I'm an amateur 42 years removed from academia so I don't always know "the language".
In a paper I am writing, I specify search limits like $\;4\le m \le 9\;$ and then show, in examples, how some members in the search set yield integers for a function: $\;k=f(m).\quad$ In an example where $f(4)=1\land f(8)=7\;$ I have been using the following notation. $$m\in\{4,8\}\implies k \in\{1,7\} $$
The "sets" can be larger or more complex such as for a cubic where there can be up to $3$ solutions for a given $m$. $$m\in \{7\}\implies k\in\{5,8,3\}\qquad\land\qquad m\in\{8\}\implies k\in\{7\}$$
Does this notation make sense at all and does it make it clear that $4$ corresponds to $1$ and that $8$ corresponds to $7$? Alternatively, is there a better way to express these "summaries"?
This notation is the briefest I could think of outside of showing ordered pairs such as $\quad(7,5)\;(7.8)\;(7,3)\quad (8,7).\quad$ What I want to emphasize is the the "search" found its objective(s) within the specified limits because the point of the bigger "thing" is that what we are looking for can be found $\textbf{without brute force}$.
$\textbf{Update:}$ In response to where this fits in the middle of a paper, here is one of a dozen or more "problems" as I have written them so far. The question remains about the best way to present the results of the search.
$$mk^3-m^3k+D=0\implies$$ \begin{equation} k_0=\sqrt{\frac{4m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\cos^{-1}{\biggl(-\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}\\ \quad k_1=\sqrt{\frac{4m^2}{3}}\cos\biggl({\biggl(\frac{1}{3}\biggr)\cos^{-1}{\biggl(\frac{3\sqrt{3}D}{2m^4}\biggr)}\biggr)}\\ \quad k_2=k_1-k_0 \end{equation}
$$\text{ for }\quad\lfloor\sqrt[4]{2D}\rfloor\approx\sqrt[4]{\frac{8D}{3}} \le m \\ \le \frac{\sqrt[3]{\sqrt{3} \sqrt{27 D^2 - 4} + 9 D}}{\sqrt[3]{2} \sqrt[3]{3^2}} + \frac{\sqrt[3]{\frac{2}{3}}}{\sqrt[3]{\sqrt{3} \sqrt{27 D^2 - 4} + 9 D}}\approx\big\lceil\sqrt[3]{D}\big\rceil$$
$$D=840\implies \lfloor\sqrt[4]{2(840)}\rfloor=6 \le m \le \lceil\sqrt[3]{840}\rceil=10\quad\text {and we find}$$ $$m\in \{7\}\implies k\in\{5,8,3\}\qquad\land\qquad m\in\{8\}\implies k\in\{7\}$$ $$F(7,5)=(24,70,74)\quad F(7,8)=(-15,112,113)\quad F(7,3)=(40,42,58)\\ F(8,7)=(15,112,113)$$
Not related to the original question. The "approximate" limits shown are what I found heuristically before I was able to determine the more complicated "exact" limits. I'm thinking of trashing the latter because the approximate limits are lower and higher and would not add much to the search. Still, the exact limits $are$ exact and omitting them might detract from the precision expected in a paper for publication.