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

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.

• 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. Commented May 2, 2023 at 21:54
• Indeed, how could I have missed that detail, @User4576283! Posting an answer now. Commented May 3, 2023 at 2:00
• 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) Commented May 5, 2023 at 9:12
• 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. Commented May 5, 2023 at 9:21
• Blimey! The (proper) divisor $n$ was missing in the original list $(1)$ from this comment, @User4576283. =) Commented May 8, 2023 at 13:37

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.)

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 $$\frac{2p^{2a}m^2}{p^{2a+1}+1} \neq 1.$$

• 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. Commented May 3, 2023 at 15:09
• I have decided to award the bounty to you. Enjoy, @User4576283. =) Commented May 13, 2023 at 2:04
• Thank you @JoseArnaldoBebitaDris for the award. I appreciate it. Commented May 14, 2023 at 3:13
• You should check out MO when you get the chance, @User4576283. ^_^ Commented May 14, 2023 at 3:27