, 2022

For $d = 1$, we take $k = 15$ and the sequence

8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9.

Two neighbouring numbers in this sequence always add up to 9, 16, or 25. For square $d > 1$, we also take $k = 15$ and the same sequence as above, except that we multiply all numbers by $d$. Two neighbouring numbers in this sequence always add up to $9d$, $16d$, or $25d$, which are all squares.

Now consider a non-square $d$. We will show that this will not satisfy the conditions. Suppose that there does exists a $k$ and a sequence $a_1d, a_2d, \dots, a_kd$, such that $\{a_1, a_2, \dots, a_k\} = \{1, 2, \dots, k\}$. Write $d = cm^2$, where $m$ is a positive integer such that $c$ is not divisible by a square greater than 1. Then for all $i$ with $1 \le i \le k-1$, we have that $a_id + a_{i+1}d$ is a square and $d = cm^2 \mid a_id + a_{i+1}d$, hence $cd = c^2m^2 \mid a_id + a_{i+1}d$, which yields that $c \mid a_i + a_{i+1}$. From this, we obtain that $a_{i+1} \equiv -a_i \mod c$ and hence $a_{i+2} \equiv -a_{i+1} \equiv a_i \mod c$. Therefore, there are at most two distinct residue classes modulo $c$ occurring among the $a_i$, namely the classes of $a_1$ and $a_2$. However, $\{a_1, a_2, \dots, a_k\} = \{1, 2, \dots, k\}$ and $k \ge 3$, and therefore we must have $c \le 2$. Because $d$ is not a square, $c = 1$ is impossible, hence $c = 2$. But then we have $a_{i+1} \equiv -a_i \equiv a_i \mod 2$, so there is at most one residue class modulo 2 occurring in the sequence, which gives a contradiction.

We conclude that the $d$ that satisfy the conditions are the squares. $\square$