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sum a_ia_{i+2} = 0 if a_i =\sqrt2 a_{i+2} - \sqrt3 a_{i+1}

Source: 2022 South Russian Girls MO - Assara Seniors p6

March 1, 2024
algebraSequencerecurrence relation

Problem Statement

There are 20222022 numbers arranged in a circle a1,a2,..,a2022a_1, a_2, . . ,a_{2022}. It turned out that for any three consecutive aia_i, ai+1a_{i+1}, ai+2a_{i+2} the equality ai=2ai+2−3ai+1a_i =\sqrt2 a_{i+2} - \sqrt3 a_{i+1}. Prove that ∑i=12022aiai+2=0\sum^{2022}_{i=1} a_ia_{i+2} = 0, if we know that a2023=a1a_{2023} = a_1, a2024=a2a_{2024} = a_2.
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