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(Preamble: This question is an offshoot of this earlier MSE post.)

The title says it all.

Is $\bigg(q^k n^2 \text{ is perfect }\bigg) \iff \bigg(D(q^k)D(n^2) = 2s(q^k)s(n^2)\bigg)$ only true for odd perfect numbers $q^k n^2$?

Here, $$D(x) = 2x - \sigma(x)$$ is the deficiency of $x$, $$s(x) = \sigma(x) - x$$ is the sum of the aliquot divisors of $x$, and $\sigma(x)$ is the sum of divisors of $x \in \mathbb{N}$, the set of positive integers.

IN RESPONSE TO A CLARIFICATION FROM mathlove

That is:

If $\gcd(y,z)=1$, is the biconditional "$yz$ is perfect $\iff D(y)D(z)=2s(y)s(z)$" always true?

Jose Arnaldo Bebita Dris
  • 8,471
  • Transferring my answer from MY ATTEMPT section (onwards) to an actual answer to this question below, in a bit... – Jose Arnaldo Bebita Dris Oct 02 '18 at 09:34
  • Thanks for fixing the title, @HenningMakholm. – Jose Arnaldo Bebita Dris Oct 02 '18 at 09:52
  • I don't get the meaning of the question. Can you make it clearer? Or I would say that it is only true for odd perfect numbers because it is about odd perfect numbers. In the answer below, you say "Therefore, the required relationship $D(q^k)D(n^2) = 2s(q^k)s(n^2)$ holds for both even and odd perfect numbers." I think this is ambiguous. I don't understand the meaning of the sentence "$D(q^k)D(n^2)=2s(q^k)s(n^2)$ holds for even perfect numbers". You might want to clarify what you want to say. – mathlove Oct 11 '18 at 15:17
  • @mathlove, for even perfect numbers, just set $q = 2^p - 1$, $k = 1$, and $n^2 = 2^{p-1}$. Of course, we then have to exclude the even perfect number $6$ because it is squarefree. – Jose Arnaldo Bebita Dris Oct 12 '18 at 03:22
  • It turns out that the formula still holds even with $6$ included. – Jose Arnaldo Bebita Dris Oct 12 '18 at 03:25
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    @mathlove, I have edited my question in response to your clarification in your last comment. – Jose Arnaldo Bebita Dris Oct 15 '18 at 05:46
  • Thanks. One question. Which do you mean (1) Is the biconditional "$yz$ is perfect and $\gcd(y,z)=1$ $\iff$ ..." always true? or (2) Suppose $\gcd(y,z)=1$. Then, is the biconditional "$yz$ is perfect $\iff$ ..." always true? – mathlove Oct 15 '18 at 12:03
  • @mathlove: Good catch! I meant (2). – Jose Arnaldo Bebita Dris Oct 15 '18 at 12:35

2 Answers2

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If $\gcd(y,z)=1$, is the biconditional "$yz$ is perfect $\iff D(y)D(z)=2s(y)s(z)$" always true?

Yes.

If $yz$ is perfect with $\gcd(y,z)=1$, then since $$\sigma(yz)=\sigma(y)\sigma(z)=2yz$$ we have $$\begin{align}D(y)D(z)&=(2y-\sigma(y))(2z-\sigma(z))\\\\&=(2y-\sigma(y))\left(2z-\frac{2yz}{\sigma(y)}\right) \\\\&=4yz-\frac{4y^2z}{\sigma(y)}-2z\sigma(y)+2yz \\\\&=4yz-2z\sigma(y)-\frac{4y^2z}{\sigma(y)}+2yz \\\\&=2(\sigma(y)-y)\left(\frac{2yz}{\sigma(y)}-z\right) \\\\&=2(\sigma(y)-y)(\sigma(z)-z) \\\\&=2s(y)s(z)\end{align}$$

If $D(y)D(z)=2s(y)s(z)$ and $\gcd(y,z)=1$, then $$\begin{align}&(2y-\sigma(y))(2z-\sigma(z))=2(\sigma(y)-y)(\sigma(z)-z) \\\\&\implies 4yz-2y\sigma(z)-2z\sigma(y)+\sigma(y)\sigma(z)=2\sigma(y)\sigma(z)-2z\sigma(y)-2y\sigma(z)+2yz \\\\&\implies 2yz=\sigma(y)\sigma(z) \\\\&\implies 2yz=\sigma(yz) \\\\&\implies \text{$yz$ is perfect}\end{align}$$

mathlove
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MY ATTEMPT

For even perfect numbers $2^{p-1}(2^p - 1)$, I get $$D(2^p - 1)D(2^{p-1}) = (2(2^p - 1) - (2^p))(1) = 2^{p+1} - 2 - 2^p = 2^p - 2$$ $$2s(2^p - 1)s(2^{p-1}) = 2(2^p - (2^p - 1))(2^p - 1 - 2^{p-1}) = 2(1)(2^{p-1} - 1) = 2^p - 2.$$

Thus, the equation $$D(2^p - 1)D(2^{p-1}) = 2s(2^p - 1)s(2^{p-1}) = 2^p - 2$$ is true.

Therefore, the required relationship $$D(q^k)D(n^2) = 2s(q^k)s(n^2)$$ holds for both even and odd perfect numbers.

Here is my question:

Does this proof suffice?

Added October 02 2018

Note that it would appear as though we have the corresponding equation $$D(q^k)D(n^2) = 2s(q^k)s(n^2) = q^k - 1$$ for odd perfect numbers. We show here that this assumption is false.

Assuming that $D(q^k)D(n^2) = 2s(q^k)s(n^2) = q^k - 1$, we get $$\frac{2(q^k - 1)}{(q - 1)}s(n^2) = q^k - 1,$$ since $s(q^k) = \sigma(q^k) - q^k = \sigma(q^{k-1})$. This simplifies to $$s(n^2) = \frac{q-1}{2}$$ or $$\frac{\sigma(n^2)}{n^2} - 1 = \frac{s(n^2)}{n^2} = \frac{q-1}{2n^2}.$$ Using the following results from this paper: $$\frac{8}{5} < \frac{\sigma(n^2)}{n^2}$$ and $$\frac{q}{n^2} \leq \frac{q^k}{n^2} < \frac{2}{3},$$ we get a contradiction, as follows: $$\frac{3}{5} = \frac{8}{5} - 1 < \frac{\sigma(n^2)}{n^2} - 1 = \frac{s(n^2)}{n^2} = \frac{q-1}{2n^2} < \frac{q}{2n^2} \leq \frac{q^k}{2n^2} < \frac{1}{2}\cdot\frac{2}{3} = \frac{1}{3}.$$

Jose Arnaldo Bebita Dris
  • 8,471