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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.)

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    $\begingroup$ 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 $y=1$. $\endgroup$
    – W-t-P
    Jul 13, 2019 at 19:48
  • $\begingroup$ @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. $\endgroup$ Jul 14, 2019 at 5:59
  • $\begingroup$ 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". $\endgroup$
    – W-t-P
    Jul 14, 2019 at 6:46
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    $\begingroup$ @W-t-P: I invite you to write out your (first) comment as an actual answer, because I think it is correct. $\endgroup$ Jul 14, 2019 at 6:59

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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.

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    $\begingroup$ 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 $$\bigg(\sigma(x)-x\bigg)\bigg(\sigma(y)-y\bigg)=0.$$ $\endgroup$ Jul 14, 2019 at 10:28
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    $\begingroup$ Gladly accepting your answer now, @W-t-P! $\endgroup$ Jul 14, 2019 at 10:29

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