Hereinafter, we shall let $\sigma(z)$ be the

*sum of divisors*of the positive integer $z$. Denote the deficiency of $z$ by $D(z) = 2z - \sigma(z)$, and the sum of aliquot divisors of $z$ by $s(z) = \sigma(z) - z$.
We shall compute here a formula for $D(x)D(y) - D(xy)$ in terms of the sum-of-aliquot-divisors function, when $\gcd(x,y)=1$.

Suppose that $\gcd(x,y)=1$.

Then we have

$$D(x)D(y) - D(xy) = (2x - \sigma(x))(2y - \sigma(y)) - (2xy - \sigma(xy))$$

$$= 4xy - 2y\sigma(x) - 2x\sigma(y) + \sigma(x)\sigma(y) - 2xy + \sigma(x)\sigma(y),$$

where we have used the condition $\gcd(x,y)=1$ in the last equation to derive $\sigma(xy)=\sigma(x)\sigma(y)$.

This gives

$$D(x)D(y) - D(xy) = 2xy - 2y\sigma(x) - 2x\sigma(y) + 2\sigma(x)\sigma(y)$$

so that we obtain

$$D(x)D(y) - D(xy) = 2y\bigg(x - \sigma(x)\bigg) - 2\sigma(y)\bigg(x - \sigma(x)\bigg)$$

which simplifies to

$$D(x)D(y) - D(xy) = 2\bigg(x - \sigma(x)\bigg)\bigg(y - \sigma(y)\bigg) = 2\bigg(\sigma(x) - x\bigg)\bigg(\sigma(y) - y\bigg) = 2s(x)s(y).$$

Here are my inquiries:

**QUESTIONS**

**(1)**Is it possible to extend the formula

$$D(x)D(y) - D(xy) = 2s(x)s(y)$$

to, say, something that uses three or more arguments (which are pairwise coprime)?

**(2)**If the answer to

*Question*

**(1)**is

*YES*, what is the closed form for the formula and how can it be proved, in general?

**POSTED ATTEMPT**

Here is my own attempt for the case of

**three**($3$) arguments.
Suppose that

$$\gcd(x,y)=\gcd(x,z)=\gcd(y,z)=1.$$

Then we have

$$D(x)D(y)D(z) - D(xyz) = (2x-\sigma(x))(2y-\sigma(y))(2z-\sigma(z))-(2xyz-\sigma(xyz))$$

$$=(4xy-2y\sigma(x)-2x\sigma(y)+\sigma(x)\sigma(y))(2z-\sigma(z))-2xyz+\sigma(x)\sigma(y)\sigma(z)$$

$$=8xyz-4yz\sigma(x)-4xz\sigma(y)+2z\sigma(x)\sigma(y)-4xy\sigma(z)+2y\sigma(x)\sigma(z)+2x\sigma(y)\sigma(z)-\sigma(x)\sigma(y)\sigma(z)-2xyz+\sigma(x)\sigma(y)\sigma(z)$$

$$=2xyz-2yz\sigma(x)-2yz\sigma(x)+2z\sigma(x)\sigma(y)$$

$$+2xyz-2xz\sigma(y)-2xz\sigma(y)+2x\sigma(y)\sigma(z)$$

$$+2xyz-2xy\sigma(z)-2xy\sigma(z)+2y\sigma(x)\sigma(z),$$

from which we obtain

$$=2yz(x-\sigma(x))-2z\sigma(x)(y-\sigma(y))$$

$$+2xz(y-\sigma(y))-2x\sigma(y)(z-\sigma(z))$$

$$+2xy(z-\sigma(z))-2y\sigma(z)(x-\sigma(x))$$

from which we get

$$=2y(x-\sigma(x))(z-\sigma(z))+2z(y-\sigma(y))(x-\sigma(x))+2x(z-\sigma(z))(y-\sigma(y)).$$

**This finally gives the formula**

$$D(x)D(y)D(z)-D(xyz)=2\bigg(xs(y)s(z)+ys(x)s(z)+zs(x)s(y)\bigg).$$

**Checking the formula for**$(x,y,z)=(3,5,7)$

**gives**

$$D(x)D(y)D(z)-D(xyz)=D(3)D(5)D(7)-D(105)=2\cdot{4}\cdot{6}-18=48-18=30$$

$$2\bigg(xs(y)s(z)+ys(x)s(z)+zs(x)s(y)\bigg)=2\bigg(3\cdot s(5)s(7)+5\cdot s(3)s(7)+7\cdot s(3)s(5)\bigg)=2\bigg(3\cdot{1}\cdot{1}+5\cdot{1}\cdot{1}+7\cdot{1}\cdot{1}\bigg)=2\cdot{15}=30.$$