The number of sumsets in a finite field

The number of sumsets in a finite field Noga Alon Andrew Granville Adrián Ubis Abstract We prove that there are 2 p/2+op) distinct sumsets A + B in F p where A, B as p 1 Introduction For any subsets A and B of a group G we define the sumset A + B := {a + b : a A, b B} There are 2 n subsets of an n element additive group G and every one of them is a sumset, since A = A+{0} for every A G However if we restrict our summands to be slightly larger, then something surprising happens when G = F p : there are far fewer sumsets: Theorem 1 Let ψx) be any function for which ψx) and ψx) x/4 as x There are exactly 2 p/2+op) distinct sumsets in F p with summands of size ψp); that is, exactly 2 p/2+op) distinct sets of the form A + B with A, B ψp) where A, B F p Green and Ruzsa [GrRu] proved that there are only 2 p/3+op) distinct sumsets A + A in F p The count in Theorem 1 cannot be decreased by restricting the size of one of the sets: Theorem 2 For any given prime p and integer k satisfying k = op), there exists A F p with A = k for which there are at least 2 p/2+op) distinct sumsets of the form A + B with B F p These results do not give a good idea of the number of distinct sumsets of the form A + B, as B varies through the subsets of F p when A has a given small size Tel Aviv University, Tel Aviv 69978, Israel and IAS, Princeton, NJ, 08540, USA Research supported by the Israel Science Foundation, by a USA-Israel BSF grant, and by the Ambrose Monell Foundation Email: nogaa@tauacil Université de Montréal, Montréal QC H3C 3J7, Canada L auteur est partiellement soutenu par une bourse de la Conseil de recherches en sciences naturelles et en génie du Canada Email: andrew@dmsumontrealca Universidad de La Rioja, Logroño 26004, Spain Supported by a Spanish MEC grant and by a Comunidad de Madrid-Univ Autónoma Madrid grant Email: adrianubis@gmailcom 1

Theorem 3 For each fixed integer k 1 there exists a constant µ k [ 2, 2] such that max A F p, A =k #{A + B : B F p} = µ p+op) k 1) We have µ 1 = 2, µ 2 := 1754877666, the real root of x 3 2x 2 + x 1 and, for each fixed integer k 3, we have 1 2 + 3 µ k k ) log k 2 + O 2) k Moreover µ k 5 5 /2 2 3 3 ) 1/5 = 1960131704 for all k 2, so that if A 2 then #{A + B : B F p } 19602 p+op) Remark: With a more involved method the constant 19602 in the last bound can be improved to 19184 see [Ubi]) We immediately deduce the following complement to Theorem 1: Corollary 1 Fix integer k 1 Let µ k = max l k µ l There are exactly µ k )p+op) distinct sumsets in F p with summands of size k The existence of µ k is deduced from the following result involving sumsets over the integers Define SA, G) to be the number of distinct sumsets A+B with B G; above we have looked at SA, F p ), but now we look at SA, {1, 2,, N}): Proposition 1 For any finite set of non-negative integers A with largest element L, there exists a constant µ A such that SA, {1, 2,, N}) = µ N+OL) A Moreover µ k = sup A Z 0 A =k By Theorem 3 or by Theorems 1 and 2 taken together) we know that µ k 2 as k In fact we believe that it does so monotonically: µ A Conjecture 1 We have µ 1 = 2 > µ 2 > µ 3 > > µ k > > 2 If this is true then µ k = µ k, evidently One can ask even more precise questions, for example for the number of distinct sumsets A + B where the sizes of A and B are given: Define S k,l G) = #{A + B : A, B G, A = k, B = l} for any integers k, l > 1 By Theorem 1 we know that if k, l as p then S k,l F p ) 2 p/2+op) We wish to determine for which values of k and l we have that S k,l F p ) 2 p/2+op) The Cauchy-Davenport Theorem [Cau] says that for any A, B F p we have A + B minp, A + B 1), hence S k,l F p ) = O1) whenever k + l > p O1) Let us see what we can say otherwise 2

Theorem 4 Let φ = 1+ 5 2 and let ψx) be any function for which ψx) as x i) If k + l p then S k,l F p ) [p/2] k+l 2) / min{k, l} If k + l p/2φ then S k,l F p ) p O1) ) [p/2] k+l If φp/3 + O1) > k + l > p/2φ then S k,l F p ) φ p k l /p If p k + l φp/3 + O1) then S k,l F p ) pφ p k l /p + 1 k l) In summary, if k + l p then S k,l F p ) p O1) max [p h)/2] ) h k+l h ii) For any integers with k, l ψp) and p k l p, we have S k,l F p ) x x such that 2 p x x k+l) In particular, if k, l ψp) then if and only if k + l p/4 k+l) 1+o1) with S k,l F p ) = 2 p/2+op) 3) Note that Theorem 4ii) cannot hold for k + l very close to p by the last estimate in Theorem 4i) The structure of sumsets has a rich history, from Cauchy [Cau] onwards, and has been studied from several different perspectives Most important are lower bounds on the size of the sumset, the lattice structure of A and B when the size of the sumset A + B is not much larger than that of A and B ie the Freiman-Ruzsa theorem), and the discovery of long arithmetic progressions in the sumset A + B when it is fairly small From our problem, many questions naturally arise: Give a precise asymptotic for the number of sumsets in F p as well as for the number of sumsets A + A in F p Which sets S have at least 2 cp representations as A + B for a given c > 0, and in particular for c = 1/2? Can one quickly identify a sumset S in F p, where S = A + B with A, B k? Perhaps though this seems unlikely) any sumset contains enough structure that is quickly identifiable? Perhaps most non-sumsets are easily identifiable in that they lack certain structure? We do know [Al] that any complement of a set of size c p log p is a sumset A + A for some A F p, for some absolute constant c > 0 Given a set S for which there exist sets A, B with A = k, B = l such that A+B = S, can one find such a pair A, B quickly? Can one quickly identify those sumsets in F p which have many representations as A+B? 3

Estimate the size of the smallest possible collection C k of sets in F p such that if S = A+B where A, B k then there exist A, B C k for which S = A + B Perhaps even something stronger than Conjecture 1 is true: for any A Z of size 1 < k <, does there exist a A such that µ A\{a} > µ A? 2 Lower bounds For a given integer k let A = {0, [p k)/2] + 1, [p k)/2] + 2,, [p k)/2] + k 1} For any subset B of {0, 1, 2,, [p k)/2]}, we see that A + B [0, p 1] and B = A + B) {0, 1, 2, [p k)/2]}, and thus the sets A + B are all distinct Hence there are at least 2 [p k)/2]+1 2 p k)/2 distinct sets A + B as B varies over the subsets of F p This implies Theorem 2, hence the lower bound in Theorem 1 when ψp) = op), and it also implies the lower bound µ k 2 in Theorem 3 Let A = {0} [x + 1,, x + k u 1] x + k u 1 + A 1 ) and B = B 1 [x + 1,, x + l v 1] {x + l v 1 + y} where A 1 [1, y] with A 1 = u, and B 1 [1, x] with B 1 = v, where u < k, y, and v < l, x Therefore B 1 = A+B) [1, x] and N+A 1 = A+B) N+[1, y]) for N = 2x+y+k+l u v 2 Also A + B [0, p 1] provided 2x + 2y + k + l u v 2 < p Therefore S k,l y x u) v) If k+l p/2φ then we select u = k 1, v = l 1, y = [pk 1)/2k+l 2)], x = p 1)/2 y This gives S k,l p O1) ) [p/2] k+l by Stirling s formula; Sk,l [p/2] k+l 2) / min{k, l} if k + l p If k + l > p/2φ then we select u = [kp + 1 k l)/ 5k + l)], v = [lp + 1 k l)/ 5k + l)], y = [φu], x = [φv] to obtain S k,l φ p k l /p + 1 k l) by Stirling s formula If k + l φp/3 + O1) then we change the above construction slightly: If instead we take B 1 [0, x 1] then there is a unique block of k + l u v 3 consecutive integers in A + B starting with 2x + 2 Now we can also consider the sums r + A) + B, for any r mod p); notice that we can identify the value of r from A + B, since the longest block of consecutive integers in A + B starts with 2x + 2 + r Hence S k,l pφ p k l /p + 1 k l) These last three paragraphs together imply the first part of Theorem 4 Now, given k p/4, select l = [p/4] k so that, by the above, there are p O1) [p/2] [p/4]) = 2 p/2 p O1) distinct sumsets A + B as A and B vary over the subsets of F p of size k and l respectively This implies the lower bound in Theorem 1 4

3 First upper bounds In this section we shall use a combinatorial argument to bound the number of sumsets A + B whenever A is small, in which case we can consider A fixed Throughout we let r C+A n) and r C A n)) denote the number of representations of n as c + a respectively, c a) with a A and c C Proposition 2 Let G be an abelian group of order n and let A G be a subset of size k 2 Then n ) n #{A + B : B G} n min min{2 n j, 2 [jk/k l+1)] } 4) 2 l k [j/l] j=0 Proof Given a set B we order the elements of B by greed, selecting any b 1 B, and then b 2 B so as to maximize A + {b 2 }) \ A + {b 1 }), then b 3 B so as to maximize A + {b 3 }) \ A + {b 1, b 2 }), etc Let B l be the set of b i such that A + {b 1, b 2,, b i } contains at least l more elements than A + {b 1, b 2,, b i 1 }, and suppose that B l + A = j By definition j = B l +A l B l, so that B l [j/l] and so there are no more than n i [j/l] i) choices for ) n n [j/l]) Next we have to determine B l Note that j/l n/2, for l 2, and so n i [j/l] i the number of possibilities for A + B given B l and hence B l + A): Our first argument: Since B l + A B + A G, the number of such sets A + B is at most the total number of sets H for which B l + A H G, which equals 2 n j Our second argument: Let C = B l + A, and let D be the set of d G for which r C A d) k + 1 l If b B \ B l then r C A b) = b + A) B l + A) k + 1 l, so that b D Hence B \ B l ) D, and so there are 2 D possible sets B \ B l, and hence B, and hence A + B Now D k + 1 l) r C A d) = A C = kj, d G so that D kj/k + 1 l), and the result follows Simplifying the upper bound: The upper bound in Proposition 2 is evidently ) n n 2 min max min{2 n j, 2 [jk/k l+1)] } 2 l k 0 j n [j/l] Now n [j/l]) 2 [jk/k l+1)] is a non-decreasing function of j, as l 2, and so the above is ) n n 2 min 2 n j 2 l k [j/l] max k l+1) 2k l+1) n j n The j + l)th term equals the jth term times n [j/l])/2 l [j/l] + 1) This is < 1 if and only if n < 2 l + 1)[j/l] + 2 l Now 2 l + 1)[j/l] + 2 l > 2l + 1) j 2l + 1) k l + 1) l l 2k l + 1) n, 5

and this is n unless l = k 4 Hence one minimizes by taking j = k l+1) n+o1) at a cost 2k l+1) of a factor of at most n Therefore our bound becomes n O1) νk n where ν k := min 2 l k ν k,l and ν k,l := 2 k l2k l + 1)) 2k l+1 k l + 1) k l+1 k l+1 2k l+1 l l2k l + 1) k l + 1)) l ) 1 2k l+1, using Stirling s formula A brief Maple calculation yields that ν k > 2 for all k 7 and ν 8 = 1982301294, ν 9 = 1961945316, ν 10 = 1942349376,, with ν k < 191 for k 12, and ν k decreasing rapidly and monotonically eg ν k < 19 for k 13, ν k < 18 for k 23, ν k < 17 for k 45, and ν k < 16 for k 117) In general taking l so that l 2 k log k/ log 2, one gets that ν k = 1 ) ) log 2 log k 2 exp 2 + o1), k which implies the upper bound in 2) of Theorem 3, as well as the upper bound implicit in Theorem 1 when min{ A, B } = op) 4 Upper bounds on S k,l F p ) using combinatorics The value of x in Theorem 4ii) must always lie in the range [p/2, p] since x k+l) 2 x Therefore if k + l = op) then the number of sumsets A + B is smaller than the number of possibilities for A and B so that S k,l F p ) p k ) ) p = l ) ) ) 1+o1) p x x 2 Ok+l) = 2 Ok+l) = k + l k + l k + l The Cauchy-Davenport Theorem states that A + B min{ A + B 1, p}, so that p ) ) p p S k,l F p ) j k + l 1 j=k+l 1 for k+l > 1/2+ɛ)p For the last part of Theorem 4, note that this is < 2 8p/17 for k+l 9p/10 Now we consider the case l, p k l p with k < op) and k For each fixed A of cardinality k and B of cardinality l, we proceed as in Proposition 2 taking l there as m here, and choosing m = ok) with m ): Hence there exists a subset B m B with A + B m = j and B m j/m p/m, and a subset D, determined by A and B m, with D kj j1 + Om/k)) and B \ B k+1 m m D Now A + B = A + B m ) A + B \ B m )) so the number of possibilities for A + B) \ A + B m ) is bounded above by the number of subsets of F p \ A + B m ), which is 2 p j, and also by the number of subsets of D with cardinality in the range [l [j/m], l], which is l ) ) D j 2 op), i l + k i=l [j/m] 6

since D j + op) and i = l + k + op) Hence the number of possible sumsets A + B is bounded by p k) 2 op), the number of possibilities for A, times p i [p/m] i) 2 op), the number of possibilities for B m, times 2 op) min{ j l+k), 2 p j }, the number of possibilities for A + B) \ A + B m ) This gives us the upper bound ) j S k,l F p ) 2 op) min{, 2 p j } = 2 1+o1))p x) 5) l + k where x is chosen as in Theorem 4ii), noting that p x p as l + k p 5 Sumsets from big sets We modify, simplify and generalize Green and Ruzsa s argument [GrRu], which they used to bound the number of sumsets A + A in F p : For a given set S, define ds := {ds : s S} Let G = Z/mZ For any A G define Âx) = a A eax/m) For a given positive integer L < m let H be the set of integers in the interval [ L 1), L 1] For a given integer d with 1 < dl < m we partition the integers in [1, m] as best as we can into arithmetic progressions with difference d and length L That is for 1 i d we have the progressions I i,k := {i + jd : kl j min{k + 1)L 1, [m i)/d]}} for 0 k [m i)/ld] We then let A L,d be the union of the I i,k that contain an element of A so that A A L,d ) Note that there are [m/l] + d such intervals I i,k Our goal is to prove the following analogy to Proposition 3 in [GrRu]: Proposition 3 If A, B Z/mZ, with α = A /m and β = B /m and m > 4L) 1+16αβL4 ɛ 2 2 ɛ 1 3, with L 3, 6) then there exists an integer d, with 1 d m/4l, such that A + B contains all those values of n for which r AL,d +B L,d n) > ɛ 2 m, with no more than ɛ 3 m exceptions In this paragraph we follow the proof of Proposition 3 in [GrRu] with the obvious modifications): Lemma 1 If A Z/mZ then there exists 1 d m/4l such that for all x Z/mZ Âx) 2 1 Ĥdx) 2L 1 ) 2 2 log 4L logm/4l) A m, with L 3, 7

Proof Sketch) Fix δ so that the right side above equals δm) 2, and hence δ 2/m Let R be the set of r Z/mZ such that Âr) δm; the result follows immediately for any x R By Parseval s inequality we have R δm) 2 r R Âr) 2 r Âr) 2 = m A, so that R δ 2 A /m Moreover, by the arithmetic-geometric mean inequality, we have ) R ) R 1 1 ) R m A Âr) 2 R Âr) 2 R Âr) 2 = R r R r R Consider the vectors v i [0, 1) R with rth coordinate ri/m mod 1) for each r R If we partition the unit interval for the rth coordinate into intervals of roughly equal length, all δm) 1/2 /4L 1) Âr) 1/2 which is 1/4L 1)), then, by the pigeonhole principle, two such vectors, with 0 i < j m/4l, lie in the same intervals since 1 + 4L 1) r R = Âr) δm 4L 1/2 4L r R Âr) δm r 1/2 ) ) 1/4 R A /m 4L) A /δ2m = m δ 2 R 4L, using the last displayed equation Therefore for d = j i we have ) 1/2 rd m 1 δm 4L 1 Âr) 4L ) R ) R /4 m A δm) 1/2 R for all r R, where t is the shortest distance from t to an integer Now Re1 et)) 2π 2 t 2 and jt j t, so that 1 Ĥdx) 2L 1 = 1 2L 1 2π2 2L 1 L 1 j= L 1) L 1 j= L 1) )) jdx 1 e m 2 jdx m 2π2 L 2 dx 3 m If x R then, by combining the last two displayed equations, this is 2π2 L 2 3 The result follows since 1 + Ĥdr) 2L 1 1 4L 1) 2 δm Âx) 2 as Ĥdx) 2L 1 8 δm 2 Âx) 2

Proof of Proposition 3 By Parseval s formula, and then Lemma 1 we have r A+Bn) r A+dH+B+dHn) 2L 1) 2 n 2 = 1 m Âx) 2 ˆBx) 2 1 x Ĥdx) 2L 1 ) 2 2 log 4L logm/4l) A x ˆBx) 2 = log 4L logm/4l) A B m ɛ2 2ɛ 3 m 3 16L 4 in this range for m Here r A+dH+B+dH n) denotes the number of representations of n as a + di + b + dj with a A, b B and i, j H) Now if g A L,d then there exists j H such that g + dj A, by definition, and hence r A+dH g) 1 Therefore r A+dH g) r AL,d g) for all g G, so that r A+dH+B+dH n) r AL,d +B L,d n) for all n Therefore if N is the set of n A + B such that r AL,d +B L,d n) > ɛ 2 m, then r A+dH+B+dH n) > ɛ 2 m and the above yields ɛ 2 m) 2 N 2L 1) ɛ2 2ɛ 3 m 3 4 16L 4 so that N ɛ 3 m Next we prove a combinatorial lemma based on Proposition 5 of [GrRu]: Proposition 4 For any subsets C, D of F p, and any m r min C, D ), there are at least min C + D, p) r m 1)p/r values of n mod p) such that r C+D n) m Proof Pollard s generalization of the Cauchy-Davenport Theorem [Pol] states that min{r, r C+D n)} r minp, C + D r) r[minp, C + D ) r] n The left hand side here is m 1)p N m ) + rn m where N m is the number of n mod p) such that r C+D n) m The result follows since p N m p Proof of upper bounds on S k,l F p ) using Fourier analysis: Suppose that L is given and d p/4l, and that M and N are unions of some of the arithmetic progressions I i,j Note that there are 2 p/l+d such sets M given d), and hence a total of e Op/L) possibilities for d, M and N We now bound the number of distinct sumsets A + B for which A L,d = M and B L,d = N in two different ways: First, since A M and B N there can be no more than ) M N ) k l M + N ) k+l 2 M + N such pairs 9

Second, select 2ɛ 1 p min M, N ) and 2ɛ 3 p max M, N ) Let Q be the values of n mod p) such that r M+N n) ɛ 2 1p Taking r = ɛ 1 p and m = ɛ 2 1p in Proposition 4, we have Q R := min M + N, p) 2ɛ 1 p By Proposition 3, A + B is given by Q less at most ɛ 3 p elements, union some subset of F p \ Q Hence the number of distinct sumsets A + B is [ɛ 3 p] ) Q 2 p Q i i=0 ) ) Q p2 p Q p2 maxp M N,0)+2ɛ 1p p [ɛ 3 p] [ɛ 3 p] as Q R > 2ɛ 3 p If M + N p/2 then the number of sumsets is 2 p/2 by the first argument Let L = [log p) 1/10 ] and ɛ 1 = ɛ 3 = 1/2L If A, B > p/l then M A > 2ɛ 1 p and N B > 2ɛ 1 p, so the second argument is applicable; therefore if M + N > p/2 then the number of sumsets is 2 p/2 L Op/L) Hence the total number of sumsets A + B with A, B > p/l is at most 2 p/2 L Op/L) which implies the upper bound in Theorem 1 taken together with the argument, for min{ A, B } = op), given at the end of section 3) Assume that l k p/log p) 1/4 with p k l p We select ɛ 1 = k/2p log log p, ɛ 3 = l/2p log log p and L = [log p) 1/20 ], so that the second argument above is applicable Taking x = M + N we have that { ) } x S k,l F p ) max min, 2 maxp x,0) 1/ɛ 3 ) Oɛ3p) = 2 1+o1))p x) 2 op) 0 x 2p k + l as in 5) This completes the proof of Theorem 4ii), combined with the results of the previous section Finally, 3) follows noting that x p/2 unless k + l p/4, in which case x p/2 6 Sumsets in finite fields and the integers Let A F p be of given size k 2, and let d = [p 1 1/k ] Consider the sets ia, the least residues of ia, a A, for 0 i p 1 Two, say ia and ja with i j mod p), must have those least residues between the same two multiples of p 1 1/k for each a A since there are < p/p 1 1/k ) k = p possibilities), and so the least residues of la, a A, with l = i j are all d in absolute value Hence the elements of d + la are all integers in [0, 2d]; and SA, F p ) = Sd + la, F p ) as may be seen by mapping A + B d + la) + lb) Hence we may assume, without loss of generality, that A is a set of integers in [0, L] where L 2p 1 1/k The case k = 2 is of particular interest since then SA, F p ) = S{0, 1}, F p ) by taking l = 1/b a), d = al when A = {a, b} We now compare SA, F p ) with SA, {1, 2,, p}) When we reduce A + B, where A {0,, L} and B {0,, p 1} are sets of integers, modulo p, the reduction only affects the residues in {0,, L 1} mod p) Hence SA, {1, 2,, p})2 L SA, F p ) SA, {1, 2,, p}) 7) 10

Now suppose A {0,, L} is a set of integers Suppose that Mr N < Mr + 1) for positive integers M, r, N We see that SA, {1, 2,, N}) SA, {1, 2,, Mr + 1)}) SA, {1, 2,, M}) r+1, the last inequality coming since the sumsets A + B with B {1, 2,, Mr + 1)} are the union of the sumsets A + B i with B i {Mi + 1, 2,, Mi + 1)} for i = 0, 1, 2,, r In particular for m A N) := SA, {1, 2,, N}) 1/N we have m A N) m A M) 1+1/r This implies that lim sup N m A N) m A M) for any fixed M, and then lim sup N m A N) = lim inf N m A N) so the limit, say µ A, exists and satisfies SA, {1, 2,, M}) µ M A 8) In the other direction we note that if B = i B i where B i {M + L)i + 1, M + L)i + 2,, M + L)i + M} then distinct {A + B i } 0 i r 1 give rise to distinct A + B Hence SA, {1, 2,, M}) r SA, {1, 2,, rm + L)}) and letting r we have Finally, by the inequalities 7), 8) and 9) we arrive at SA, {1, 2,, M}) µ M+L A 9) SA, F p ) = µ p A eol) = µ p A eop1 1/k) = µ p+op) A This proves Proposition 1, as well as the first part of Theorem 3 61 Precise bounds when k = 2 By the previous section we know that µ 2 = µ {0,1} Now S is a sumset of the form {0, 1} + B if and only if, when one writes the sequence of 0 s and 1 s given by s n = 1 if n S, otherwise s n = 0 if n S, there are no isolated 1 s Let C n be the number of sequences of 0 s and 1 s of length n such that there are no isolated 1 s, so that S{0, 1}, {1, 2,, N}) = C N+1 We can determine C n+1 by induction: If the n + 1)th element added is a 0 then it can be added to any element of C n If the n + 1)th element added is a 1 then the nth digit must be a 1, and then we either have an element of C n 1 or the next two digits are 1 and 0 followed by any element of C n 3 Hence C n+1 = C n + C n 1 + C n 3 with C 1 = 1, C 2 = 2, C 3 = 4, C 4 = 7 In fact it is easily checked, by induction, that C n+1 = 2C n C n 1 + C n 2 which is explained by the fact that x 4 x 3 x 2 1 = x+1)x 3 2x 2 +x 1), where the higher degree polynomials are characteristic polynomials for the recurrence sequence), and hence C n cµ n 2 for some constant c > 0, with µ 2 as in Theorem 3, implying a strong form of the first part of Theorem 3 for k = 2 By a more precise analysis we could even estimate the number of sets C = {0, 1} + B with either C or B of given size 11

62 Precise bounds when k = 3 It is not hard to generalize the procedure for the case {0, 1} to any A Z finite, namely to prove that µ A is the root of a polynomial with integer coefficients and degree smaller than 2 2L+1 when A {0, 1,, L}) In the special case of three elements is enough to deal with A = {0, a, b} for a, b coprime positive integers We can show that µ {0,a,b} µ as a + b, where we define µ = lim p #{B + {0, 0), 1, 0), 0, 1)} : B F p F p } 1/p2, which one can prove exists, and is < µ 2 Therefore either µ 3 = µ {0,a,b} for some a and b or µ 3 = µ Maple experimentation leads us to guess that µ 3 = µ {0,1,4} = 16863, a root of an irreducible polynomial of degree 21 All this is detailed in Chapter 3 of the third author s PhD thesis [Ubi] 63 Lower bounds on µ k That µ k 2 follows by choosing A = 1 2A with A Z any finite set Let A k = {1, 3,, 3 k 1 }, and write B {1, 2,, 3n} as B = 3B 0 3B 1 1) with B 0, B 1 {1, 2,, n} Since A k+1 = 1 3A k we have B + A k+1 ) \ 3Z = 3B 1 + A k ) 1) 3B 0 + 1), which shows that SA k+1, {1, 2,, 3n}) SA k, {1, 2,, n}) 2 n, and so µ Ak+1 2 1 3 µ 1 3 A k Since µ A1 = 2, an induction argument implies µ k µ Ak 2 1/2+31 k /2, which gives the lower bound for µ k in 2) 7 A non-trivial bound for fixed k 2 Let A be any set of given size k 2 in F p For any two distinct elements a, b A we can map x x a)/b a) so that 0, 1 A, and this will not effect the count of the number of sumsets containing A The number of sumsets C = A + B with B F p is obviously bounded above by { # B : B 2p } { + # C : C 3p } 5 5 { +# C : B : 2p 5 < B < C < 3p 5 } and B + {0, 1} C 12

The first two terms have size 2p p [2p/5]), the third requires some work: We observe that such C must have at least 2p/5 pairs of consecutive elements; so if c is the smallest integer 1 that belongs to C then we suppose that C = m k=1 c + I k) and C = m k=1 c + J k) where I 1, J 1, I 2, J 2,, I m, J m is a partition of {0,, p 1} into non-empty set of integers from intervals taken in order Any such set partition will do provided, for i k = I k and j k = J k, we have each i k, j k 1, m 3p 5 i k m + 2p 5, k=1 since C = m k=1 i k and m k=1 i k 1) B, and m k=1 i k + m k=1 j k = p Now there are p possible values for c, and the number of possible sets of values of i k such that m ) k=1 i k = x is, and of jk is ) p x 1 Therefore the number of possible such C is x 1 m 1 m 1 p 2 p m p/5 m p/5 2p/5+m x 3p/5 x 1 m 1 ) ) p 2 p 2 p 3 2m 2 [2p/5 2] ) p x 1 m 1 ) ) p p 3 [2p/5] Note that a c ) b) d a+c b+d) follows from defining a b) to be the number of ways of choosing b elements from a) Hence the number of sumsets A + B is p 3 p [2p/5]) = c p+op) where c = 5 5 /2 2 3 3 ) 1/5 = 1960131704 This implies the bound µ k c for all k 2 of Theorem 3; and we deduce the last part of Theorem 3 immediately from this taken together with Theorem 1 References [Al] N Alon Large sets in finite fields are sumsets J Number Theory, 126, 110 118, 2007 [Cau] A Cauchy Recherches sur les nombres J École Polytech, 9 99 116, 1813 [GrRu] B Green, I Z Ruzsa Counting sumsets and sum-free sets modulo a prime Studia Sci Math Hungar, 41 285 293, 2004 [Pol] J M Pollard A generalisation of the theorem of Cauchy and Davenport J London Math Soc 2), 8 460 462, 1974 [Ubi] A Ubis Cuestiones de la Aritmética y del Análisis Armónico PhD thesis, Universidad Autónoma de Madrid, Madrid, 2006 English translation available at the web address http://wwwuames/gruposinv/ntatuam/downloads/phdubispdf) 13