#A6 INTEGERS 15A (2015) ON REDUCIBLE AND PRIMITIVE SUBSETS OF F P, I. Katalin Gyarmati 1.

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1 #A6 INTEGERS 15A (015) ON REDUCIBLE AND PRIMITIVE SUBSETS OF F P, I Katalin Gyarmati 1 Deartment of Algebra and Number Theory, Eötvös Loránd University and MTA-ELTE Geometric and Algebraic Combinatorics Research Grou, Budaest, Hungary gykati@cs.elte.hu András Sárközy 1 Deartment of Algebra and Number Theory, Eötvös Loránd University, Budaest, Hungary sarkozy@cs.elte.hu Received: 10/18/13, Revised: /3/15, Acceted: 3/18/15, Published: 6/15/15 Abstract A set A F is said to be reducible if it can be reresented in the form A = B + C with B, C F, B, C. If there are no sets B, C with these roerties then A is said to be rimitive. First three criteria are resented for rimitivity of subsets of F. Then the distance between a given set A F and the closest rimitive set is studied. Dedicated to the memory of Paul Erdős on the occasion of the 100th anniversary of his birthday. 1. Introduction We will need Definition 1. Let G be an additive semigrou and A, B 1,..., B k subsets of G with If B i for i = 1,,..., k. (1.1) A = B 1 + B + + B k, then this is called an (additive) k-decomosition of A, while if a multilication is defined in G and (1.1) and A = B 1 B... B k (1.) 1 Research artially suorted by ERC-AdG.8005, Hungarian National Foundation for Scientific Research, grants no. K10091 and NK Research artially suorted by ERC-AdG.8005, Hungarian National Foundation for Scientific Research, grants no. K10091 and NK

2 INTEGERS: 15A (015) hold, then (1.) is called a multilicative k-decomosition of A. (A decomosition will always mean a non-trivial one, i.e., a decomosition satisfying (1.1).) In 1948 H.H. Ostmann [16], [17] introduced some definitions on additive roerties of sequences of non-negative integers and studied some related roblems. The most interesting definitions are: Definition. A finite or infinite set C of non-negative integers is said to be reducible if it has an (additive) -decomosition C = A + B with A, B. If there are no sets A, B with these roerties then C is said to be rimitive (or irreducible). Definition 3. Two sets A, B of non-negative integers are said to be asymtotically equal if there is a number K such that A \ [K, +1) = B \ [K, +1) and then we write A B. Definition 4. An infinite set C of non-negative integers is said to be totally rimitive if every C 0 with C 0 C is rimitive. Ostmann also formulated the following beautiful conjecture: Conjecture 1. The set P of the rimes is totally rimitive. If A is an infinite set of non-negative integers, then let A(n) denote its counting function: A(n) = {a : a ale n, a A}. Insired by Ostmann s work, Turán asked the following question: is it true that if A is any infinite set of non-negative integers then one can change at most o(a(n)) elements of it u to n so that the new set A 0 should be totally rimitive? Sárközy [4] gave an a rmative answer to this question (Theorems A, B and C will be resented here in a slightly simlified form): Theorem A There is a ositive absolute constant c such that if A is an infinite set of non-negative integers then one can change elements of it so that the number of A(n) the elements changed in [0, n] is less than c for every n > n 0 and the new set A 0 is totally rimitive. (log log A(n)) 1/ Answering a question of Erdős, Sárközy [5] also roved: Theorem B There is a ositive absolute constant c 0 such that if A is an infinite set of non-negative integers, and its comlement A = {0, 1,,... } \ A satisfies! A(n) = {a : 0 ale a ale n, a / A} < c 0 n (log log n) 1/3 (log n) 4 (1.3) for n 3, then A is reducible.

3 INTEGERS: 15A (015) 3 Erdős also conjectured that if we change o(n 1/ ) elements of the set of squares u to n, then the new set is always totally rimitive. Sárközy and Szemerédi [7] roved this conjecture in the following slightly weaker form: Theorem C If " > 0 and we change o(n 1/ " ) elements of the set of the squares u to n then we get a totally rimitive set. Volkmann [8], [9] Wirsing [30] and Sárközy [19], [0] estimated the Lebesgue measure, res. Hausdor dimension of the oint set assigned to reducible sets. Hornfeck [15], Hofmann and Wolke [14], Elsholtz [5], [6], [7] and Puchta [18] roved artial results toward Ostmann s Conjecture 1 on the totally rimitivity of the set P of the rimes. Elsholtz [8] also studied multilicative decomositions of shifted sets P 0 + {a} with P 0 P. So far we have surveyed the aers written on decomositions of sets of integers. Sárközy [6] roosed to study analogous roblems in finite fields. Observe that the notions of additive and multilicative decomositions, reducibility and rimitivity can be extended from integers to any semigrou, in articular, to the additive grou of F and multilicative grou of F for any rime ; in the rest of the aer we will use these definitions in this extended sense. First (insired by Erdős s roblem and Theorem C on the set of squares) it was conjectured in [6] that for every rime the set of the modulo quadratic residues is rimitive. (We will identify F with the set of the residue classes modulo and, as it is customary, we will not distinguish between residue classes and the integers reresenting them.) This conjecture is still oen but artial results have been roved by Sárközy [6], Shkredov [] and Sharlinski [3]. Dartyge and Sárközy [3] conjectured that the set of modulo rimitive roots is rimitive. This conjecture is also still oen but artial results have been roved by Dartyge and Sárközy [3] and Sharlinski [3]. Sárközy [1] also studied multilicative decomositions of the shifted set of the modulo quadratic residues. By Theorem B every infinite set A of non-negative integers satisfying (1.3) is reducible, and by Theorem A, the uer bound in (1.3) for A(n) cannot be relaced by O n. In a recent aer Gyarmati, Konyagin and Sárközy [11] studied (log log n) 1/ the analogue of these results in finite fields: they estimated the cardinality f() of the largest rimitive subset of F. Note that earlier Green, Gowers and Green [1], [13], and Alon [1] had studied a closely related roblem: they estimated the cardinality g() of the largest subset A of F which cannot be reresented in form B + B = A. Clearly f() ale g(). Imroving on results of Gowers and Green, Alon roved that /3 c 1 (log ) < g() < c 1/ 1/3 log.

4 INTEGERS: 15A (015) 4 In [11] we roved that f() is much smaller than this: for > 0 we have log log c 3 (log ) < f() < c 1/ 4 log. (1.4) Alon, Granville and Ubis [] estimated the number of distinct sumsets A + B in F under various assumtions on the cardinality of A and B. Among others they roved that there are /+o() distinct sumsets A + B in F with A, B! 1 as! 1. They also roved Theorem D There are less than (1.960) +o() reducible subsets of F. (So that almost all of the subsets of F are rimitive.) In this aer our goal is to continue the study of the reducible and rimitive subsets of F and the connection between them. First in Section we will resent three criteria for rimitivity of a subset A of F. Then in Section 3 we will show that these criteria are indeendent: neither of them follows from any of the others. In Section 4 we will show that any small subset of F can be made rimitive by adding just one element. Finally, in Section 5 we will discuss a roblem on the finite field analogue of Theorem A. (In the sequel of this aer we will extend the notions of reducibility and rimitivity, and we will study these extended notions.). Three Criteria for Primitivity in F. Ostmann and others have given several criteria for rimitivity of sequences of integers, but no rimitivity criteria are known in F. Thus we will resent three criteria of this tye, then we will illustrate their alicability, and we will also study the connection between them. Theorem 1. Assume that A = {a 1, a,..., a t } F and there are i, j with 1 ale i < j ale t such that and a i + a j a k / A for every k with 1 ale k ale t, k 6= i, k 6= j (.1) a i a j + a k / A for every k with 1 ale k ale t, k 6= j. (.) Then A is rimitive. Corollary 1. If is a rime of form = 4k + 1 and A F is defined by a a 1 A = {0, 1} [ {a F : = 1, = 1, a 6= 1, a 6= }, then A is rimitive.

5 INTEGERS: 15A (015) 5 Corollary. If A = {a 1, a,..., a t } F is a Sidon set, then it is rimitive. (A set A = {a 1, a,..., a t } is called Sidon set if the sums a i +a j with 1 ale i < j ale t are distinct.) Proof of Theorem 1. Assume that contrary to the statement of the theorem A is a set satisfying the assumtions, however, there are B F, C F with A = B + C, B, C. (.3) It follows from a i A, a j A and (.3) that there are b u B, b v B, c x C and c y C with a i = b u + c x (.4) and Now we have to distinguish two cases. a j = b v + c y. (.5) Case 1. Assume that b u 6= b v and c x 6= c y. Then by (.3), (.4) and (.5) we have a i + a j = (b u + c x ) + (b v + c y ) = (b u + c y ) + (b v + c x ) = a r + a s (.6) with a r = b u + c y A and a s = b v + c x A. Then by (.4) and c x 6= c y, and a i 6= a r (.7) a j 6= a r (.8) by (.5) and b u 6= b v. (.6), (.7) and (.8) contradict (.1) (with a r in lace of a k ). Case. Assume that b u = b v (.9) or c x = c y ; we may assume that (.9) holds. Then (.5) can be rewritten as By B there is a b B with Then by (.3) we have and a j = b u + c y. It follows from (.4), (.5), (.9), (.11) and (.1) that b 6= b u. (.10) a = b + c x A (.11) a q = b + c y A. (.1) a i a j = (b u b v ) + (c x c y ) = c x c y = a a q

6 INTEGERS: 15A (015) 6 whence where by (.5), (.9), (.10) and (.1) a i a j + a q = a A (.13) a q = b + c y 6= b u + c y = b v + c y = a j. (.14) (.13) and (.14) contradict (.) which comletes the roof of Theorem 1. Proof of Corollary 1. By the construction of the set A we have 0 A and 1 A. We will show that (.1) and (.) in Theorem 1 hold with a i = 0, a j = 1; in other words, we have and 1 a k / A for every a k 6= 0, 1 (.15) 1 + a k / A for every a k 6= 1. (.16) Consider first (.15). By the construction of A, it follows from a k A, a k 6= 0, a k 6= 1 that = 1 and = 1. Then by = 4k + 1 we have ak 1 ak = ak ak = ak 1 = 1. This imlies by the definition of A that 1 a k A may hold only if 1 a k = 0 or 1 a k = 1 whence a k = 1 or a k = 0. But it is assumed in (.15) that a k 6= 0, 1, thus, indeed, (.15) holds. Now consider (.16). It follows from a k A and a k 6= 1 that either ak 1 = 1 whence 1 + a k / A or we have a k = 0 whence 1 + a k = 1 which again does not belong to A so that (.16) holds. Proof of Corollary. If A = 1 or, then A is rimitive trivially. If A = t >, then a i, a j in the theorem can be chosen as any two distinct elements of A, e.g., we may take a i = a 1 and a j = a. Also, (.1) and (.) in Theorem 1 hold trivially by the definition of Sidon sets which roves the rimitivity of A. Theorem. If A F is of the form A = {0} [ A 0 with A 0 (/3, /3), (.17) and then A is rimitive. A > 4, (.18) Proof of Theorem. Assume that contrary to the statement of the theorem (.17) holds, however there are sets B F, C F with A = B + C, B, C. (.19)

7 INTEGERS: 15A (015) 7 Since 0 A, thus it follows from (.18) that there are b 0 B, c 0 C with 0 = b 0 + c 0. Write B 0 = B + { b 0 } and C 0 = C + { c 0 } so that 0 B 0 and 0 C 0 and, by (.19), B 0 + C 0 = B + C = A, B 0 = B, C 0 = C. (.0) Reresent every non-zero element of B 0 and C 0 by an integer from the interval (0, ) and let B 0 = {0, b 0 1,..., b 0 r} and C 0 = {0, c 0 1,..., c 0 s} with 0 < b 0 1 < < b 0 r < and 0 < c 0 1 < < c 0 s < (.1) where r 1 and s 1, and by (.18) and (.0), It follows that or s (r + 1)(s + 1) = B 0 C 0 A > 4. r (.) ; we may assume that (.) holds. Then by (.0) and (.1) we have for i = 1,,..., r and b 0 i = b 0 i + 0 B 0 + C 0 = A, 0 < b 0 i < (.3) c 0 1 = c B 0 + C 0 = A, 0 < c 0 1 <. (.4) By the construction of A it follows from (.3) and (.4) that and by (.0) we have 3 < b0 i + c 0 1 < 4 3, (.5) b 0 i + c 0 1 B 0 + C 0 = A. (.6) But it follows from the construction of A that it has only a single element in the interval, namely (= 0). Thus by (.5) and (.6) we have 3, 4 3 b 0 i + c 0 1 = 0 for i = 1,,..., r. By (.) this holds for both i = 1 and i = so that b c 0 1 = 0 = b + c 0 1 whence b 0 1 = b 0 which contradicts (.1) and this comletes the roof of Theorem.

8 INTEGERS: 15A (015) 8 Theorem 3. Let A F and for d F denote the number of solutions of a a 0 = d, a A, a 0 A by f(a, d). If then A is rimitive. max f(a, d) < A 1/, (.7) df Note that Corollary also follows from this criterion trivially. (In the sequel of this aer we will also aly this criterion for roving a stronger result along these lines.) Proof of Theorem 3. Assume that contrary to the statement of the theorem there are B F, C F with We may assume that By (.8) and (.9) we have whence A = B + C, B, C. (.8) B C. (.9) A = B + C ale {(b, c) : b B, c C} = B C ale B It follows from (.7) and (.30) that A 1/ ale B. (.30) max f(a, d) < B. (.31) df On the other hand, let c and c 0 be two distinct elements of C. Then by (.8), for every B we have a = b + c A and a 0 = b + c 0 A. For this air (a, a 0 ) we have a a 0 = (b + c) (b + c 0 ) = c c 0, and for di erent b values we get di erent solutions of a a 0 = c c 0, a A, a 0 A. (.3) It follows that the number of solutions of (.3) is at least as large as the number of b s: f(a, c c 0 ) B. (.33) Since c 6= c 0 we have c c 0 6= 0 thus (.3) contradicts (.31) and this comletes the roof of Theorem 3. Now we will rove that Theorem 3 is shar in the range 0 < A 1/ (and in the next section we will also show that if a set A F satisfies the assumtions in Theorem 3 then we must have A 1/ ):

9 INTEGERS: 15A (015) 9 Theorem 4. If is large enough and k is a ositive integer with then there is a set A F such that and A is reducible. k 0 < k < 1 1/4, (.34) A = k, (.35) max f(a, d) = A 1/ (.36) df Proof of Theorem 4. Write m = k. By theorems of Erdős and Turán [9], [10] and Chowla [4] the cardinality of the maximal Sidon set selected from {1,,..., N} is (1 + o(1))n 1/. Thus for k large enough there is a Sidon set m 1/ B = {b 1, b,..., b k } {1,,..., m 1} with B = k =. (.37) Let C = {c 1, c,..., c k } = (m) B = {mb 1, mb,..., mb k } and A = B + C. (.38) Then clearly A is reducible. Moreover, every a A can be written in the form a = b i + c j = b i + mb j (.39) with some i, j {1,,..., k}, and by (.34) here we have 0 < b i < m, 0 < b j < m and 0 < b i + mb j < m + m(m 1) < m = 8k 4 <. (.40) (.39) and (.40) determine b i and c j uniquely, thus we have A = B + C = B C = k (.41) which roves (.35). Finally, consider a d F with f(a, d) > 0 so that there are a, a 0 with Let a be of the form (.39) and Then we have a a 0 = d, a A, a 0 A. a 0 = b i 0 + c j 0 = b i 0 + mb j 0. d = a a 0 = (b i b i 0) + m (b j b j 0) = u + mv (.4)

10 INTEGERS: 15A (015) 10 with and, by (.34), 0 ale b i b i 0 = u < m (.43) 0 ale b j b j 0 = v < m (.44) d ale u + mv = b i b i 0 + m b j b j 0 < m + m(m 1) < m = 8k 4 <. (.45) By (.43), (.44) and (.45), d determines u and v in (.4) uniquely. If u = b i b i 0 6= 0 and v = b j b j 0 6= 0 then by the Sidon roerty of B the air u, v determines b i, b i 0, b j and b j 0, and thus also a and a 0 uniquely so that we have f(a, d) = 1. If u = b i b i 0 = 0 and v = b j b j 0 6= 0 then i = i 0 can be chosen in k ways while v determines j and j 0 uniquely, thus, by (.41), f(a, d) = k = A 1/. (.46) Similarly, if u = b i b i 0 6= 0 and v = b j b j 0 = 0 then j = j 0 can be chosen in k ways while i, i 0 are uniquely determined thus again (.46) holds and this also roves (.36). 3. Comarison of Three Criteria Let F 1, F and F 3 denote the family of the subsets A of F that satisfy the assumtions in Theorems 1, and 3, resectively, and let L 1, L and L 3 denote the cardinality of the largest subset belonging to F 1, F and F 3, resectively. First we will estimate F 1, F, F 3, L 1, L and L 3. Theorem 5. We have (i) and (ii) and F 1 / O(1) (3.1) L 1 = + O(1). (3.) F = /3+O(1) (3.3) L = + O(1). 3 (3.4) (iii) F 3 ale ex (1 + o(1)) /3 log (3.5) and L 3 ale (1 + o(1)) /3. (3.6)

11 INTEGERS: 15A (015) 11 Proof of Theorem 5. (i) Write 3 B = b : We will show that if A 0 B then ale b ale 3, b, b 6= 0,. A = {0, 1} [ A 0 F 1. (3.7) It su ces to rove that such a set A satisfies (.1) and (.) in Theorem 1 with a i = 1, a j = 0. For these values of a i and a j conditions (.1) and (.) become and 1 a / A for a A, a 6= 0, 1 (3.8) 1 + a / A for a A, a 6= 0. (3.9) Indeed, if a A, a 6= 0, 1 then by a A 0 B we have 1 ale 1 a, 1 + a ale 1 and a A 0 B is even so that 1 a and 1 + a are odd; thus 1 a, 1 + a / A 0 whence (3.8) and (3.9) follow; if a = 1 then 1 + a = = / A 0 thus again (3.9) holds. Since clearly B = O(1), it follows that A 0 B (and also A in (3.7)) can be chosen in / O(1) ways, which roves (3.1). Taking A 0 = B in (3.7) we get that A = {0, 1} [ B F 1. Thus, clearly we have L 1 {0, 1} [ B B = O(1). (3.10) In order to give an uer bound for L 1 consider a set A which satisfies the assumtions in Theorem 1 with some fixed a i, a j. Then by (.1), for any air a, a 0 F with a i + a j a = a 0 only at most one of a and a 0 may belong to A. There are at most +1 such airs (including the air (a, a 0 ) with a = a 0 ), and every element of F belongs to one of these airs. Thus A is at most +1 (=the number of airs) + (to also count a i and a j ) = + O(1), so that L ale + O(1) which, together with (3.10), roves (3.). (ii) The number of the sets A of form (.17) is equal to the number of the sets A 0 F with A 0 (/3, /3) which is clearly /3 /3+O(1) = /3+O(1)

12 INTEGERS: 15A (015) 1 which roves (3.3). The maximal cardinality of a set A of form (.17) is at most A ale {0} + A 0 ale 1 + {a : /3 ale a < /3} = 3 + O(1) which roves (3.4). (iii) If A F 3 then (.7) holds so that X A 1/ = ( 1) A 1/. (3.11) df f(a, d) < X df On the other hand, clearly we have X {(a, a 0 ) : a, a 0 A, a df f(a, d) = X df It follows from (3.11) and (3.1) that a 0 = d} = {(a, a 0 ) : a, a 0 A, a 6= a 0 = A ( A 1). (3.1) A 1/ ( A 1) < 1. (3.13) Now assume that contrary to (3.6) there is an " > 0 such that for infinitely many rimes there is an A F 3 with A > (1 + ") /3. (3.14) It follows from (3.13) and (3.14) that (1 + ") 1/ 1/3 (1 + ") /3 1 < 1 whence (1 + ") 3/ (1 + ") 1/ /3 < 1 But for! 1 the limit of the left hand side is (1 + ") 3/ (> 1) while the limit of the right hand side is 1, thus for large enough this inequality cannot hold, and this contradiction roves (3.6). It follows from (3.6) that XL 3 F 3 ale {A : A F, A ale L 3 } = = ex (1 + o(1)) /3 log k=1 L 3 1. ale = L3+1 L 3 which roves (3.5) and this comletes the roof of Theorem 5. (We remark that with a little work it could be also shown that (3.5) and (3.6) hold with equality sign but we do not need this here.)

13 INTEGERS: 15A (015) 13 Note that comaring (3.1), (3.3) and (3.5) we can see that Theorem 1 covers more rimitive sets than Theorem and Theorem 3, and both Theorem 1 and Theorem cover much more rimitive sets than Theorem 3. Moreover, by (3.), (3.4) and (3.6) there are much larger rimitive sets covered by Theorems 1 and than by Theorem 3. In site of this Theorem 3 seems to be at least as useful and alicable as the other two theorems since it covers almost all the thin subsets A of F (almost all the subsets A with A /3 ); on the other hand, e.g. a set A satisfying Theorem must have a very secial structure: aart from 0 A, it must lie comletely in the interval (/3, /3). Now we will show that Theorems 1, and 3 are indeendent. Proosition 1. For large enough Theorems 1, and 3 are indeendent: for either of the three criteria there is an A F which satisfies the conditions in it but which does not satisfy the conditions in the other two theorems. Proof of Proosition 1. By (3.1), (3.3) and (3.5) in Theorem 1 for large enough there are much more subsets A F satisfying the assumtions in Theorem 1 than the ones in Theorems and 3, and there are much more subsets satisfying the assumtions in Theorem than the ones in Theorem 3. There are (many) Sidon sets A with A (0, /3) and A > 1; these sets A satisfies the assumtions in Theorem 3 but not (.17) in Theorem. With a little work it could be shown that almost all the subsets A F with A = 1 n/3 satisfy the inequality ale f(a, d) < A 1/ for every d F ; such a subset A satisfies the assumtions in Theorem 3 but not (.) in Theorem 1. Finally, consider the set A = {0} [ {a : /3 < a < /3}. For large enough this set satisfies the assumtions in Theorem. On the other hand, for any a i, a j A we also have a i A since A also contains the negative of every element of it; take a k = a i in (.). Then a i a j + a k = a i a j a i = a j A (since the negative of a j also belongs to A) so that (.) in Theorem 1 does not hold. 4. Making Primitive Set From a Small Subset of F by Adding a Single Element By Theorem D almost all the subsets of F are rimitive. But how are the few reducible subsets distributed in the sace formed by the subsets of F? Are there

14 INTEGERS: 15A (015) 14 balls of not very small radius in this sace such that every subset belonging to them is reducible or the oosite of this is true: for any fixed subset A F there is a rimitive subset very close to it? First we will show that for small subsets of F this is so in a very strong sense (while for any subset A the roblem will be studied in Section 5). Theorem 6. Let be a rime with and let A F, 0 < A < > 3 (4.1) 1/ 3 1. (4.) Then there is an x F \ A such that the set A [ {x} is rimitive. Proof of Theorem 6. Fix some a A. If there is an x F \ A (4.3) such that the assumtions (.1) and (.) in Theorem 1 hold with the set A x = A [ {x} in lace of A and with a and x in lace of a i and a j, resectively, then by Theorem 1 this set A x is rimitive, which roves our claim. Thus, if contrary to the statement of the theorem there is no x satisfying (4.3) for which A x is rimitive, then for all these x values either (.1) and (.) fails with a = a i, a j = x, i.e., there is either an a k with a + x a k A or an a 0 k with a x + a 0 k A so that either or x A + A + { a} (4.4) x A A + {a} (4.5) must hold. But by (4.1) and (4.) the total number of x values satisfying (4.4) and (4.5) is at most A + A + A A ale 1 A ( A + 1) + A ( A 1) + 1 = 1 A (3 A 1) + 1 = 1 A (3 A + 1) + 1 A < 1 A A ale 1 A 6 A < A = F \ A

15 INTEGERS: 15A (015) 15 which contradicts the fact that every x satisfying (4.3) must also satisfy one of (4.4) and (4.5), and this comletes the roof of Theorem 6. It is a natural question to ask: what can one say from the oosite side? More recisely, let h() denote the greatest integer h such that for every A F with A ale h one can find an x F \ A for which the set A [ {x} is rimitive. Then by Theorem 6 for > 3 we have " # 1/ 3 1 ale h(). (4.6) On the other hand, it follows trivially from our result [11] in (1.4) that ale h() < c 4 log. This uer bound can be imroved easily to h() < 1 + O(1). (4.7) Proosition. Let 5 and define A F by A = [ 1 4 ] [ k=0 Then any set B F with A B is reducible. {4k, 4k + 1}. Proof of Proosition. Clearly, if 5, B F and A B then B has a reresentation B = {0, 1} + C with C so that, indeed, B is reducible. It follows from this roosition that for this set A we have ale 1 h() < A ale 4 which roves (4.7). There is a large ga between the lower bound (4.6) and the uer bound (4.7); it is not clear which one is closer to h(). Of course, the set A constructed in Proosition ossesses a much stronger roerty than the one needed for h() < A so that robably the uer bound (4.7) obtained in this way is far from the value of h(), but the lower bound (4.6) also seems to be far from h().

16 INTEGERS: 15A (015) Making Primitive Set From Any Subset of F by Changing Relatively Few Elements By Theorem D above (the result of Alon, Granville and Ubis []) there are only a few reducible subsets in F. Moreover, our results and methods oint to direction that the reducible sets are not well-distributed in the sense that there are less reducible sets among the small subsets of F than the large ones. This exlains that we have been able to show that from any small subset of F one can make a rimitive set by adding a single element but, on the other hand, we have not been able to rove such a result for larger subsets. Now we will rove that if instead of adding just one element we may change more (but still relatively few) elements of the given subset then we may make a rimitive set also from larger subsets. Theorem h 7. Let 3 be a rime and A a subset of F. Then by removing at 3+ most i 5 A elements of A and adding at most two elements of F \ A one can form a set B which is rimitive. Corollary 3. If 3 is a rime and A is a subset of F with 5 1 A < 1/, then by adding at most two elements of F \ A one can form a set B which is rimitive. (We remark that it follows already from Theorem 6 with a constant factor c slightly smaller than the one in the uer bound here that if A < c 1/ then by adding just one element of F \ A to A we can get a rimitive set B.) In order to formulate another consequence of Theorem 7 we need one more definition: Definition 5. If A, B are subsets of F then their distance d(a, B) is defined as the cardinality of their symmetric di erence (in other words, d(a, B) is the Hamming distance between A and B). It follows trivially from Theorem 7 that Corollary 4. If set B F such that 3 is a rime and A is a subset of F then there is a rimitive d(a, B) ale " 3 + # 5 A +. Proof of Theorem 7. We will use the following lemma.

17 INTEGERS: 15A (015) 17 Lemma 1. Let 3 be a rime, A F. Suose that there are u, v F for which 3v + u u / A + A, v / A A, / A. (5.1) Then adding at most two elements of F \ A to A one can form a set B which is rimitive. Proof of Lemma 1. Let A = {a 1, a,..., a s }. For some fixed u, v satisfying (5.1) define a s+1 and a s+ by Then by (5.1) a s+1 = u + v, a s+ = u v. In other words a s+1 a s+1 + a s+ = u / A + A, a s+1 a s+ = v / A A, a s+ + a s+1 = u + 3v / A. a s+1 + a s+ a k / A for 1 ale k ale s, a s+1 a s+ + a k / A for 1 ale k ale s + 1. Using Theorem 1 with i = s + 1, j = s + we get that A [ {a s+1, a s+ } is rimitive, and this comletes the roof of the lemma. Now we return to the roof of the theorem. Clearly Theorem 7 is trivial for A 3 5 h 3+ 5 A i (in this case A, and then removing A 1 elements from A we get a set which contains only one element, and thus it is reducible), thus we may assume A < 3 5. (5.) First we rove that there exist a set A 0 A and an element u F such that Indeed, for d F let u / A 0 + A 0 and A 0 A A. (5.3) h(a, d) def = {(a, a 0 ) : a + a 0 = d, a, a 0 A}. Clearly, X X h(a, d) = df X 1 = X 1 = A. df a,a 0 A a,a 0 A a+a 0 =d

18 INTEGERS: 15A (015) 18 On the other hand, we have whence Let u F be an element with min df h(a, d) ale X df h(a, d) = A min h(a, d) ale A df. (5.4) h(a, u) = min h(a, d) def = t, (5.5) df and (a 1, a 0 1), (a, a 0 ),..., (a t, a 0 t) the solutions of the equation By (5.4) and (5.5) we have a + a 0 = u with a, a 0 A. t = h(a, u) ale A. For A 0 = A \ {a 1, a,..., a t } the equation cannot be solved; thus a + a 0 = u, a, a 0 A 0 ( A) u / A 0 + A 0. This roves (5.3). Consider a set A 0 and an element u F for which (5.3) holds. We will rove that there exists a set A 00 A 0 and an element v F with and v / A 00 A 00, u + 3v / A 00 (5.6) A 00 A 0 A A A. (5.7) Since A 00 A 0 by (5.3) trivially holds. u / A 00 + A 00 (5.8)

19 INTEGERS: 15A (015) 19 Again, for d F we define f(a 0, d) def = {(a, a 0 ) : a a 0 = d, a, a 0 A 0 }. Let G = {v F : u+3v A 0 }. Since u is fixed (see (5.3)), we have Clearly, X df \G f(a 0, d) = X X df \G a,a 0 A a a 0 =d G ale A 0 ale A. (5.9) 1 ale X df X a,a 0 A a a 0 =d On the other hand, by (5.9) and (5.10) we have ( A ) min df \G f(a0, d) ale ( It follows from this and (5.) that min df f(a0, d) ale \G Let v F \ G be an element with 1 = X a,a 0 A G ) min df \G f(a0, d) ale A A ale A 3 5 f(a 0, v) = 1 = A 0 ale A. (5.10) X df \G ale f(a 0, d) ale A. (5.11) A. (5.1) min df f(a0, d) def = s, (5.13) \G and (b 1, b 0 1), (b, b 0 ),..., (b s, b 0 s) the solutions of the equation By (5.1) and (5.13) we have b b 0 = v with b, b 0 A 0. s = f(a 0, v) ale A. (5.14) For the equation cannot be solved, thus A 00 = A 0 \ {b 1, b,..., b s }, (5.15) b b 0 = v, b, b 0 A 00 ( A) v / A 00 A 00. (5.16)

20 INTEGERS: 15A (015) 0 By v F \ G and the definition of G we have u+3v / A 0. Since A 00 A 0 we have u + 3v / A 00. (5.17) (5.6) and (5.7) follow from (5.14), (5.15), (5.16) and (5.17). Thus we have constructed a set A 00 A and u, v F for which u / A 00 + A 00, v / A 00 A 00, u + 3v / A 00, A 00 A A. Using Lemma 1 wee see that it is ossible to add at most two elements of F \A 00 to A 00 so that we get a rimitive set B. This comletes the roof of Theorem Generalizations In order to kee our resentation more transarent and the discussions simler, we have decided to stick to F in this aer. However, we remark that all but one of our results can be generalized easily: Theorems 1, 3, 6, 7 and Corollaries, 3, 4 can be extended to any Abelian grous, Theorems, 4, 5 and Proositions 1, to cyclic grous (and Corollary 1 is the only result whose roof goes through only in F ). Acknowledgement. We would like to thank tha anonymous referee for suggesting us an idea to sharen Theorem 6 and also to shorten its roof. References [1] N. Alon, Large sets in finite sets are sumsets, J. Number Theory 16 (007), [] N. Alon, A. Granville and A. Ubis, The number of sumsets in a finite field, Bull. Lond. Math. Soc. 4 (010), [3] C. Dartyge and A. Sárközy, On additive decomositions of the set of rimitive roots modulo, Monatsh. Math. 169 (013), [4] S. Chowla, Solution of a roblem of Erdős and Turán in additive number theory, Proc. Nat. Acad. Sci. India 14 (1944), 1-. [5] C. Elsholtz, A remark on Hofmann and Wolke s additive decomositions of the set of rimes, Arch. Math. (Basel) 76 (001), [6] C. Elsholtz, The inverse Goldbach roblem, Mathematika 48 (001), [7] C. Elsholtz, Additive decomosability of multilicatively defined sets, Funct. Arox. Comment. Math. 35 (006),

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