Set Addition Theorems ´ Eric Balandraud eric.balandraud@imj-prg.fr Journées Hadamard 2015 Introduction Theorem (Lagrange). For any integer n ∈ N, there are (a,b,c,d) ∈ N4, such that: n=a2+b2+c2+d2. It has been conjectured by Waring that for any power k, there exists an integer (denote g(k) is minimal one), such that all integers can be written as a sum of g(k) k-th powers. This has been proved by Hilbert. Lagrange theorem asserts that g(2)=4, few values are known. Conjecture (Goldbach). For any even integer greater or equal to 4, n∈2.(N\ {0,1}, there are two prime numbers (p,q)∈P2, such that: n=p+q. These two problems deal with sums whose terms are multiplicatively defined. Additive number theory would consider any set of terms (forget about the multiplicative definition of these particular problems) and try to prove that ifthesetislargeenough,thensowillbethesetsofsums,uptoapointwhereit can cover all the integers. Additive combinatorics would consider similar ques- tionsanddevelopmentsontheintegers,residuesmoduloaprimeoranyabelian group. The interested reader may appreciate the following references and the references therein: • “Additive Number Theory, Inverse problems and the Geometry of Sum- sets”, M.B. Nathanson, GTM 165, Springer-Verlag (1996). • “Additive Combinatorics”, T. Tao and V.H. Vu, Cambridge Studies in advanced mathematics 105, Cambridge University Press. 1 First Session Let A and B be two non-empty finite subsets of Z, Z/pZ or any abelian group, we define their sumset: A+B ={a+b|a∈A, b∈B}. Its cardinality is invariant by translation, indeed |t+X| = |X| whatever is the set X, so that |(t+A)+(s+B)|=|A+B|. Theorem 1. Let A and B be two non-empty subsets of Z, then: |A+B|≥|A|+|B|−1. One has equality if and only if • either min{|A|,|B|}=1, • or A and B are arithmetical progressions with same difference. Proof. Denote A={a b2 ⊕ ⊕ ··· ⊕ ⊕ = > . . . . . . . . . = > b(cid:96)−1 ⊕ ⊕ = > b(cid:96) ⊕ ⊕ So these sequences are termwise equal. The second to d-th equalities are: a −a =b −b , i∈[2,d]. i i−1 2 1 This proves that A is an arithmetical progressions with difference b −b . And 2 1 the d-th to the one before the last equalities are: a −a =b −b , j ∈[2,(cid:96)]. d d−1 j j−1 ThisprovesthatB isanarithmeticalprogressionwithdifferencea −a . So, k k−1 A and B are arithmetical progressions with same difference. 3 Another topic of interest in additive combinatorics is the restricted sumset defined by: A+˙B ={a+b|a∈A, b∈B, a(cid:54)=b}. This can be generalized for sums of more than two terms in the following way: For a given set A⊂Z, and h∈N, one consider the h-fold restricted sumset: h∧A={a +···+a |a ∈A, a (cid:54)=a }. 1 h i i j Itscardinalityisalsoinvariantbytranslation,indeedh∧(t+A)=h.t+h∧A, so that |h∧(t+A)|=|h∧A|. Thesetsh∧Aand(|A|−h)∧Aaresymmetric,indeed(|A|−h)∧A=(cid:0)(cid:80) a(cid:1)− a∈A h∧A, so |(|A|−h)∧A|=|h∧A|. Theorem 2. Let A be a non-empty subsets of Z, and h∈[0,|A|], then: |h∧A|≥1+h(|A|−h). One has equality if and only if • either h∈{0,1,|A−1|,|A|}, • or h = 2, A = {a < a < a < a } and its 4 elements satisfy a +a = 1 2 3 4 1 4 a +a , 2 3 • or h∈[2,|A|−2], |A|≥5 and A is an arithmetical progression. Proof. Denote A={a 0, j ∈[0,h−1], i,j+1 i,j i+h−j i+h−(j+1) 4 and s =s . i,h i+1,0 To summarize: s 4,ifwesupposethat(cid:12)(cid:80)d {0,a }(cid:12)=1+(d+(d−1)+ (cid:12) i=1 i (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) ···+1). Theinequality(cid:12)(cid:80)d {0,a }(cid:12)≥(cid:12)(cid:80)d−1{0,a }(cid:12)+dhastobeanequality (cid:12) i=1 i (cid:12) (cid:12) i=1 i (cid:12) (cid:12) (cid:12) and necessarily (cid:12)(cid:80)d−1{0,a }(cid:12) = 1+((d−1)+···+1). From the induction (cid:12) i=1 i (cid:12) hypothesis, one has a =i.a , and: i 1 d−1 (cid:88) (cid:115) d(d−1)(cid:123) {0,a }= 0, .a . i 2 1 i=1 Therefore, d (cid:88){0,a }=(cid:115)0,d(d−1)(cid:123).a (cid:91)(cid:115)ad,ad + d(d−1)(cid:123).a . i 2 1 a a 2 1 1 1 i=1 If ad is not an integer or ad > d(d−1), these two sets are disjoint, one get (cid:12) a1 (cid:12) a1 2 (cid:12)(cid:80)d {0,a }(cid:12)=d(d−1)+2. So1+d(d+1) =d(d−1)+2whatimpliesd=1or (cid:12) i=1 i (cid:12) 2 d=2, contradiction. Otherwise ad is an integer strictly greater than d−1 and a(cid:12)1 (cid:12) aad1 ≤ d(d2−1). But then, one has (cid:12)(cid:12)(cid:80)di=1{0,ai}(cid:12)(cid:12) = d(d2−1) + aad1 +1. This yields the equality: 1+ d(d+1) = d(d−1) + ad +1 what naturally implies ad =d. 2 2 a1 a1 Suppose that the hypothesis A ∪ (−A) = ∅ is not satisfied, consider the (cid:104) (cid:105) example A = [−a,a], so |A| = 2a+1. One has Σ(A) = −a(a+1),a(a+1) and 2 2 |Σ(A)| = a(a+1)+1 = |A|−1|A|+1 +1 = |A|2−1 +1 < 1+ |A|(|A|+1) and the 2 2 4 2 conclusion of the theorem does not hold. 8 Second Session As a sumset can be defined on any group, one can consider the same problems in a larger group. To work on a finite group, naturally the size of any sumset will be bounded by the above by the size of the group, and no lower bound can exceed its cardinality. Lemma 1 (PrehistoricalLemma). Let (G,+) be a finite group (non necessarily abelian), A and B be two non-empty subsets of G, if |A| + |B| > |G| then A+B =G. Proof. Let x ∈ G, then the two sets x − B and A have, by the pigeonhole principle,acommonelement. Sotherearea∈Aandb∈B suchthatx−b=a, which gives x = a + b ∈ A + B. Since this holds for any x ∈ G, we have G=A+B. In any group, another kind of structure, the finite subgroups, will enable very small sumsets. Indeed: Exercise 1. Let A be a non-empty finite subset of any group (G,+) (non necessarily abelian), one has |A+A|=|A| if and only if A is a coset of a finite subgroup of G. InthegroupsZ/pZ,wherepisaprimenumber,theadditiveresultsarevery similar, even if the group is finite. These groups share with Zd the property to have no proper non-trivial finite subgroups. Sincethereisnoorderrelationcompatiblewiththeaddition,oneneedsnew tools to produce addition results. DefinitionoftheDysone-transform: ConsidertwosetsAandBofanabelian group G, and e∈G, one considers: A(e)=A∪(B+e) B(e)=B∩(A−e). One has A(e)+B(e)⊂A+B and |A(e)|+|B(e)|=|A|+|B|. Indeed: |A|+|B|=|A|+|B+e| =|A∪(B+e)|+|A∩(B+e)| =|A∪(B+e)|+|(A−e)∩B| =|A(e)|+|B(e)|. 9

Set Addition Theorems

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