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

Question

(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?

Answer 1

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 $$\begin{align}D(y)D(z)&=(2y-\sigma(y))(2z-\sigma(z))\\\\&=(2y-\sigma(y))\left(2z-\frac{2yz}{\sigma(y)}\right) \\\\&=4yz-\frac{4y^2z}{\sigma(y)}-2z\sigma(y)+2yz \\\\&=4yz-2z\sigma(y)-\frac{4y^2z}{\sigma(y)}+2yz \\\\&=2(\sigma(y)-y)\left(\frac{2yz}{\sigma(y)}-z\right) \\\\&=2(\sigma(y)-y)(\sigma(z)-z) \\\\&=2s(y)s(z)\end{align}$$

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

Answer 2

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: $$\frac{3}{5} = \frac{8}{5} - 1 < \frac{\sigma(n^2)}{n^2} - 1 = \frac{s(n^2)}{n^2} = \frac{q-1}{2n^2} < \frac{q}{2n^2} \leq \frac{q^k}{2n^2} < \frac{1}{2}\cdot\frac{2}{3} = \frac{1}{3}.$$