Preamble: I apologize in advance if what I am asking for in this question, I could get an answer easily so myself, for example by coding a short Mathematica script. It is just that I have not yet studied the rudiments of Mathematica (or even Python) scripting, so I am not there yet.
Let $\sigma(z)$ be the sum of the divisors of $z \in \mathbb{N}$. Denote the deficiency of $z$ by $D(z) := 2z - \sigma(z)$ and the sum of the aliquot parts of $z$ by $s(z) := \sigma(z) - z$.
Here is my question:
Does there exist a natural number $N = xy$ satisfying $D(x)D(y) = 2s(x)s(y)$ with $\gcd(x,y)=1$?
I would also be interested in answers to the more general question:
Does there exist a natural number $N = xy$ satisfying $D(x)D(y) = 2s(x)s(y)$?
MY TRY
I note that, in general, the deficiency and sum-of-aliquot-parts functions are not weakly multiplicative. That is: $$D(xy) \neq D(x)D(y)$$ and $$s(xy) \neq s(x)s(y)$$ when $\gcd(x,y)=1$.
In fact, I know that when $\gcd(x,y)=1$, then both the inequalities $$D(xy) \leq D(x)D(y)$$ and $$s(xy) < s(x)s(y)$$ hold.
Of course, $D(xy)$ and $s(xy)$ are related by the equation $$D(xy) + s(xy) = xy.$$ We also have the equations $$D(x) + s(x) = x$$ $$D(y) + s(y) = y.$$ Multiplying the last two equations, and equating I obtain: $$D(x)D(y) + s(y)D(x) + s(x)D(y) + s(x)s(y) = xy = D(xy) + s(xy).$$
We are given that $D(x)D(y)=2s(x)s(y)$. So we substitute and obtain $$3s(x)s(y) + s(y)D(x) + s(x)D(y) = xy$$ $$3s(x)s(y) + s(y)\bigg(x - s(x)\bigg) + s(x)\bigg(y - s(y)\bigg) = xy$$ $$s(x)s(y) + xs(y) + ys(x) = xy$$ $$s(y)(s(x) + x) + s(x)(s(y) + y) = s(x)s(y)+xy.$$
It is here where I get stuck. I feel that I am just going around in circles here. Note that the condition $$s(y)(s(x) + x) + s(x)(s(y) + y) = s(x)s(y)+xy$$ is derived independently of the divisibility constraint $$\gcd(x,y)=1.$$
POSTING WHAT FOLLOWS AS AN ANSWER TO THIS QUESTION
Added April 12, 2018 Now I get it! This last condition $$s(y)(s(x) + x) + s(x)(s(y) + y) = s(x)s(y)+xy$$ together with the divisibility constraint $$\gcd(x,y)=1$$ simplifies to $$\sigma(xy)=\sigma(x)\sigma(y)=2xy$$ which implies that $N = xy$ must be a perfect number.
Added April 12, 2018 Now I get it! This last condition $$s(y)(s(x) + x) + s(x)(s(y) + y) = s(x)s(y)+xy$$ together with the divisibility constraint $$\gcd(x,y)=1$$ simplifies to $$\sigma(xy)=\sigma(x)\sigma(y)=2xy$$ which implies that $N = xy$ must be a perfect number.