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\prod_{i = 1}^{2006} (a_{i}^2-i) is a mutliple of 3 when a_i is 1-2006

Source: KJMO 2006 p1

May 1, 2019
algebranumber theorypermutationscombinatoricsmultipleProduct

Problem Statement

a1,a2,...,a2006a_1, a_2,...,a_{2006} is a permutation of 1,2,...,20061,2,...,2006. Prove that ∏i=12006(ai2−i)\prod_{i = 1}^{2006} (a_{i}^2-i) is a multiple of 33. (00 is counted as a multiple of 33)
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