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(a_n^2+b_n^2)(a_{n+1}^2+b_{n+1}^2), (a_n^2+a_{n+1}^2)(b_n^2+b_{n+1}^2)

Source: SRMC 2015

September 2, 2018
Perfect SquareArithmetic Progressionnumber theorySequencesSequence

Problem Statement

Let {an}n≥1\left\{ {{a}_{n}} \right\}_{n \geq 1} and {bn}n≥1\left\{ {{b}_{n}} \right\}_{n \geq 1} be two infinite arithmetic progressions, each of which the first term and the difference are mutually prime natural numbers. It is known that for any natural nn, at least one of the numbers (an2+an+12)(bn2+bn+12)\left( a_n^2+a_{n+1}^2 \right)\left( b_n^2+b_{n+1}^2 \right) or (an2+bn2)(an+12+bn+12)\left( a_n^2+b_n^2 \right) \left( a_{n+1}^2+b_{n+1}^2 \right) is an perfect square. Prove that an=bn{{a}_{n}}={{b}_{n}}, for any natural nn .
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