Poser - Futility Closet

Poser

Does the sequence of squares contain an infinite arithmetic subsequence?

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Consider three squares in arithmetic progression, $a_{1}^{2}$ , $a_{2}^{2}$ , and $a_{3}^{2}$ . Then $a_{3}^{2} - a_{2}^{2} = a_{2}^{2} - a_{1}^{2}$ , or (a₃ – a₂)(a₃ + a₂) = (a₂ – a₁)(a₂ + a₁). We know that (a₂ + a₁) < (a₃ + a₂), so (a₂ – a₁) must be greater than (a₃ – a₂). Now suppose that $a_{1}^{2}, a_{2}^{2}, a_{3}^{2}, ...$ is an infinite arithmetic progression. Then (a₂ – a₁) > (a₃ – a₂) > (a₄ – a₃) > … . That’s impossible, since there’s no infinite decreasing sequence of positive integers. So the answer is no. From Arthur Engel’s excellent Problem-Solving Strategies, 2008.