# Square-root cancellation for the signs of Latin squares

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## Abstract

Let , thus establishing the conjecture that the number of even and odd Latin squares, while conjecturally not equal in even dimensions, are equal to leading order asymptotically. Two proofs are given: both proceed by applying a differential operator to an exponential integral over SU(

*L*(*n*) be the number of Latin squares of order*n*, and let*L*^{even}(*n*) and*L*^{odd}(*n*) be the number of even and odd such squares, so that*L*(*n*)=*L*^{even}(*n*)+*L*^{odd}(*n*). The Alon-Tarsi conjecture states that*L*^{even}(*n*) ≠*L*^{odd}(*n*) when*n*is even (when*n*is odd the two are equal for very simple reasons). In this short note we prove that$$\left| {{L^{even}}\left( n \right) - {L^{odd}}\left( n \right)} \right| \leqslant L{\left( n \right)^{\frac{1}{2} + o\left( 1 \right)}}$$

*n*). The method is inspired by a recent result of Kumar-Landsberg.### Mathematics Subject Classification (2000)

05B15 05A16## Preview

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© János Bolyai Mathematical Society and Springer-Verlag Berlin Heidelberg 2017