In this paper, the conception of the difference sequence and $\infty$-difference length of a subset of groups is introduced. As an important case, subsets of the additive group of integers are considered as follows:
Let $0\in A\subseteq \mathbb{Z}$ and put $A_0:=A$, $A_n:=A_{n-1}-A_{n-1}$, for all $n\in \mathbb{Z}^+$. Note that $A-A=\{a_1-a_2: a_1,a_2\in A\}$, $A\subseteq A_1\subseteq A_2\subseteq
\cdots \subseteq A_n\subseteq \cdots \subseteq \bigcup_{n=1}^\infty A_n$, and $\bigcup_{n=1}^\infty A_n$ is a subgroup of $(\mathbb{Z},+)$.
Now, there are two special questions:
(a) if $A=\{2^m:m\in \mathbb{Z}^+\}\cup\{0\}$, then is there any positive integer $N$ such that $A_N=A_{N+1}$? (if yes, what is the least such $N$?)
(b) what about $A=\{m^k:m\in \mathbb{Z}\}$, where $k\geq 3$ is a fixed integer?
Remark.
We know that if the following properties hold, then the answer of (a) is negative:
(1) for every even integer $n$, there exists an integer $k=2^q$ and $a_1,\cdots,a_k\in A$ such that
$$
n=a_1+\cdots+a_{\frac{k}{2}}-(a_{\frac{k}{2}+1}+\cdots+a_k);
$$
(2) denoting by $k(n)$ the least $k$ obtained from (1), the set of all $k(n)$
where $n$ runs over all even integers, is unbounded above.
Best Answer
Sketch of a proof for (a): $A_0$ has $m$ elements smaller than or equal to $2^m$. You can form $m^2$ pairs of them, so $A_1$ has (at most) $m^2$ elements with an absolute value smaller than or equal to $2^m$ (the larger elements don't play a role, which can be seen by looking at the binary expansion). That means $A_n$ has at most $m^{2^n}$ elements with an absolute value smaller than or equal to $2^m$.
Every positive even number $s$ can be written (uniquely) as the sum of powers of 2 (again, look at the binary expansion): $s = \sum_{i=1}^{t}{2^{s_i}}$ with $t$ and each $s_i$ positive integers. Now $-(2^m) \in A_1$ for every positive integer $m$, and we can write $s = 2^{s_t} - (-(2^s_{t-1})) - (-(2^s_{t-2})) ... - (-(2^s_1))$ so $s \in A_t$. A similar sum/difference works for negative even numbers. So if an $N$ would exist as described in (a), $A_N$ should contain all even numbers.
Now take $m = N^N$, then $A_N$ has $N^{N2^N}=2^{N \log_2 N 2^N}$ elements with an absolute value smaller than or equal to $2^{N^N}$, while there are $2^{N^N}+1$ even numbers, which is (much) more.
Conclusion: such an $N$ does not exist.