13, 2021).

The identity is:

Proposition:  If $N = q^k n^2$ is an odd perfect number with special prime $q$, then

$$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q}.$$


Proof:

Our starting point is the following blog post, where it is proved that

$$\gcd(n^2, \sigma(n^2)) = 2(1 - q)n^2 + q\sigma(n^2).$$

However, note that we have

$$\frac{\sigma(n^2)}{q^k} = \gcd(n^2, \sigma(n^2)).$$

These equations are equivalent to

$$2(1 - q)n^2 + q\sigma(n^2) = \frac{\sigma(n^2)}{q^k}.$$

Factoring out $qn^2$ on the LHS, we obtain

$$qn^2 \Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q^k}.$$

Multiplying both sides of the last equation by $q^{k-1}$, we get

$$N\cdot\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg) = \frac{\sigma(n^2)}{q}.$$

(Note that the RHS of the last equation is an odd integer.)

This concludes our proof.

QED.

In particular, we have

$$\frac{N}{\sigma(n^2)/q} = \frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)}.$$

But we also know from the following MSE post that

$$I(n^2) - \frac{2(q - 1)}{q} = \frac{2(q - 1)}{q\bigg(q^{k+1} - 1\bigg)}.$$

This means that we obtain

$$\frac{1}{\Bigg(I(n^2) - \frac{2(q - 1)}{q}\Bigg)} = \frac{q\bigg(q^{k+1} - 1\bigg)}{2(q - 1)} = \frac{q\sigma(q^k)}{2}.$$

But $\sigma(q^k) \equiv k + 1 \pmod 4$, since $q \equiv 1 \pmod 4$, and since $k \equiv 1 \pmod 4$, then $$\sigma(q^k) \equiv k + 1 \equiv 2 \pmod 4.$$

This finding implies that $\sigma(n^2)/q$ divides $N = q^k n^2$, which is almost a proof of the desired divisibility constraint $\sigma(n^2)/q \mid n^2$, as the latter constraint would imply $k = 1$.