Orthogonal Latin Squares Based on Groups by Anthony B. Evans

Author:Anthony B. Evans
Language: eng
Format: epub, pdf
ISBN: 9783319944302
Publisher: Springer International Publishing

It follows that and thus St = 0, for t = 1, …, q − 2. For t = q − 1,

Next assume that the sequence {x1, …, xq} of elements of GF(q) satisfies St = 0 for t = 1, …, q − 2, and Sq−1 = q − 1. By Theorem 9.1, for each i there exists a unique polynomial fi of degree at most q − 1 over GF(q) that satisfies fi(x) = 0 if x≠xi and fi(xi) = 1. By Theorem 9.3, fi(x) = 1 − (x − xi)q−1. Set . Then f is a polynomial of degree at most q − 1 over GF(q) and, as f(a) = |{i∣xi = a}|⋅ 1, {x1, …, xq} is a permutation of the elements of GF(q) if f(x) = 1. As direct computation yields

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