Problem 5

Is it possible to choose

1983

distinct positive integers, all less than or equal to

10^5

, no three of which are consecutive terms of an arithmetic progression? Justify your answer.

Step 2 of 4: Exclude an arithmetic progression

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2a_j=a_i+a_k\Rightarrow a_i=a_k

Detailed analysis

Suppose $i<j<k$ and $2a_j=a_i+a_k$ . Every $a_n$ has only digits $0,1$ in base $3$ , so $2a_j$ has only digits $0,2$ and no carries. Also $a_i+a_k$ has no carries because each summand has only digits $0,1$ . At each digit, a $0$ in $2a_j$ forces both summands to have $0$ , and a $2$ forces both to have $1$ . Hence $a_i=a_k$ , contradicting injectivity.