This question is a follow-up to these previous posts: MSE1 and MSE2.
Let $x, y$ be positive integers. We call $\sigma(x)$ the sum of the divisors of $x$.
Let the deficiency function $D(x)$ be defined by $$D(x) = 2x - \sigma(x).$$
Now here is my question:
If $x \mid y$, does it follow that $D(xy) = D(x)D(y)$? If the implication does not hold in general, what additional assumptions on $x, y \in \mathbb{N}$ must be made in order for the equation to hold?
So for example, take $x = 2$ and $y = 4$, so that: $$D(8) = 1 = 1\cdot1 = D(2)D(4).$$
If I take $x = 3$ and $y = 9$, then $$D(27) = 14 = 2\cdot7 \neq 2\cdot5 = D(3)D(9).$$
If $x = 3$ and $y = 15$, we get $$D(45) = 12 = 2\cdot6 = D(3)D(15).$$
Lastly, if $x = 3$ and $y = 21$, we obtain $$D(63) = 22 = 2\cdot{11} \neq 2\cdot{10} = D(3)D(21).$$
Comments are most welcome. Thanks!
Added February 13 2017
It turns out we may say something about $x$, $y$, $D(x)$, and $D(y)$ when $x \mid y$.
If $x \mid y$, then by a property of the abundancy index $I(z)=\sigma(z)/z$, we have $$I(x) \leq I(y)$$ $$2 - I(y) \leq 2 - I(x)$$ $$\dfrac{2y-\sigma(y)}{y} \leq \dfrac{2x-\sigma(x)}{x}$$ $$\dfrac{D(y)}{y} \leq \dfrac{D(x)}{x}.$$
Additionally, note that we have $x^2 \mid xy \mid y^2$ if $x \mid y$, so that $$I(x^2) \leq I(xy) \leq I(y^2)$$ $$2 - I(y^2) \leq 2 - I(xy) \leq 2 - I(x^2)$$ $$\dfrac{2y^2 - \sigma(y^2)}{y^2} \leq \dfrac{2xy - \sigma(xy)}{xy} \leq \dfrac{2x^2 -\sigma(x^2)}{x^2}$$ $$\dfrac{D(y^2)}{y^2} \leq \dfrac{D(xy)}{xy} \leq \dfrac{D(x^2)}{x^2}.$$
I am tempted to call the ratio $D(z)/z$ as the deficiency index, moving forward!