Prove there are 2022 pairwise coprime intergers in $A$ or provide a counterexample An infinite sequence of positive integers $A = \{a_n\}_{n \geq 1}$ satsifies $2022 \geq a_{i + 1} - a_i \geq 1$. If there exists no infinite subsequence $B = \{b_n\}_{n{\geq 1}}$ of $A$ with $2022 \geq b_{i + 1} - b_i \geq 1$, such that all terms in $B$ have a common factor $k > 1$, then there exists 2022 pairwise coprime integers in $A$.
I don't understand this question, so I don't know where to start. If possible, can someone tell me how I should approach this? I tried listing some examples, and it all worked out, but I don't know how to do this question.
 A: What you need to do for this kind of "unorthodox" question is to get familiar enough with the question so that you "understand" it. Check each part of the condition. Can I feel a pattern? Are there some implications? Check the target. How can I connect the target to the conditions? What are the simple cases and extreme cases? You just continue to grope in the dark room, finding/imagining/trying all possibilities.
While I was trying to construct a counterexample, it occurs to me it could happen the only choice of subsequence $B$ with $2022 \geq b_{i + 1} - b_i \geq 1$ is $A$ itself, or a contiguous subsequence of $A$. That leads to my answer below.

Here is a counterexample.
Let $a_i=2010i+\begin{cases}2&\text{if }i\text{ is odd}\\5&\text{if } i\text{ is even}\end{cases}$.
$$2010+2,\, 2010\times2+5,\, 2010\times3+2,\, 2010\times4+5,\,\cdots$$
Since any two non-adjacent elements of $A$ are more than $2022$ apart, a subsequence of $A$ with adjacent elements no more than $2022$ apart must be a contiguous subsequence of $A$.

*

*Any $3$ consecutive elements of $A$ are relative prime, since any element of $A$ is not divisble by $3$, any element at even index is not divisble by $2$ and the difference of the adjacent differences between adjacent elements is $\pm6$.

*Elements of $A$ at odd indices are divisible by $2$. Elements of $A$ at even indices are divisible by $5$.

