, 2021

It is straightforward to check that $n = 3, 5$ indeed satisfy the condition.

Below we prove that $n \le 5$. Suppose $a_0, a_1, \dots, a_k$ satisfying the condition and from now on, $\equiv$ works in modulo $n$. Since $a_{i_0} = 0$ for some $i_0$, so $a_i^2 \equiv c(i - i_0)$ for some constant $c$. In particular, $1 = 1^2 \equiv c(i_1 - i_0)$ for some $i_1$, and hence $(i_1 - i_0)(a_i^2 - a_j^2) \equiv (i_1 - i_0)(i-j)c \equiv i-j \ne 0$ unless $i-j = 0$. Therefore, $n$ must be a prime number, for if $n = ab$ for some $a \ge b > 1$, then $a_i = (a+b)/2, a_j = (a-b)/2$ for some $i \ne j$ and $a_i^2 - a_j^2 \equiv 0$. In particular, $\{a_i^2\}$ is the set of quadratic residues.

WLOG assume $i_0 \ne k$, otherwise, reverse the sequence so that $i_0 = 0$. Since $c = a_{i_0+1}^2$, $c$ is a quadratic residue, thus, $\{a_i^2\}, \{c^{-1}a_i^2\}, \{-i_0, i_0+1, \dots, k-i_0\}$ are all the sets of quadratic residues and equivalent modulo $n$. If $i_0 > 0$, then $-1$ is a quadratic residue, then we must have $k = 2i_0$ and since $i_0 < k < n - i_0$, $k$ cannot be a quadratic residue, thus 2 cannot be a quadratic residue. If $i_0 = 0$, then $k$ is a quadratic residue, and $k < 2k = n - 1 = 2i_0$ is not a quadratic residue, thus 2 cannot be a quadratic residue. Therefore, we must have $i_0 \le 1$, $n = 2k+1 \le 2(2i_0) + 1 \le 5$.