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Problems in Additive Number Theory, V: Affinely Inequivalent Mstd Sets

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Problems in Additive Number Theory, V: Affinely Inequivalent Mstd Sets WN North-Western European Journal of Mathematics M EJ Problems in additive number theory, V: Affinely inequivalent mstd sets Melvyn B. Nathanson1 Received: November 9, 2016/Accepted: June 26, 2017/Online: July 25, 2017 Abstract An mstd set is a finite set of integers with more sums than differences. It is proved that, for infinitely many positive integers k, there are infinitely many affinely inequivalent mstd sets of cardinality k. There are several related open problems. Keywords: mstd sets, sumsets, difference sets. msc: 11B13, 05A17, 05A20, 11B75, 11P99. 1 Sums and differences In mathematics, simple calculations often suggest hard problems. This is certainly true in number theory. Here is an example: 3 + 2 = 2 + 3 but 3 2 , 2 3: − − This leads to the following question. Let A be a set of integers, a set of real num- bers, or, more generally, a subset of an additive abelian group G. We denote the cardinality of the set A by A . Define the sumset j j A + A = a + a0 : a;a0 A 2 and the difference set A A = a a0 : a;a0 A : − − 2 For all a;a0 G with a , a0, we have a + a0 = a0 + a because G is abelian. However, 2 a a0 , a0 a if G is a group, such as R or Z, with the property that 2x = 0 if and only− if x =− 0. It is reasonable to ask: In such groups, does every finite set have the property that the number of sums does not exceed the number of differences? Equivalently, is A + A A A for every finite subset A of G? ≤ − The answer isj “no.”j Aj set withj more sums than differences is called an mstd set. 1Department of Mathematics, Lehman College (CUNY), Bronx, NY 10468, USA 121 Problems in additive number theory M. B. Nathanson As expected, most finite sets A of integers do satisfy2 A + A < A A . For example, if j j j − j A = 0;2;3 f g then A + A = 0;2;3;4;5;6 and A A = 3; 2; 1;0;1;2;3 f g − f− − − g with A + A = 6 < 7 = A A : j j j − j It is also easy to construct finite sets A for which the number of sums equals the number of differences. For example, if A is an arithmetic progression of length k in a torsion-free abelian group, that is, a set of the form A = a + id : i = 0;1;2;:::;k 1 (1) f 0 − g for some d , 0, then the number of sums equals the number of differences: A + A = a + id : i = 0;1;2;:::;2k 2 f 0 − g A A = a + id : i = (k 1); (k 2);:::; 1;0;1;:::;k 2;k 1 − f 0 − − − − − − − g and A + A = A A = 2k 1: j j j − j − In an abelian group G, the set A is symmetric if there exists an element w G such that a A if and only if w a A. For example, the arithmetic progression2(1) is symmetric2 with respect to w− =2 2a + (k 1)d. We can prove that every finite 0 − symmetric set has the same number of sums and differences. More generally, for 0 j h, consider the sum-difference set ≤ ≤ 8 9 > h j h > >X− X > (h j)A jA = < a a : a A for i = 1;:::;h=: − − > i − i i 2 > :> i=1 i=h j+1 ;> − For h = 2 and j = 0, this is the sumset A+A. For h = 2 and j = 1, this is the difference set A A. − 2Cf. Hegarty and Miller, 2009, “When almost all sets are difference dominated”; Martin and O’Bryant, 2007, “Many sets have more sums than differences”. 122 1. Sums and differences Lemma 1 – Let A be a nonempty finite set of real numbers with A = k. For j j j 2 0;1;2;:::;h , there is the sum-difference inequality f g (h j)A jA h(k 1) + 1: j − − j ≥ − Moreover, (h j)A jA = h(k 1) + 1 j − − j − if and only if A is an arithmetic progression. Proof. If A is a set of k real numbers, then hA h(k 1) + 1. Moreover, hA = h(k 1) + 1 if and only if A is an arithmetic progressionj j ≥ 3−. j j − For every number t, the translated set A = A t satisfies 0 − (h j)A0 jA0 = (h j)A jA (h 2j)t − − − − − − and so (h j)A0 jA0 = (h j)A jA : − − j − − j Thus, after translating by t = min(A), we can assume that 0 = min(A). In this case, we have (h j)A ( jA) (h j)A jA: − [ − ⊆ − − Because (h j)A is a set of nonnegative numbers and jA is a set of nonpositive numbers, we− have − (h j)A ( jA) = 0 − \ − f g and so (h j)A jA (h j)A + jA 1 j − − j ≥ j − j j− j − (h j)(k 1) + 1 + j(k 1) + 1 1 ≥ − − − − = h(k 1) + 1: − Moreover, (h j)A jA = h(k 1) + 1 if and only if both (h j)A = (h j)(k 1) + 1 and jA =j j(k− 1) +− 1, or,j equivalently,− if and only if A isj an− arithmeticj − progression.− Thisj− completesj − the proof. 3Nathanson, 1996, Additive Number Theory: Inverse Problems and the Geometry of Sumsets, Theo- rem 1.6. 123 Problems in additive number theory M. B. Nathanson Theorem 1 – Let A be a nonempty finite subset of an abelian group G. If A is symmetric, then (h j)A jA = hA (2) j − − j j j for all integers j 0;1;2;:::;h . In particular, for h = 2 and j = 1, 2 f g A A = A + A : j − j j j Thus, symmetric sets have equal numbers of sums and differences. Note that the nonsymmetric set A = 0;1;3;4;5;8 f g satisfies A + A = [0;16] 14;15 and A A = [ 8;8] 6 n f g − − n f± g and so A + A = A A = 15: j j j − j This example4 shows that there also exist non-symmetric sets of integers with equal numbers of sums and differences. Proof. If j = 0, then (h j)A jA = hA. If j = h, then (h j)A jA = hA. Equation (2) holds in both cases. Thus,− we− can assume that 1 j −h 1.− − ≤ ≤ − Let A be a symmetric subset with respect to w G. Thus, a A if and only if 2 2 w a A. For every integer j, define the function fj : G G by fj (x) = x + jw. For − 2 Z ! all j;` we have fj f` = fj+`. In particular, fj f j = f0 = id and fj is a bijection. 2 Ph − Let x = a hA, and let a0 = w a A for i = 1;:::;h. If 1 i j h, then i=1 i 2 i − i 2 ≤ ≤ ≤ 0 h 1 BX C f j (x) = B aiC jw − @B AC − i=1 h j h X− X = a (w a ) i − − i i=1 i=h j+1 − h j h X− X = a a0 (h j)A jA i − i 2 − − i=1 i=h j+1 − and so hA (h j)A jA : j j ≤ j − − j 4Due to Marica, 1969, “On a conjecture of Conway”. 124 1. Sums and differences Ph j Ph Let y = − a a (h j)A jA. For h j + 1 i h, let a0 = w a A. i=1 i − i=h j+1 i 2 − − − ≤ ≤ i − i 2 Then − 0 h j h 1 BX− X C B C fj (y) = B ai aiC + jw @B − AC i=1 i=h j+1 − h j h X− X = a + (w a ) i − i i=1 i=h j+1 − h j h X− X = a + a0 hA i i 2 i=1 i=h j+1 − and so (h j)A jA hA : j − − j ≤ j j Therefore, (h j)A jA = hA and the proof is complete. j − − j j j Let A be a nonempty set of integers. We denote by gcd(A) the greatest common divisor of the integers in A. For real numbers u and v, we define the interval of integers [u;v] = n Z : u n v . If u ;v ;u ;v are integers, then [u ;v ]+[u ;v ] = f 2 ≤ ≤ g 1 1 2 2 1 1 2 2 [u1 + u2;v1 + v2]. Theorem 2 – Let A be a finite set of nonnegative integers with A 2 such that 0 A j j ≥ 2 and gcd(A) = 1. Let a∗ = max(A). There exist integers h1, C, and D and sets of integers C∗ [0;C + D 1] and D∗ [0;C + D 1] such that, if h 2h1, then the sum-difference set has⊆ the structure− ⊆ − ≥ ja∗ + (h j)A jA = C∗ [C + D;ha∗ (C + D)] (ha∗ D∗) − − [ − [ − for all integers j in the interval [h ;h h ]. Moreover, 1 − 1 (h j)A jA = (h j0)A j0A j − − j − − for all integers j;j [h ;h h ]. 0 2 1 − 1 Proof. Because A [0;a∗], we have hA [0;ha∗] for all nonnegative integers h. By a fundamental⊆ theorem of additive⊆ number theory5, there exists a positive integer h0 = h0(A) and there exist nonnegative integers C and D and sets of integers C [0;C 2] and D [0;D 2] such that, for all h h0, the sumset hA has the rigid structure⊆ − ⊆ − ≥ hA = C [C;ha∗ D] (ha∗ D): (3) [ − [ − 125 Problems in additive number theory M.
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