2006
DOI: 10.1007/s10623-006-8153-3
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Latin Squares without Orthogonal Mates

Abstract: In 1779 Euler proved that for every even n there exists a latin square of order n that has no orthogonal mate, and in 1944 Mann proved that for every n of the form 4k + 1, k ≥ 1, there exists a latin square of order n that has no orthogonal mate. Except for the two smallest cases, n = 3 and n = 7, it is not known whether a latin square of order n = 4k + 3 with no orthogonal mate exists or not. We complete the determination of all n for which there exists a mate-less latin square of order n by proving that, wit… Show more

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Cited by 22 publications
(30 citation statements)
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“…If n = 14, then T is given by (6,5,11), (7,6,13), (8,4,12), (9, 1, 10), (10,7,3), (11,8,5), (12, 9, 7), (13, 10, 9)}.…”
Section: Transversals Coinciding On a Single Entrymentioning
confidence: 99%
“…If n = 14, then T is given by (6,5,11), (7,6,13), (8,4,12), (9, 1, 10), (10,7,3), (11,8,5), (12, 9, 7), (13, 10, 9)}.…”
Section: Transversals Coinciding On a Single Entrymentioning
confidence: 99%
“…After submitting this paper we received a preprint [6] in which Evans has independently obtained our Corollary 1.1 using a variation of our technique.…”
Section: Acknowledgmentsmentioning
confidence: 99%
“…The general case for n = 4t +3 remained unsettled until two independently obtained results were simultaneously published in 2006. Those results, by Evans [11], and by Wanless and Webb [18], each conclude the existence of a bachelor latin square of order n for all n / ∈ {1, 3}. The actual result due to Wanless and Webb [18] is stated below in Theorem 1.6.…”
Section: Corollary 15mentioning
confidence: 86%
“…Other new general constructions of confirmed bachelor latin squares are described in [10]. We refer the reader to that article for a comparison of the known constructions for bachelor latin squares [10][11][12]18].…”
Section: Discussionmentioning
confidence: 99%
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