# A solution for $A-A=\mathbb{Q}\setminus \{\pm1,\pm 2,\cdots,\pm m\}$

Let $$m$$ be a given natural number and consider the additive group of rational numbers $$\mathbb{Q}$$. We are looking for a subset $$A\subseteq \mathbb{Q}$$ such that $$A-A=\mathbb{Q}\setminus \{\pm1,\pm 2,\cdots,\pm m\}$$.

For $$m=1$$, one can see the answer in Some co-finite subsets of rational numbers.

A candidate solution is of the form $$A= \left\{ m+r_m+\sum_{k=m}^n r_k : n \ge m-1\right\} \cup B$$, where $$\Bbb{Q} \cap (m, +\infty) = \{ r_k \}_{k \ge m}$$ and $$B$$ is an appropriate subset of $$\Bbb{Q} \cap (-m,m)$$ (but we are not sure about existence of such $$B$$).

Note that $$A-A=\{a_1-a_2: a_1,a_2\in A\}$$.