Additive Latin Transversals We prove that for every odd prime p, every k p and every two subsets A = {a1, . . . , ak} and Summary: Additive Latin Transversals Noga Alon Abstract We prove that for every odd prime p, every k p and every two subsets A = {a1, . . . , ak} and B = {b1, . . . , bk} of cardinality k each of Zp, there is a permutation Sk such that the sums ai + b(i) (in Zp) are pairwise distinct. This partially settles a question of Snevily. The proof is algebraic, and implies several related results as well. 1 Introduction In this note we prove several results in Additive Number Theory, using the algebraic approach called Combinatorial Nullstellensatz in [1]. Other results in Additive Number Theory proved using this approach appear in [1] and in its many references, including, for example, [2], [3], [4]. Our first result here is the following theorem. Theorem 1.1 Let p be an odd prime, and let A and B be two subsets of cardinality k each of the finite field Zp. Then there is a numbering {a1, . . . , ak} of the elements of A and a numbering {b1, . . . , bk} of those in B such that the sums ai + bi (in Zp) are pairwise distinct. This partially settles a question of Snevily, who conjectured that the above is in fact true even when the field Zp is replaced by any Abelian group of odd order. Since the above theorem is trivial for k = p (as in this case we can simply take ai = bi), its assertion follows from the following more general result. Theorem 1.2 Let p be a prime, suppose k < p, let (a1, . . . , ak) be a sequence of not necessarily Collections: Mathematics