Is there a name for this property/theorem of divisibility in modular arithmetic?

Question

Statement: Given $S = az \pm b$ and $T = ak \pm b$, and $a \neq b$:

$$S \mid T \iff k \equiv \pm z \pmod S$$

This holds true only if the three instances of $\pm$ multiply to $+$, i.e., all are $+$ or exactly two are $-$. (Proof at the end.)

I assume I have not discovered something even vaguely new; in fact, this seems like it ought to be a somewhat basic theorem or property, but in going through some online texts and discussions, I don't see it anywhere. It's possible that it's related to the CRT, since we have

$$S \equiv \pm b \pmod a \\ T \equiv \pm b \pmod a \\ k \equiv \pm z \pmod S$$

But since $S$ and $T$ are different integers in the same residue class and the same moduls, but their divisibility is related to the... I guess here, $z$ and $k$ are the quotients by Eucliden division(?)... this doesn't seem to fit with the normal applications of the CRT.

Does anyone know of a name for this theorem or property, or have a reference to it/about it?

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Proof: Take all of the instances of $\pm$ to be positive. We have:

$$S = az + b \\ T = ak + b$$

And the claim is that $S \mid T \iff k \equiv +z \pmod S$. Taking this equivalence between $k$ and $z$, we can rewrite $k$ as:

$$k = nS + z = n(az+b) + z$$

for some integer $n$. Then

$$T = ak+b = a(n(az+b)+z)+b \tag{1}$$ $$T = a^2nz+anb+az+b = an(az+b) + (az+b) = (az+b)(an+1) \tag{2}$$

The RHS of $(1)$ contains all three of the signs of interest. Replacing any two with negative signs gives the equivalents:

$$T = a(n(az-b)-z)+b = (az-b)(an-1)\\ T = a(n(az-b)+z)-b = (az-b)(an+1)\\ T = a(n(az+b)-z)-b = (az+b)(an-1)$$

Taking all three or any one sign as negative leads to a mismatch, requiring $S$ to be both $az+b$ and $az-b$.

Answer 1

$\!\bmod az-b\!:\ \color{#c00}{b\equiv az}\,$ so $\,ak-\color{#c00}b\equiv ak-\color{#c00}{az}\equiv a(k-z),\,$ so $\,ak-b\equiv 0\iff \color{#90f}a(\color{#0a0}{k-z})\equiv 0.\,$ This is not generally equivalent to $\,\color{#0a0}{k\equiv z}\,$ except when $\,\color{#90f}a\,$ is cancellable (= invertible), i.e. when $\,1 = (a,\,az-b) = (a,b).\,$ Then $\, b/a\equiv ba^{-1}$ is the unique root of $\,ax\equiv b,\,$ so since $\,k\,$ and $z$ are roots we deduce $\,k\equiv z\,$ by uniqueness.

There is no standard name for such simple inferences from basic congruence laws, but it can be interpreted as uniqueness of (multivalued) modular fractions, as explained briefly here (it shows that the reduced fraction is unique w.r.t. a reduced modulus, i.e. when we reduce the fraction by cancelling its gcd, we must also cancel the gcd from the modulus too, i.e. cancel it everywhere). But this is not a standard topic, since almost all ENT authors omit treatment of modular fractions at such introductory levels.