# Are there (restricted) instances when the deficiency and sum-of-proper-divisors functions are multiplicative?

A function $$f : \mathbb{N} \rightarrow \mathbb{Q}$$ is said to be multiplicative if $$f(ab) = f(a)f(b)$$ whenever $$\gcd(a,b)=1$$.

It is known that the sum-of-divisors function $$\sigma(x) = \sum_{d \mid x}{d}$$ is multiplicative. It follows that the abundancy index function $$I(x) = \frac{\sigma(x)}{x}$$ is also multiplicative.

It is also known that the deficiency $$D(x) = 2x - \sigma(x)$$ and the sum-of-proper-divisors $$s(x) = \sigma(x) - x$$ functions are, in general, not multiplicative.

My question is:

Are there (restricted) instances when the deficiency and sum-of-proper-divisors functions are multiplicative?

MOTIVATION FOR THE QUESTION

When $$yz$$ is a perfect number for $$\gcd(y,z)=1$$, then we know that $$\sigma(yz) = \sigma(y)\sigma(z) = 2yz.$$

It turns out that we can also show that $$D(y)D(z) = 2s(y)s(z),$$ if $$yz$$ is a perfect number with $$\gcd(y,z)=1$$.

(Of course, I do not hope to show that $$D(yz)=D(y)D(z)$$ if $$yz$$ is perfect and $$\gcd(y,z)=1$$, since $$yz$$ is perfect implies that $$D(yz)=0$$. I just want to know if further simplified expressions may be obtained for either $$D(y)D(z)$$ or $$s(y)s(z)$$ (with $$\gcd(y,z)=1$$), whether or not $$yz$$ is perfect.)

• At least, for $s(x)$ the answer is negative. For, $s(x)s(y)=s(xy)$ would be equivalent to $(\sigma(x)-x)(\sigma(y)-y)=\sigma(xy)-xy$, which by multiplicativity of $\sigma$ can be written as $y\sigma(x)+x\sigma(y)=2xy$, and hence is equivalent to $$\frac{\sigma(x)}{x}+\frac{\sigma(y)}y=2.$$ However, each of the two summands in the LHS exceeds $1$, except if $x=1$ and / or $y=1$. Jul 13, 2019 at 19:48
• @W-t-P: Thank you for your comment. Your argument appears to show that $s(x)s(y)=s(xy)$ for $\gcd(x,y)=1$ only holds when $x=1$ and $y=1$. Of course, we know that $\gcd(1,y)=\gcd(x,1)=1$, $s(1)s(y)=s(1\cdot{y})=s(y)$, and $s(x)s(1)=s({x}\cdot{1})=s(x)$. Note that $s(1)=\sigma(1)-1=1-1=0$. So your logical connective should be an AND instead of an OR. Jul 14, 2019 at 5:59
• You are right, the argument shows that for $s(xy)=s(x)s(y)$ with $(x,y)=1$ to hold, it is necessary and sufficient that $x=y=1$. On the other hand, formally, "and / or" can be replaced by just "or". Jul 14, 2019 at 6:46
• @W-t-P: I invite you to write out your (first) comment as an actual answer, because I think it is correct. Jul 14, 2019 at 6:59

None of these functions is multiplicative on any reasonable domain; in fact, for $$x$$ and $$y$$ coprime, we have $$s(xy)=s(x)s(y)$$ if and only if $$x=y=1$$, and $$D(xy)=D(x)D(y)$$ if and only if $$\min\{x,y\}=1$$.
Sufficiency is readily verified. For necessity, notice that $$s(x)s(y)=s(xy)$$ is equivalent to $$(\sigma(x)−x)(\sigma(y)−y)=\sigma(xy)−xy$$, which by multiplicativity of $$\sigma$$ is further equivalent to $$yσ(x)+xσ(y)=2xy$$, and eventually can be written as $$\frac{\sigma(x)}x + \frac{\sigma(y)}y=2.$$ Taking into account that $$\sigma(x)\ge x$$ and $$\sigma(y)\ge y$$, with equalities if and only if $$x=1$$ and $$y=1$$, respectively, we conclude that $$x=y=1$$.
Similarly, for $$x$$ and $$y$$ coprime, the equality $$D(xy)=D(x)D(y)$$ can be equivalently rewritten as $$(\sigma(x)-x)(\sigma(y)-y)=0,$$ which implies $$x=1$$ or $$y=1$$ since if we had $$x,y>1$$, then both factors in the LHS were strictly positive.
• Just double-checking: $$D(xy) = D(x)D(y) \implies 2xy - \sigma(xy) = (2x-\sigma(x))(2y-\sigma(y)) = 4xy - 2x\sigma(y) - 2y\sigma(x) + \sigma(x)\sigma(y) \implies 2xy - 2x\sigma(y) - 2y\sigma(x) + 2\sigma(x)\sigma(y) = 2\bigg(x(y - \sigma(y)) - \sigma(x)(y - \sigma(y)\bigg) = 0$$ from which it (indeed) follows that $$\bigg(\sigma(x)-x\bigg)\bigg(\sigma(y)-y\bigg)=0.$$ Jul 14, 2019 at 10:28