Existance of a nonempty subset $X$ of $\mathbb{Z}$: $\forall x \in X, k \in \mathbb{N}, \exists ! y \in X$ satisfying $\mid x-y \mid 2^k$

Question

Define the set $A = \{2^k \mid k \in \mathbb{Z}_+\}$

Does there exist a nonempty subset $X$ of $\mathbb{Z}$ satisfying the following condition: For all $a \in A$ and $x \in X$, there exists a unique element $y \in X$ such that $\mid x-y \mid = a$?

In other words, for all $x \in X$ and $k \in \mathbb{Z}_+$ we can find a unique $y_k \in X$ (dependent on $k$) such that $\mid x - y_k\mid = 2^k$.

At first I assumed that such a subset exist, and its elements are of the form $2^n - 2^m$ for positive integers $m, n$, and try to show that each integer can be expressed in this form for a unique pair $(m,n)$, but I couldn't do it yet. The thing that matters, is that I have no clue whether my assumption was right or wrong; therefore maybe I'm trying to prove a wrong clause.

Can anyone help me with this? All contributive help will be aprreciated. Thank you.

Answer 1

HINT: Take $X$ to be the set of even integers that are not divisible by $3$. In case you run into too much trouble, there’s a further hint in the spoiler-protected block.

For any $x\in X$ and $k\in\Bbb Z^+$, exactly one of $x+2^k$ and $x-2^k$ is divisible by $3$.