Maths Olympiad Prep

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Algebra Difficulty 5.1 AIME, harder Prove it Romania

Let kk be a positive integer greater than 11 and let (an)n1(a_n)_{n \ge 1} be a sequence of pairwise distinct positive integers. Show that the set M={a1k,a2k,a3k,}M = \{a_1^k, a_2^k, a_3^k, \dots\} does not contain an infinite arithmetic sequence.

Solution

Assume, by way of contradiction, that MM contains the infinite arithmetic sequence {b1k,b2k,b3k,}\{b_1^k, b_2^k, b_3^k, \dots\}, where 1b1<b2<b3<1 \le b_1 < b_2 < b_3 < \dots.
If we denote by rr its common difference, we have bn+1kbnk=rb_{n+1}^k - b_n^k = r, for all n1n \ge 1.
Observe that
r=bn+1kbnk=(bn+1bn)(bn+1k1+bn+1k2bn++bn+1bnk2+bnk1). r = b_{n+1}^k - b_n^k = (b_{n+1} - b_n) (b_{n+1}^{k-1} + b_{n+1}^{k-2}b_n + \dots + b_{n+1}b_n^{k-2} + b_n^{k-1}).
The expression in the second parenthesis is clearly increasing, hence the sequence bn+1bnb_{n+1} - b_n is decreasing, and since all its terms are positive integers, we reached a contradiction.

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