2
$\begingroup$

Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of the positive integer $x$. Denote the aliquot sum of $x$ by $s(x)=\sigma(x)-x$ and the deficiency of $x$ by $d(x)=2x-\sigma(x)$. Finally, denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.

A number $P$ is said to be perfect if $\sigma(P)=2P$. On the other hand, if a number $A$ satisfies $\sigma(A)=2A-1$, then $A$ is said to be almost perfect. (The only known examples for $A$ are the powers of two, including $1$.) Lastly, a number $D$ is said to be deficient-perfect if the deficiency $d(D)$ of $D$ divides $D$.


Even Perfect Numbers

The three smallest examples for $P$ are $$6=2\cdot{3}={2^1}\cdot\left(2^2 - 1\right)$$ $$28=4\cdot{7}={2^2}\cdot\left(2^3 - 1\right)$$ $$496={16}\cdot{31}={2^4}\cdot\left(2^5 - 1\right).$$

The Euclid-Euler Theorem states that an even number $E$ is perfect if and only if $E$ has the form $$E={2^{p-1}}\cdot\left(2^p - 1\right) = \frac{M_p (M_p + 1)}{2},$$ where $M_p = 2^p - 1$ is a Mersenne prime. (Note that if $M_p$ is prime, then $p$ is also prime. The converse does not necessarily hold.)

Only $51$ Mersenne primes are known, corresponding to an equal number of (even) perfect numbers. (See this GIMPS page.) It is currently unknown whether there are finitely many such numbers $E$. It has been conjectured that there are infinitely many even perfect numbers.

Note that we have the equations $$1=\gcd\left(M_p,\frac{M_p + 1}{2}\right)=\sigma\left(\frac{M_p + 1}{2}\right)/M_p=\frac{M_p + 1}{\sigma(M_p)}=\frac{d\left(\frac{M_p + 1}{2}\right)}{s(M_p)}=\frac{2s\left(\frac{M_p + 1}{2}\right)}{d(M_p)}$$ so that $$d\left(\frac{M_p + 1}{2}\right)d(M_p)=2s\left(\frac{M_p + 1}{2}\right)s(M_p).$$

(Note that, trivially, we have $(M_p + 1)/2 < M_p$.)


Odd Perfect Numbers

Much less is known about odd perfect numbers $O$. We do not know an actual value for even just a single divisor of $O$, apart from the trivial factor $1$. This is because we do not have an example for $O$, unlike the case for $E$. (In a letter to Mersenne dated November $15$, $1638$, Descartes showed that $$\mathscr{D} = {{3}^2}\cdot{{7}^2}\cdot{{11}^2}\cdot{{13}^2}\cdot{22021} = 198585576189$$ would be an odd perfect number if $22021$ were prime.)

Nonetheless, Euler did prove that $O$ must necessarily have the so-called Eulerian form $$O = q^k n^2$$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

Note that we have the equations $$1 < \gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{d(n^2)}{s(q^k)}=\frac{2s(n^2)}{d(q^k)} \tag{*}$$ so that $$d(q^k)d(n^2)=2s(q^k)s(n^2).$$

(It is known that $q^k < n^2$.)


Odd Perfect Numbers Must Have a Deficient-Perfect Divisor

Remark 1. We may define a deficient-perfect number $D$ to be a positive integer having a proper divisor $e$ satisfying $$\sigma(D) = 2D - e.$$ The divisor $e$ is said to be the deficient divisor of $D$.

Remark 2. It is also easy to show that the deficient divisor $e$ of a deficient-perfect number $D$ is the greatest common divisor of $D$ and $\sigma(D)$. That is, we have $$e=\gcd(D,\sigma(D))$$ if $σ(D) = 2D − e$ and $e | D$, $e \neq D$.

Holdener and Rachfal (2019) proved the following assertion:

THEOREM 1. Every odd perfect number has a deficient-perfect divisor. In particular, if $O = q^k n^2$ is an odd perfect number given in Eulerian form, then the divisor $D_O = q^j n^2$ is deficient-perfect, with deficient divisor $$e_O=\frac{2q^j n^2}{q^{j+1} + 1},$$ where $j=(k-1)/2$.

In particular, by Remark 1 and Remark 2, we obtain $$\gcd\left(D_O,\sigma(D_O)\right)=e_O=2D_O-\sigma(D_O)=d(D_O).$$

Following in the footsteps of this short ResearchGate article, titled "Gcd in Odd perfect number" by Devansh Singh (B.Tech, M.E., I.E.T. Lucknow), we proceed as follows:

From the Eulerian form $O=q^k n^2$, we get $$O={q^{(k+1)/2}}\cdot\left(q^{(k-1)/2} n^2\right)={q^{(k+1)/2}}\cdot{D_O}.$$ Since $\gcd\left(q^{(k+1)/2},q^{(k-1)/2} n^2\right)=1$ if and only if $k=1$, then we are unable to tie up, in full generality, the value of $$\sigma(O/q^k)/q^{j+1}=\sigma(n^2)/q^{(k+1)/2}$$ with that of $$\gcd(D_O,\sigma(D_O))=\gcd\left(q^{(k-1)/2} n^2,\sigma(q^{(k-1)/2} n^2)\right),$$ by using Singh's approach.


This is getting too long already, so let me ask my question (as is) in the title:

QUESTION. Does an odd perfect number have a divisor (other than $1$) which must necessarily be almost perfect?


MY ATTEMPT

Note from the definitions that, if a number is almost perfect, then it is automatically deficient-perfect.

Suppose to the contrary that the deficiency $d(D_O)$ of the deficient-perfect divisor $D_O=q^j n^2$ of an odd perfect number $O=q^k n^2$ satisfies $d(D_O)=1$ (where $j=(k-1)/2$). (Notice that $D_O$ is a square.)

In other words, assume that $D_O$ is almost perfect. Then we have, by a criterion in this paper, the following inequality $$\frac{2D_O}{D_O + 1}< I(D_O) < \frac{2D_O + 1}{D_O + 1}$$ since $D_O > 1$ and $D_O$ is almost perfect.

Using the fact that the abundancy index function $I$ is multiplicative, we obtain $$I(D_O)=I(q^j n^2)=I(q^j)I(n^2).$$

If $k=1$, then using the equations and inequality in $(*)$, we obtain $$1<\gcd(n^2,\sigma(n^2))=d(n^2),$$ which contradicts our earlier assumption that $d(D_O)=1$. Thus, it suffices to consider $k \neq 1$. Since $k \equiv 1 \pmod 4$ holds, then $k \geq 5$. It follows that $j=(k-1)/2 \geq 2$.

In particular, we have $$I\left((qn)^2\right) = I(q^2)I(n^2) \leq I(D_O) < \frac{2q^j n^2 + 1}{q^j n^2 + 1} \tag{1}$$ and $$\frac{2q^j n^2}{q^j n^2 + 1} < I(D_O) = I(q^j)I(n^2) = \left(\frac{q^{j+1} - 1}{q^j (q - 1)}\right)\cdot{I(n^2)} < \left(\frac{q}{q - 1}\right)\cdot\left(\frac{2q}{q + 1}\right) = \frac{2q^2}{q^2 - 1}. \tag{2}$$

Inequality $(2)$ results to $$0 < 2\left(q^j n^2 + q^2\right)$$ which is trivial.

On the other hand, we derive the following estimate from Inequality $(1)$: $$I(q^2)I(n^2)=\left(\frac{\sigma(q^2)}{q^2}\right)\cdot{I(n^2)}=\left(\frac{q^2 + q + 1}{q^2}\right)\cdot{I(n^2)}$$ $$>\left(\frac{q^2 + q + 1}{q^2}\right)\cdot\left(\frac{2(q-1)}{q} + \frac{1}{qn^2}\right).$$

We now test whether the inequality $$\left(\frac{q^2 + q + 1}{q^2}\right)\cdot\left(\frac{2(q-1)}{q} + \frac{1}{qn^2}\right) < \frac{2q^j n^2 + 1}{q^j n^2 + 1} \tag{3}$$ indeed holds. Using some help from WolframAlpha, Inequality $(3)$ holds when $q < n$.


To conclude, $D_O = q^j n^2$ might be an almost perfect divisor of the odd perfect number $O=q^k n^2$, where $j=(k-1)/2$, as we were not able to derive a contradiction from assuming that $D_O$ is indeed almost perfect, and then using our criterion for almost perfect numbers.

I will stop here for the time being, as I have got some more work to do.

$\endgroup$
6
  • 1
    $\begingroup$ Correct me if I am wrong but I thought Holdener and Rachfal had already proven that $D_O$ is deficient perfect but not almost perfect? You have already correctly pointed out using different notation that $D_O$ is not almost perfect because Holdener and Rachfal had proven that its deficiency is not 1 but rather $\frac{p^{2a}m^2}{p^{2a+1}/2}$ where $p^{4a+1}m^2$ is an odd perfect number. $\endgroup$ Commented May 2, 2023 at 21:54
  • $\begingroup$ Indeed, how could I have missed that detail, @User4576283! Posting an answer now. $\endgroup$ Commented May 3, 2023 at 2:00
  • $\begingroup$ Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Consider the following (proper) divisors of $N$, which are necessarily deficient, since $N$ is perfect. Note that this list is not exhaustive: $$q, q^{(k-1)/2}, q^{(k+1)/2}, nq, n q^k , n^2, n^2 q^{(k-1)/2}, n^2 q^{(k+1)/2}. \tag{1}$$ From the list in $(1)$, it suffices to consider the subset $$q^{(k-1)/2}, n^2, n^2 q^{(k-1)/2}, \tag{2}$$ of (proper divisors) of $N$ which might be almost perfect, as an odd almost perfect number has to be a square. (continued) $\endgroup$ Commented May 5, 2023 at 9:12
  • $\begingroup$ This is because the (odd) numbers in the list $(1) \setminus (2)$ are $$q, q^{(k+1)/2}, nq, nq^k, n^2 q^{(k+1)/2}, \tag{3}$$ all of which are necessarily non-squares because of $q$ is a prime number satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$. Henceforth, one can now use the criterion $$\frac{2A}{A+1} < I(A)=\frac{\sigma(A)}{A} < \frac{2A+1}{A+1}$$ which holds if and only if $1 < A$ is almost perfect. $\endgroup$ Commented May 5, 2023 at 9:21
  • $\begingroup$ Blimey! The (proper) divisor $n$ was missing in the original list $(1)$ from this comment, @User4576283. =) $\endgroup$ Commented May 8, 2023 at 13:37

2 Answers 2

1
$\begingroup$

Assume to the contrary that the deficient divisor $e_O$ satisfies $$1 = e_O = \frac{2q^j n^2}{q^{j+1} + 1}. \tag{1}$$ (Note that this condition holds if and only if $d(D_O)=2D_O-\sigma(D_O)=1$.)

At once, we see that $$(1) \iff q^{j+1} + 1 = 2q^j n^2$$ $$\iff q + \left(\frac{1}{q}\right)^j = 2n^2, \tag{2}$$ where one notices that the left-hand side of Equation $(2)$ is not an integer (since $j \geq 2$), which contradicts the fact that the right-hand side of Equation $(2)$ must be an integer.

Thus, $D_O$ is not almost perfect.

(I would like to qualify that I have not re-read Holdener and Rachfal's paper, so I am guessing their proof might be very similar to this one.)

$\endgroup$
1
+100
$\begingroup$

By the way, just to clarify, I do not know for sure if Holdener and Rachfal proved that $$\frac{2p^{2a}m^2}{p^{2a+1}+1} \neq 1$$ in their paper. In fact, I don't think they did.

Your method is correct. There is another way of proving that $$\frac{2p^{2a}m^2}{p^{2a+1}+1} \neq 1,$$ where $$p^{4a+1} m^2$$ is an odd perfect number. Although this method is longer, it is useful in explaining why $$\frac{2p^{2a}m^2}{p^{2a+1}+1} \neq 1.$$

Here is a proof that $$\frac{2p^{2a}m^2}{p^{2a+1}+1}\neq 1.$$

First notice that $$\dfrac{2p^{2a}m^2}{p^{2a+1}+1}=\dfrac{{2p^{2a}}\cdot{\dfrac{\sigma(p^{4a+1})}{2}}\cdot{i(q)}}{p^{2a+1}+1} \quad (*)$$ where $i(q)=\gcd(m^2,\sigma(m^2))$.

Notice that $$\sigma(p^{4a+1})={p^{2a+1}}\cdot{\sigma(p^{2a})}+\sigma(p^{2a})=\left(p^{2a+1}+1\right)\cdot\sigma(p^{2a}).$$

Replacing $\sigma(p^{4a+1})$ with $$(p^{2a+1}+1)\sigma(p^{2a})$$ in Equation $(*)$ above we get: $$\dfrac{{2p^{2a}}\cdot{\dfrac{\sigma(p^{4a+1})}{2}}\cdot{i(q)}}{p^{2a+1}+1}=\frac{{p^{2a}(p^{2a+1}+1)\sigma(p^{2a})}\cdot{i(q)}}{p^{2a+1}+1}={p^{2a}}\cdot{\sigma(p^{2a})}\cdot{i(q)}$$

Therefore we establish the following important result: $$\frac{2p^{2a}m^2}{p^{2a+1}+1}={p^{2a}}\cdot{\sigma(p^{2a})}\cdot{i(q)}.$$

Now it has been proven by various researchers including Dris that $i(q)>1$.

This concludes the proof that $$\frac{2p^{2a}m^2}{p^{2a+1}+1} \neq 1.$$

$\endgroup$
4
  • 1
    $\begingroup$ Just to let you know, I have un-accepted your response because I realized that my inquiry in the original post remains unanswered, @User4576283. Nonetheless, a bounty is on your way for the effort you put in for crafting this answer. $\endgroup$ Commented May 3, 2023 at 15:09
  • $\begingroup$ I have decided to award the bounty to you. Enjoy, @User4576283. =) $\endgroup$ Commented May 13, 2023 at 2:04
  • 1
    $\begingroup$ Thank you @JoseArnaldoBebitaDris for the award. I appreciate it. $\endgroup$ Commented May 14, 2023 at 3:13
  • $\begingroup$ You should check out MO when you get the chance, @User4576283. ^_^ $\endgroup$ Commented May 14, 2023 at 3:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .