Is it possible to compute $\sigma(AB)$ if $\gcd(A, B) = C > 1$?

Question

Let $\sigma$ be the classical sum-of-divisors function.

I know, for one, that $\sigma$ is weakly multiplicative (that is, $\sigma(xy) = \sigma(x)\sigma(y)$ whenever $\gcd(x, y) = 1$).

Well, of course I also know how to bound $\sigma(AB)$ if $\gcd(A, B) > 1$:

We have the bounds $A\sigma(B) < \sigma(AB) < \sigma(A)\sigma(B)$.

I also have the bound $B\sigma(A) < \sigma(AB)$.

My question is this:

Is it possible to compute $\sigma(AB)$ in terms of $C = \gcd(A, B) > 1$?

I am not too sure if the equation $$\sigma(AB) = \frac{1}{C}\cdot\sigma(A)\sigma(B)$$ works.

Or something like the following perhaps?

$$\sigma\left(\frac{AB}{C}\right) = \sigma\left(\frac{A}{C}\right)\sigma\left(\frac{B}{C}\right)$$

Answer 1

Note that you can always write $\sigma(AB/C^2)=\sigma(A/C)\sigma(B/C)$ whenever $C=gcd(A,B)$ since $gcd(A/C,B/C)=1$