A note on the number of additive triples in subsets of integers

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1 A note on the number of additive triples in subsets of integers Katherine Staden Mathematics Institute and DIMAP University of Warwick Coventry CV4 7AL, UK Abstract An additive triple in a set A of integers is a subset {x, y, z} A such that x + y = z. In this note, we precisely count the both the minimum and maximum number of additive triples among all subsets of the first n positive integers of a given size. These quantities were determined by Huczynska in [4]. We further determine the extremal subsets. 1 Introduction An additive triple, or Schur triple, or sum in a set A is a subset {x, y, z} A such that x+y = z. Note that x, y, z need not be distinct. Given n m 1, write [m, n] := {m,..., n} and [n] := [1, n]. The size and structure of the largest sum-free subsets of [n] is well known. Indeed, every such subset has size n/ and the unique extremal subsets are O = {1, 3, 5,..., n/ 1} and [ n/ + 1, n]; and additionally [n/, n 1] if n is even. Stability results of Deshouillers, Freiman, Sós and Temkin [3] and more recently Tran [7] show that every large sum-free subset must resemble one of these examples. In this note we are interested in the more general question of counting the number of sums in a set of a fixed size a [n]. Given A [n], define { ( ) s(a) := A ({x, y}, z) A : x + y = z} to be the number of additive triples in A. Let s (a) be the minimum of s(a) and s + (a) the maximum over all subsets A [n] of size a. So s (a) > 0 if and only if a > n/. We prove that in fact, when a > n/, we have s (a) = (a n/ )(a n/ ), and, moreover, there is a unique extremal set (i.e. a set which attains this bound). The situation for s + (a) is only slightly more complicated, and there may be multiple extremal sets, but they have a very similar structure. Given a set X of natural numbers, write b X := {bx : x X}. Our main result is the following. Theorem 1 For all positive integers n a, Supported by ERC grant

2 (i) s (a) = max{0, (a n/ )(a n/ )}, and if a > n/ then the unique extremal set is [n a + 1, n]. (ii) s + (a) = a /4 and the only extremal sets are of the form l [a], or if a is odd additionally l ([a 1] {a + 1}), for l N. Notice that, if a n/, there is more than one (usually many) sum-free subsets of [n], so we do not provide a characterisation for minimising sets here. These conclusions are not particularly surprising. One might expect that, in order to obtain a minimising set of size a > n/, one should add appropriate elements to a sum-free set of largest size n/. These elements should be as large as possible since the number of additive triples is limited by edge effects. This is indeed the case, but note that adding a single new element to the other sum-free set of largest size, the set of odd integers, gives rise to many additive triples. For the maximisation problem, [a] is a natural candidate, and it is clear that dilations have no effect. Note that here n is redundant; its only effect is to upper bound l. The questions addressed in this paper are instances of the so-called supersaturation problem which has been studied in several contexts in discrete mathematics. It was initiated by Erdős in 1955 [] who asked for the minimum number of triangles one can guarantee in a graph of given order and size. In fact a special case was already addressed by Rademacher in 1941 (unpublished), but the problem was only solved asymptotically more than sixty years later by Razborov [5]. In contrast, the maximisation problem for triangles is relatively simple. Indeed, determining the maximum number of triangles in a graph of given order and size is a consequence of the Kruskal-Katona theorem on set systems. See Chapter VII in [1] for more details on the history and recent progress on this fascinating problem in graph theory. Samotij and Sudakov [6] initiated the study of exact supersaturation in finite abelian groups. That is, given an (additive) abelian group Γ, and a [ Γ ], they asked for the minimum number of (ordered) additive triples in subsets A of Γ of size a. They were able to solve the problem completely in the case when Γ = Z p for prime p, and in the hypercube Z n for all natural numbers n. They obtained other partial results and explained why this problem seems rather hard to solve for general abelian groups Γ. Theorem 1 shows that, in the integer setting, the situation is very much simpler. The proof of Theorem 1 The proof of Theorem 1 is elementary. Let us briefly describe the main idea for (i) (the proof of (ii) is very similar). We choose an a [n] and corresponding hypothetical worst counterexample A of size a. This means that, when compared to the conjectured extremal subset [n a + 1, n] of the same size, the set A contains the fewest number of additive triples. Then every element of A must lie in few additive triples, or we could remove it and find a worse counterexample. Then it is simply a matter of considering an element of A which should lie in many additive tuples (the smallest element of A is a natural choice), and showing that it must

3 in fact lie in too many. First we do some easy calculations to determine s([n a + 1, n]) and s([a]). Proposition For all positive integers n a we have ( n ) ( s([n a + 1, n]) = a a a s([a]) =, 4 and if a is odd, then s([a 1] {a + 1}) = s([a]). n ), Proof. For each i [a], the number of additive triples in [n a + 1, n] whose smallest element is n i + 1 is min{0, [n i + 1, i 1] } = min{0, i n 1}. So s([n a + 1, n]) = i [a] min{0, i n 1} = a i= n/ +1 (i n 1) = a n/ (a n 1 + ( n/ + 1) n 1) = (a n/ )(a n/ ), as required. For each i [ a/ ], the number of additive triples in [a] whose smallest element is i is [i, a i] = a + 1 i. If i [ a/ + 1, a] then there are no such additive triples. So s([a]) = i [ a/ ] (a + 1 i) = a/ (a + 1 a/ + a 1) = a /4. The final assertion follows from the fact that, when a is odd, both a and a + 1 lie in the same number of additive triples whose other members lie in [a 1]. Given a set A [n] and y A, we write s A (y) := s(a) s(a \ {y}) for the number of additive triples in A which contain y. Given B A \ {y}, write s A (y, B) = s A (y) s A\B (y) for the number of additive triples in A which contain y and at least one element from B. Proof of Theorem 1. We first prove (i). Suppose that (i) is not true for some fixed n N. For all y [n], write I y := [n y + 1, n]. Among all a [n] consider only those for which the difference s(i a ) s (a) 0 is maximised. Among these, choose the smallest such a. Recall that we are assuming a n/ +1. Then there is a set A [n] of size a such that s(a) s(i a ) and A I a. We claim that s A (y) a n + 1 for all y A. (1) 3

4 Indeed, for all y A, we have by the choice of a that So as required. s(a) s(i a ) = s (a) s(i a ) s (a 1) s(i a ) s(a \ {y}) s(i a 1 ). s A (y) = s(a) s(a \ {y}) s(i a ) s(i a 1 ) () = s Ia (n a + 1) = [n a + 1, a 1] = a n + 1, Now let b be such that min(a) = n b+1. So b a+1. We have that s Ib (n b+1) = b n+1. For every x I b \ A, there are at most two additive triples {n b + 1, x, z}: namely taking z = n b x and z = x (n b + 1). Thus So s Ib (n b + 1, I b \ A) ( I b \ A ) = (b a). (3) s A (n b + 1) s Ib (n b + 1) (b a) = b n + 1 (b a) = a n + 1. (4) Thus we have equality in (), (3) and (4), so s(a) = s (a) and s(a ) = s (a 1) where A := A \ {n b + 1}. That is, A is extremal for parameter a 1. Since a is minimal, we have A := I a 1. But then a n + 1 = s Ia 1 {n b+1}(n b + 1) = 1 + [n a +, b 1] = b n + a 1, so b = a, a contradiction to A I a. This completes the proof of (i). We now turn to (ii), and begin by considering some small values of a. The statement is vacuous when a = 1. For a =, we have for x < y that s({x, y}) {x + x} {x, y} 1 with equality if and only if y = x. Suppose now that a = 3, let x < y < z and let A := {x, y, z}. The potential sums are x + x = y, x + x = z, x + y = z and y + y = z. At most two of these can hold simultaneously, with equality if and only if y = x and z = x + y = 3x; or y = x and z = y = 4x. Thus A = x [3] or x {1,, 4}, as required. Suppose that (ii) is not true for some fixed n N. Among all a [n] consider only those for which the difference s + (a) s([a]) 0 is maximised. Among these, choose the smallest such a. By the above discussion, we may assume that a 4. Then there is a set A [n] of size a such that s(a) = s + (a), but s(a) s([a]) and for every l [n] we have A l [a] and A l ([a 1] {a + 1}). We claim that a s A (y) for all y A. Indeed, for all y A, we have by the choice of a that s(a) s([a]) = s + (a) s([a]) s + (a 1) s([a 1]) s(a \ {y}) s([a 1]). (5) So s A (y) = s(a) s(a \ {y}) s([a]) s([a 1]) = s [a] (a) = a, (6) 4

5 as required. Now let b := max(a). So b a + 1. We have that s [b] (b) = b/. Now {1, b 1} {, b }... { b/, b/ } is a partition of [b 1], and each part destroys one additive tuple in [b] containing b. But at least [b] \ A / = (b a)/ of these parts intersect with [b] \ A. Thus s [b] (b, [b] \ A) (b a)/. Then s A (b) = s [b] (b) s [b] (b, [b] \ A) b b a = a. Thus we have equality in (5) and (6), so s(a) = s + (a) and s(a ) = s + (a 1) where A := A\{b}. That is, A is extremal for parameter a 1. Since a is minimal, there is some l N for which A = l [a 1], or A = l ([a ] {a}) if a 1 is odd. Since every sum of elements in A is a multiple of l, we must have b l N or otherwise s A (b) = 0. Thus we may assume without loss of generality that l = 1. Let G be the graph with vertex set A in which {x, y} is an edge whenever {x, y, b} is an additive triple. Using the fact that b > max(a ), we see that G has s A (b) = a/ edges. Note that G contains at most one loop (an edge of the form {x, x}), with equality if and only if b is even and b/ A. The non-loop edges form a matching, so there are either a/ or a/ 1 non-isolated vertices. Since G has a 1 vertices, it has at most one isolated vertex, with equality if and only if a is odd and there is a single loop. First consider the case when A = [a 1]. If 1 has some neighbour x in G, then b = x + 1 a, a contradiction. So 1 is the sole isolated vertex in G. Thus in [, a 1], the smallest element must be adjacent to the largest a 1 in G, so b = a + 1. Thus A = [a 1] {a + 1}, which is again a contradiction. Finally consider the case when A = [a ] {a}, in which case a is even. So G consists of a/ edges and no vertex is isolated. Since a 1 3, we see that {1, a} and {, a } must be distinct edges. So 1 + a = b = + a =, a contradiction. This completes the proof of (ii). Acknowledgements. I am grateful to Oleg Pikhurko for helpful discussions. References [1] B. Bollobás, Extremal graph theory, Academic Press, London, [] P. Erdős, Some theorems on graphs, Riveon Lematematika 9 (1955), [3] J. Deshouillers, G. Freiman, V. Sós and M. Temkin, On the structure of sum-free sets II, Astérisque, 58, (1999), [4] S. Huczynska, Beyond sum-free sets in the natural numbers, Elec. J. Combin. 1(1) (014), P1.1 [5] A. Razborov, On the minimal density of triangles in graphs, Combin. Probab. Comput. 17(4) (008),

6 [6] W. Samotij and B. Sudakov, The number of additive triples in subsets of abelian groups, Math. Proc. Cambridge Phil. Soc. 160 (016), [7] T. Tran, On the structure of large sum-free sets in integers, preprint,