We prove a conjecture of Adamaszek generalizing the seating couples problem
to the case of $2n$ seats. Concretely, we prove that given a positive integer
$n$ and $d_1,\ldots,d_n\in(\mathbb{Z}/2n)^*$ we can partition $\mathbb{Z}/2n$
into $n$ pairs with differences $d_1,\ldots,d_n$.Comment: 3 page

Costa, Iván Sadofschi

Kohen, Daniel

English

arXiv

Daniel Kohen

Iván Sadofschi Costa

Crossref

Discrete Mathematics

On a generalization of the seating couples problem

We prove a conjecture of Adamaszek generalizing the seating couples problem to the case of 2n seats. Concretely, we prove that given a positive integer n and d1,…,dn∈(Z/2n)× we can partition Z/2n into n pairs with differences d1,…,dn.Fil: Kohen, Daniel. Consejo Nacional de Investigaciones Científicas y Técnicas. Oficina de Coordinación Administrativa Ciudad Universitaria. Instituto de Investigaciones Matemáticas "Luis A. Santaló". Universidad de Buenos Aires. Facultad de Ciencias Exactas y Naturales. Instituto de Investigaciones Matemáticas "Luis A. Santaló"; ArgentinaFil: Sadofschi Costa, Iván. Universidad de Buenos Aires. Facultad de Ciencias Exactas y Naturales. Departamento de Matemática; Argentina. Consejo Nacional de Investigaciones Científicas y Técnicas; Argentin

Sadofschi Costa, Iván

CONICET Digital

ON A GENERALIZATION OF THE SEATING COUPLES PROBLEMDANIEL KOHEN AND IVÁN SADOFSCHI COSTAAbstract. We prove a conjecture of Adamaszek generalizing the seating couples prob-lem to the case of 2n seats. Concretely, we prove that given a positive integer n andd1, . . . , dn ∈ (Z/2n)× we can partition Z/2n into n pairs with differences d1, . . . , dn.1. IntroductionPreissmann and Mischler [6] proved the following result, confirming a conjecture ofR. Bacher.Theorem 1.1. Let p = 2n + 1 be an odd prime. Suppose we are given n elementsd1, . . . , dn ∈ (Z/p)×. Then there exists a partition of Z/p−{0} into pairs with differencesd1, . . . , dn.We gave a simpler proof of this theorem using the Combinatorial Nullstellensatz [4].Karasev and Petrov, independently, gave a proof of Theorem 1.1 along the same lines andprovided further generalizations [3]. In this work, they also conjectured two generalizationsof Theorem 1.1, replacing p by an arbitrary integer N . The conjecture in the case that Nis even is originally due to Adamaszek.Conjecture 1.2 ( [3, Conjecture 1]). Let N = 2n + 1 be a positive integer. Suppose weare given n elements d1, . . . , dn ∈ (Z/N)×. Then there exists a partition of Z/N − {0}into pairs with differences d1, . . . , dn.We will prove the conjecture when N is even:Theorem 2.5 ( [3, Conjecture 2]). Let N = 2n be a positive integer. Suppose we aregiven n elements d1, . . . , dn ∈ (Z/N)×. Then there exists a partition of Z/N into pairswith differences d1, . . . , dn.While finishing this paper we found out that, in his master’s thesis [5], T.R. Mezeisuggests a possible way to solve the conjecture that is similar to ours. Furthermore, heshows that Theorem 2.5 holds whenever N = 2p for p a prime number.Acknowledgments: We would like to thank the referee for the useful comments andsuggestions.2. The even caseWe recall the following version of the Cauchy-Davenport theorem.2010 Mathematics Subject Classification. 11B13, 05B10.Key words and phrases. seating, couples, Cauchy-Davenport, partition, sumset.DK was partially supported by a CONICET doctoral fellowship.12 DANIEL KOHEN AND IVÁN SADOFSCHI COSTATheorem 2.1 ([1, Theorem 1]). If A and B are nonempty subsets of Z/N where 0 ∈ B,and gcd(b,N) = 1 for all b ∈ B \ {0}, then|A+B| ≥ min{N, |A|+ |B| − 1}.Suppose that we have a partition as in Theorem 2.5. Since the di are odd numbers,each pair contains exactly one even number. Therefore, if Theorem 2.5 holds there mustexist signs si such thats1d1 + . . .+ sndn = 1− 2 + 3− . . .+ (2n− 1)− 2n = n mod N.Theorem 2.2. Let N = 2n and let d1, . . . , dn ∈ (Z/N)×. Then there exists s1, . . . , sn ∈{1,−1} such thats1d1 + . . .+ sndn = n mod 2n.Proof. It is enough to prove that there exists I ⊂ {1, . . . , n} such that∑i∈I2di = d1 + . . .+ dn + n mod 2nSince di is odd for every i, d1 + . . .+ dn +n is even and therefore our task is equivalent tofinding I such that ∑i∈Idi =d1 + . . .+ dn + n2mod n.Let Ai = {di, 0}. Applying Theorem 2.1 inductively, we see that#(A1 + . . .+An) ≥ min{n,∑#Ai − (n− 1)}= n,concluding the proof. Remark 2.3. Theorem 2.1 was stated in full strength for the benefit of the reader. However,in the previous proof we only needed to use this result in the case |B| = 2, which followsfrom the fact that A+ b = A and gcd(b,N) = 1 imply that A = Z/N .The last ingredient is the following theorem due to Hall.Theorem 2.4 ( [2]). Let A be an abelian group of order n and a1, . . . , an be a numberingof the elements of A. Let d1, . . . , dn ∈ A be elements such that d1 + . . . + dn = 0. Thenthere are permutations σ, τ ∈ Sn such thatai − aσ(i) = dτ(i)Theorem 2.5. Let N = 2n be a positive integer. Suppose we are given n elementsd1, . . . , dn ∈ (Z/N)×. Then there exists a partition of Z/N into pairs with differencesd1, . . . , dn.Proof. First, using Theorem 2.2, we may assume that d1 + . . . + dn = n mod 2n. Nowit is enough to find a numbering a1, . . . , an of the odd numbers in Z/N and σ ∈ Snsuch that 2i − ai = dσ(i) mod 2n for every i ∈ {1, . . . , n}, for then the partition in pairs{2, a1}, {4, a2}, . . . , {2n, an} works.Equivalently, we need to find a numbering b1, . . . , bn of the even numbers in Z/N suchthat 2i− bi = dσ(i) + 1 mod N for some σ ∈ Sn. Now since di + 1 is even for all i, this isthe same as finding a permutation c1, . . . , cn of {1, . . . , n} such thati− ci =dσ(i) + 12mod n,ON A GENERALIZATION OF THE SEATING COUPLES PROBLEM 3for some σ ∈ Sn. If we verify thatd1 + 12+ . . .+dn + 12= 0 mod nthis will follow from Theorem 2.4. But this holds, since d1 + . . . + dn = n mod 2n andtherefore (d1 + 1) + . . .+ (dn + 1) = 0 mod 2n, proving thatd1 + 12+ . . .+dn + 12= 0 mod n.References[1] I. Chowla. A theorem on the addition of residue classes: application to the number Γ(k) in Waring’sproblem. Q. J. Math., os-8(1):99–102, 1937.[2] M. Hall. A combinatorial problem on abelian groups. Proc. Amer. Math. Soc., 3:584–587, 1952.[3] R. N. Karasev and F. V. Petrov. Partitions of nonzero elements of a finite field into pairs. Israel J.Math., 192(1):143–156, 2012.[4] D. Kohen and I. Sadofschi. A new approach on the seating couples problem. arXiv:1006.2571.[5] T. R. Mezei. Seating couples and Tic-Tac-Toe. Master’s thesis, Eötvös Loránd University, 2013.[6] E. Preissmann and M. Mischler. Seating couples around the King’s table and a new characterizationof prime numbers. Amer. Math. Monthly, 116(3):268–272, 2009.Departamento de Matemática, Facultad de Ciencias Exactas y Naturales, Universidad deBuenos Aires and IMAS, CONICET, ArgentinaE-mail address: dkohen@dm.uba.arDepartamento de Matemática, Facultad de Ciencias Exactas y Naturales, Universidad deBuenos AiresE-mail address: isadofschi@dm.uba.ar

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On a generalization of the seating couples problem

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