Given the set

$$
M=\left\{-2^{2022},-2^{2021}, \ldots,-2^{2},-2,-1,1,2,2^{2}, \ldots, 2^{2021}, 2^{2022}\right\}
$$

Let $T$ be a subset of $M$ where adjacent numbers have the same difference when the elements are ordered by size.

![](https://cdn.mathpix.com/cropped/2024_04_17_b77a219ebee9137e97ffg-5.jpg?height=1140&width=1145&top_left_y=104&top_left_x=453)

Figure 1: Problem 3, Solution 1

(a) Determine the largest number of elements that such a set $T$ can contain.

(b) List all sets $T$ with the largest possible number of elements.

(Walther Janous)
(a) We first prove that in a set $T$ at most two elements can have the same sign. For example, let $2^{a}b$ and $cc$. Since $a>b$, thus $a \geq b+1$, it follows from this that $2^{b+1}$ is a divisor of $2^{a}$. Therefore, $2^{b+1}$ must also divide $2^{c}$. Thus, $c \geq b+1$. Consequently, our equation is equivalent to

$$
2^{a-b-1}-2^{c-b-1}=1
$$

with $a-b-1>c-b-1 \geq 0$. In the case of $c-b-1>0$, the left-hand side expression would be divisible by 2, an obvious contradiction. Therefore, $c=b+1$. The last equation then reads

$$
2^{a-b-1}-1=1 \Longleftrightarrow 2^{a-b-1}=2 \Longleftrightarrow a-b-1=1 \Longleftrightarrow a=b+2 .
$$

Thus, the elements of $T$ are $-2^{b+2}<-2^{b}<2^{b+1}<2^{d}$, but it must also hold that

$$
2^{b+1}-\left(-2^{b}\right)=2^{d}-2^{b+1} \Longleftrightarrow 3 \cdot 2^{b}=2^{b+1}\left(2^{d-b-1}-1\right)
$$

which is an obvious contradiction. In a similar manner, it follows that if one starts from two positive numbers, $T$ can contain at most one negative number. Thus, it is shown that 3 is the maximum number of elements that a set $T$ can contain.

(b) From the above proof, it follows immediately that the sets $T$

- either consist of the three numbers $-2^{b+2},-2^{b}$, and $2^{b+1}$
- or $-2^{b+1}, 2^{b}$, and $2^{b+2}$

where $0 \leq b \leq 2020$ must hold.

Remark. If one replaces the base 2 with any prime number $p$, i.e., considers the corresponding more general question for the initial sets

$$
M_{p}=\left\{-p^{2022},-p^{2021}, \ldots,-p^{2},-p,-1,1, p, p^{2}, \ldots, p^{2021}, p^{2022}\right\}
$$

one arrives at the following result:

- For $p=2$, we obtain the sets $T$ determined above.
- For $p=3$, the sets $T=\left\{-3^{b+1},-3^{b}, 3^{b}, 3^{b+1}\right\}$ with $0 \leq b \leq 2021$ result.
- For $p \geq 5$, $T$ can no longer contain more than two elements, i.e., exactly two.

(Walther Janous)