In [this](https://www.worldscientific.com/doi/10.1142/S0219498820501017) 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.



Difference sequences of sets of integers
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.


