Putnam 2022, Problem B-4

Find all integers $n$ with $n \geq 4$ for which there exists a sequence of distinct real numbers $x_1, \ldots, x_n$ such that each of the sets $\{x_1, x_2, x_3\}, \{x_2, x_3, x_4\}, \ldots, \{x_{n-2}, x_{n-1}, x_n\}, \{x_{n-1}, x_n, x_1\}, \{x_n, x_1, x_2\}$ forms a $3$-term arithmetic progression when arranged in increasing order.

Solution:
The values of \( n \) in question are the multiples of 3 starting with 9. Note that we interpret “distinct” in the problem statement to mean “pairwise distinct” (i.e., no two equal). See the remark below.  

We first show that such a sequence can only occur when \( n \) is divisible by 3. If \( d_1 \) and \( d_2 \) are the common differences of the arithmetic progressions \( \{x_m, x_{m+1}, x_{m+2}\} \) and \( \{x_{m+1}, x_{m+2}, x_{m+3}\} \) for some \( m \), then \( d_2 \in \{d_1, 2d_1, d_1/2\} \). By scaling we may assume that the smallest common difference that occurs is 1; in this case, all of the common differences are integers. By shifting, we may assume that the \( x_i \) are themselves all integers. We now observe that any three consecutive terms in the sequence have pairwise distinct residues modulo 3, forcing \( n \) to be divisible by 3.