Team Selection Tests, 2021

For each integer $x$, let $[x]$ denote the remainder of $x$ divided by $p$. For each subset $X$ of $\mathbb{Z}$ and integers $a, b$, denote
$$aX + b := \{[ax + b] \mid x \in X\}.$$
Every integer is coprime with $p$, has an inverse modulo $p$, therefore, if $a$ is coprime with $p$, it's easy to verify that $X$ is special if and only if $aX + b$ is special. Our solution is based on the following lemmas.

Lemma 1. The set $S$ of $n \ge 3$ natural numbers that are at most $\frac{p}{3}$ is special.

Proof. Let $i, j$ be the two largest numbers and $k$ be the smallest in $S$. Choose $c = i + j - k > 0$, therefore, for any triple $(x, y, z) \in S^3$ with distinct elements, we have
$$0 \ge x - y + z - c > -\frac{p}{3} - c > -\frac{p}{3} - \frac{2p}{3} = -p.$$
Hence,
$$p \mid x - y + z - c \iff x - y + z = c \iff \{x, z\} = \{i, j\} \text{ and } y = k.$$

Lemma 2. If $p > 5.6^{n-2}$, for any set $S$ of $n \ge 3$ natural numbers, there exists integers $a, b$, where $a$ is coprime with $p$, such that all elements of $aS + b$ is at most $\frac{p}{3}$.

Proof. Assume that $0 \in S$, since we can choose arbitrary integer $b_0$ such that $0 \in S + b_0$. For each $i \in \mathbb{Z}$, let $S_i = \left[ \frac{pi}{6}, \frac{p(i+1)}{6} \right) \cap \mathbb{Z}$.

Consider $S$ as a $n-1$-tuple $(x_1, x_2, \dots, x_{n-1})$, where $x_i \in S$ and $x_i \ne 0$. Each integer $a$ corresponds to $n-2$-tuple $(a_1, a_2, \dots, a_{n-2})$, where $a_i$ is the index $k$ such that $[ax_i] \in S_k$. By Pigeonhole principle, there exists the set $A$ with 6 integers $a$, corresponding to the same $n-2$-tuple. By the same argument, there exist $a_1, a_2 \in A$ such that
$$[a_1x_{n-1}] - [a_2x_{n-1}] \in \left(-\frac{p}{6}, \frac{p}{6}\right).$$
Choose $a = a_1 - a_2$, then $[ax] \in \left[0, \frac{p}{6}\right) \cup \left(\frac{5p}{6}, p\right)$ for all $x \in S$.
It's easy to verify that if $b = \lfloor \frac{p}{6} \rfloor$ then $aS + b \subset \left[0, \frac{p}{3}\right]$.
Since $p > 6^{n-1} - 2^n + 1$ then $p \ge 6^{n-1} - 2^n + 3 > 5.6^{n-2}$, holds for all integer $n \ge 3$. Hence, our proof is completed. $\square$