(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?
Best Answer
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 $$\begin{align}&(2y-\sigma(y))(2z-\sigma(z))=2(\sigma(y)-y)(\sigma(z)-z) \\\\&\implies 4yz-2y\sigma(z)-2z\sigma(y)+\sigma(y)\sigma(z)=2\sigma(y)\sigma(z)-2z\sigma(y)-2y\sigma(z)+2yz \\\\&\implies 2yz=\sigma(y)\sigma(z) \\\\&\implies 2yz=\sigma(yz) \\\\&\implies \text{$yz$ is perfect}\end{align}$$