Can you choose $1983$ pairwise distinct positive integers less than $100000$, such that no three are in arithmetic progression?
The answer is yes. We will construct a sequence of integers that satisfies the requirements. Take the following definition of $a_n$: $a_n$ in base 3 has the same digits as $n$ in base 2. For example, since $6 = 110_2$ we have $a_6 = 110_3 = 12$. First, we will prove that no three $a_n$'s are in an arithmetic progression. Assume for sake of contradiction, that $i < j < k$ are numbers such that $2a_j = a_i + a_k$. Consider the base 3 representations of both sides of the equation. Since $a_j$ consists of just 0's and 1's in base three, $2a_j$ consists of just 0's and 2's base 3. No whenever $2a_j$ has a 0, both $a_i$ and $a_k$ must have a 0, since both could only have a 0 or 1 in that place. Similarly, whenever $2a_j$ has a 2, both $a_i$ and $a_k$ must have a 1 in that place. This means that $a_i = a_k$, which is a contradiction. Now since $1983 = 11110111111_2$, $a₁₉₈₃ = 11110111111_3 < 3^{11}/2 < 10^5$. Hence, the set $\{a_1, a_2, ...a_{1983}\}$ is a set of 1983 integers less than $10^5$ with no three in arithmetic progression.