ON SOME PROBLEMS OF GYARMATI AND SÁRKÖZY. Le Anh Vinh Mathematics Department, Harvard University, Cambridge, Massachusetts

Transcription

1 #A42 INTEGERS 12 (2012) ON SOME PROLEMS OF GYARMATI AND SÁRKÖZY Le Anh Vinh Matheatics Departent, Harvard University, Cabridge, Massachusetts Received: 12/3/08, Revised: 5/22/11, Accepted: 7/1/12, Published: 7/13/12 Abstract In a recent paper, for large (but otherwise unspecified) subsets A,, C, D of F, Gyarati and Sárközy (2008) showed the solvability of the euations a + b = cd, and ab + 1 = cd with a A, b, c C, d D. They asked whether one can extend these results to every k N in the following way: for large subsets A,, C, D of F, there are a 1,..., a k, a 1,..., a k A, b 1,..., b k, b 1,..., b k with a i +b j, a i b j +1 CD (for 1 i, j k). In this paper, we give an affirative answer to this uestion. 1. Introduction In [6] and [5], Sárközy proved that if A,, C, D are large subsets of Z p, ore precisely, A D p 3, then the euation respectively a + b = cd, (1) ab + 1 = cd, (2) can be solved with a A, b, c C and d D. Gyarati and Sárközy [4] generalized the results on the solvability of euation (1) to finite fields. Using bounds of ultiplicative character sus, Shparlinski [7] extended the class of sets which satisfy this property. Furtherore, Garaev [2, 3] considered the euations (1) and (2) over soe special sets A,, C, D to obtain new results on the su-product proble in finite fields. At the end of [4], Gyarati and Sárközy proposed soe open probles related to the above euations. They asked whether one can extend the solvability of the euations (1) and (2) in the following way: for every k N, there are c = c(k) > 0 and 0 = 0 (k) such that if > 0, A,, C, D F, A D > 4 c then there are a 1,..., a k, a 1,..., a k A, b 1,..., b k, b 1,..., b k with a i + b j, a i b j + 1 CD

2 INTEGERS: 12 (2012) 2 for 1 i, j k. In this paper, we give an affirative answer to this uestion. More precisely, our results are the following. Theore 1. Let k N. If A,, C, D F with A D 4 1 2(k+2), then there are a 1,..., a k A, b 1,..., b k with a i + b j CD for 1 i, j k. Theore 2. Let k N. If A,, C, D F with A D 4 1 2(k+2), then there are a 1,..., a k A, b 1,..., b k with a i b j + 1 CD for 1 i, j k. In [4], Gyarati and Sárközy also studied the solvability of other (higher degree) algebraic euations with solutions restricted to large subsets of F. They considered the following euations: and a + b = f(c, d), a A, b, c C, d D; ab = f(c, d), a A, b, c C, d D, with f(x, y) F [x, y], A,, C, D F. We generalize Theores 1 and 2 in this direction. We have the following result for the su proble. Theore 3. Suppose that f(x, y) F [x, y], and f(x, y) is not of the for g(x) + h(y). We write f(x, y) in the for f(x, y) = g i (x)y i, i=0 with g i (x) F [x], and let I denote the greatest i value with the property that g i (x) is not identically constant. Assue that (I, ) = 1. For every k N, if A,, C, D F with A D 4 1 4(k+2), then there are a1,..., a k A, b 1,..., b k with a i + b j f(c, D) for 1 i, j k (where f(c, D) = {f(c, d) : c C, d D}). efore forulating the next theore, we need to take soe definitions fro [4]. Definition 4. A polynoial n F (x, y) = G i (y)x i = H j (x)y j F [x, y] i=1 is said to be priitive in x if (G 0 (y),..., G n (y)) = 1, and it is said to be priitive in y if (H 0 (x),..., H (x)) = 1. Definition 5. Every polynoial f(x, y) F [x, y] can be written uniuely (apart fro constant factors) in the for j=0 f(x, y) = F (x)g(x)h(x, y) where H(x, y) is priitive in both x and y. The polynoial H(x, y) (uniuely deterined up to constant factors) is called the priitive kernel of f(x, y).

3 INTEGERS: 12 (2012) 3 We now can state an analog of Theore 3 for the product proble. Theore 6. Suppose that f(x, y) F [x, y] and the priitive kernel H(x, y) of f(x, y) is not of the for c(k(x, y)) d. For every k N, if A,, C, D F with A D 4 1 4(k+2), then there are a1,..., a k A, b 1,..., b k with a i b j f(c, D) for 1 i, j k. 2. Pseudo-Randoness of Restricted-Su Graphs For any a A, c C, denote by N (a) the set of all b F such that a + b cd, and let N (a) = N (a). The following key estiate says that the cardinalities of the N D (a) s are close to when, D are large. Lea 7. For all subsets A,, C, D of F, we have N 2 (a) D < D. (a,c) F 2 Proof. For any set X, let X( ) denote the characteristic function of X. Let χ be any non-trivial additive character of F. We have N (a) = = (b,d) F 2,a+b cd=0 (b)d(d) 1 χ(s(a + b cd))(b)d(d) (b,d) F 2,s F = D Therefore N (a) D (a,c) F 2 = (b,d) F 2,s F χ(s(a + b cd))(b)d(d). χ(s(a + b cd))(b)d(d) 2 (a,c) F 2 = 1 2 (b,d) F 2,s F χ((s s )a)χ(sb s b )χ(c(s d sd))(b)d(d)(b )D(d ) a,c,b,b,d,d F = s,s F χ(s(b b ))(b)d(d)(b ) b,d,b F,s=s F = R 1 + R 2, (3)

4 INTEGERS: 12 (2012) 4 where R 1 is taken over b = b and R 2 is taken over b = b (the third line follows fro the orthogonality in a and c. Consider the second line as a su over a, then c iplies that all suands vanish unless s = s and d = d ). We have R 1 = χ(s(b b ))(b)d(d)(b ) and R 2 = b=b,d F,s=s F = ( 1) (b)d(d) = ( 1) D, b,d F (4) χ(s(b b ))(b)d(d)(b ) = b=b,d F,s=s F b,d F,s F,t=1 F,b =tb = b,d F,t=1 (b)d(d)(tb) χ(sb(1 t))(b)d(d)(tb) < 0. (5) The lea follows iediately fro (3), (4) and (5). The following result is an easy corollary of Lea 7. Corollary 8. For all subsets A,, C, D of F and c C, let N (A, ) be the nuber of pairs (a, b) A such that a + b cd. Then there exists c 0 C such that N c0,d (A, ) D A < D A. Proof. y the pigeon-hole principle, there exists c 0 C such that N c 0,D (a) D 2 1 N 2 (a) D < D. a A a A,c C y the Cauchy-Schwartz ineuality, N c0,d (A, ) D A N c0,d (a) D a A A N c 0,D (a) D a A D A. 2

5 INTEGERS: 12 (2012) 5 As a conseuence, for any two large subsets A, of F, there are any pairs (a, b) A with a + b CD. Corollary 9. For all subsets A,, C, D of F, let N C,D (A, ) be the set of pairs (a, b) A such that a + b CD. Then N C,D (A, ) D A D A. Proof. It follows iediately fro Corollary 8. Note that Corollaries 8 and 9 can be derived directly fro Theore 1 in [4]. However, Theore 1 in [4] is also an easy corollary of Lea 7 above. Theore 10. (cf. Theore 1 in [4]) For any subsets A,, C, D F, denote by N(A,, C, D) the nuber of solutions of E. (1). Then we have A D N(A,, C, D) < A D. Proof. y Lea 7, we have a A,c C N 2 (a) D (a,c) F 2 N 2 (a) D < D. y the Cauchy-Schwartz ineuality, A D N(A,, C, D) (a,c) F 2 N A (a) D a A,c C A D. N 2 (a) D 3. Pseudo-Randoness of Restricted-Product Graphs For any a A, c C, let T (a) be the set of all b F such that ab + 1 cd, and let T (a) = T (a). The following key estiate says that the cardinalities of the T c D (a) s are close to when, D are large.

6 INTEGERS: 12 (2012) 6 Lea 11. For all subsets A,, C, D of F, we have T 2 (a) D < D. (a,c) F 2 Proof. For any set X, let X( ) denote the characteristic function of X. Let χ be any non-trivial additive character of F. We have T (a) = Therefore (a,c) F 2 = 1 2 = (b,d) F 2,ab cd+1=0 (b)d(d) 1 χ(s(ab cd + 1))(b)D(d) (b,d) F 2,s F = D + 1 T 2 (a) D (b,d) F 2,s F χ(s(ab cd + 1))(b)D(d). χ(s(ab cd + 1))(b)D(d) 2 (a,c) F 2 = 1 2 (b,d) F 2,s F χ(a(sb s b ))χ(c(s d sd))χ(s s )(b)d(d)(b )D(d ) a,c,b,b,d,d F s,s F = 1 2 (R 1 + R 2 ), (6) where R 1 is taken over s = s and R 2 is taken over s = s. We have R 1 = χ(as(b b ))χ(cs(d d ))(b)d(d)(b )D(d ) a,c,b,b,d,d F,s=s F = ( 1) 2 D, (7) where the last line follows fro the orthogonality in a and then c. Considering the su over a and then over b, this iplies that all suands with b = b or d = d vanish. Now we copute R 2. R 2 = χ(as(b tb))χ(cs(d td))χ(s(1 t))(b)d(d)(b )D(d ) = a,c,b,b,d,d F s F,t=1 F a,c,b =tb,d =td F,s F,t=1 (b)d(d)(b )D(d ) < 0, (8)

7 INTEGERS: 12 (2012) 7 where the last line follows fro the orthogonality in a and c. y considering the su over a, and then over b, this iplies that all suands with b = tb or d = td vanish. The lea follows iediately fro (6), (7) and (8). Corollary 12. For all subsets A,, C, D of F and c C, let T (A, ) be the set of pairs (a, b) A such that ab + 1 cd. Then there exists c 0 C such that T c0,d (A, ) D A < D A. Proof. The proof of this corollary is siilar to that of Corollary 8, except that we use Lea 11 instead of Lea 7. We also have an analog of Corollary 9 in the shifted-product proble. Corollary 13. For all subsets A,, C, D of F, let N C,D (A, ) be the set of pairs (a, b) A such that ab + 1 CD. Then T C,D (A, ) D A D A. Siilarly as in the previous section, slightly weaker (but still useful) versions of Corollaries 12 and 13 can be derived directly fro Theore 2 in [4]. 4. Proof of Theores 1 We now give a proof of Theore 1.1. The key tool is the following lea. Lea 14. Suppose that A,, C, D of F with A, k D. D Then there are a 1,..., a k A, b 1,..., b k such that a i + b j CD for all 1 i, j k. Proof. The proof proceeds by induction on k. The base case, k = 1, follows iediately fro Corollary 9. Suppose that the theore holds for all l < k. Fro Corollary 9, we have N C,D (A, ) D A D D A = (1 + o(1)) A.

8 INTEGERS: 12 (2012) 8 y the pigeon-hole principle, there exists a 1 A such that N C,D (a 1, ) (1 + o(1)) D k 1 D. (9) D Let 1 be the set of all b such that a 1 + b CD. Fro Corollary 9, we have N C,D (A, 1 ) D A D 1 A 1 = (1 + o(1)) D A 1. y the pigeon-hole principle, there exists b 1 1 such that N C,D (A, b 1 ) (1 + o(1)) D A Let A 1 be the set of all a A such that a + b 1 CD. = 1 \{b 1 }. It follows fro (9) and (10) that k 1 A, D. D k 1 D. (10) D Set A = A\{a 1 } and Thus, by the induction hypothesis, there are a 2,..., a k A, b 2,..., b k such that a i + b j CD for all 2 i, j k. We also have a 1 + b i, a j + b 1 CD for all i, j = 1,..., k. This copletes the proof of the lea. Let c = c(k) = 1 2(k+2) and 1. Then A,,, D 1 c. It follows that k D (1+c)/2+ck 1 c A,. (11) D Therefore, Theore 1 follows iediately fro Lea 14. Note that the upper bound for the left hand side of (11) can be estiated by 1/2+kc. This can iprove the bound of Theore 1 to A D 4 1 2k Proof of Theore 2 Siilar to the previous section, we can obtain the following result fro Corollary 13. Lea 15. Suppose that A,, C, D of F with k D A,. D Then there are a 1,..., a k A, b 1,..., b k such that a i b j + 1 CD for all 1 i, j k.

9 INTEGERS: 12 (2012) 9 Let c = c(k) = 1 2(k+2) and 1, then A,,, D 1 c. It follows that k D (1+c)/2+ck 1 c A,. D Theore 2 now follows fro Lea Proof of Theore 3 We write f(x, y) in the for f(x, y) = g i (x)y i, i=0 where g i (x) F [x]. Let I denote the greatest i value with the property that g i (x) is not identically constant: g I (x) c, and either I = or g I+1 (x),..., g n (x) are identically constant. Since f(x, y) is not of the for g(x) + h(y), I > 0. Denote the degree of the polynoial g I (y) by D so that D > 0. Assue that (I, ) = 1. The following theore is due to Gyarati and Sárközy [4]. Theore 16. (cf. Theore 3 in [4]) If A,, C, D F, and the nuber of solutions of a + b = f(c, d), a A, b, c C, d D, is denoted by N, then we have N A D 1/2 (D + (I 1) ) A D 1/2. The following result is an analog of Corollary 9. Corollary 17. For all subsets A,, C, D F, let N C,D f (A, ) be the nuber of pairs (a, b) A such that a + b f(c, D). Then N C,D f (A, ) D A 1 (D + (I 1) 1/2 ) D A. Proof. For any c C, let Nf c (A,, D) denote the nuber of triples (a, b, d) A D such that a + b = f(c, d). y the pigeon-hole principle and Theore 16, there exists c 0 C such that D N c0 f (A,, D) A (D + (I 1) 1/2 ) D A. esides, for any fixed a, b and c 0, f(c 0, d) a b is a polynoial of degree on d. Therefore, the nuber of d such that a + b = f(c 0, d) is at ost. The corollary follows.

10 INTEGERS: 12 (2012) 10 As a conseuence, we have the following lea. Lea 18. Suppose that A,, C, D F with A, 1 (D + (I 1) 1/2 ) D k. D Then there are a 1,..., a k A, b 1,..., b k such that a i + b j f(c, D) for all 1 i, j k. Proof. The proof of this lea is siilar to that of Lea 14, except that we use Corollary 17 instead of Corollary 9 Let c = c(k) = 1 4(k+2) and 1, then A,,, D 1 c. It follows that 1 (D + (I 1) 1/2 ) D k 3/4 c/2+kc 1 c A,. D Theore 3 now follows fro Lea Proof of Theore 6 Using ultiplicative character sus, Gyarati and Sárközy [4] proved the following theore. Theore 19. r (cf. Theore 4 in [4]) Suppose that f(x, y) F [x, y] and that the priitive kernel H(x, y) of f(x, y) is not of the for c(k(x, y)) d. Write f(x, y) = F (x)g(y)h(x, y) in a uniue way up to constant factors. Let r, s, n, be the degrees of F, G, f(x, y) in x, f(x, y) in y, respectively. If A,, C, D F and the nuber of solutions of ab = f(c, d), a A, b, c C, d D, is denoted by N, then we have N A D < 4n1/2 3/4 ( A D ) 1/2 + 7(r + s + n + (n) 1/2 ) 5/2. Siilar to the previous sections, we have the following corollary. Corollary 20. For all subsets A,, C, D F, let N C,D f (A, ) be the nuber of pairs (a, b) A such that ab = f(c, D). Then N C,D f (A, ) D A 4n1/2 3/4 D 7(r + s + n + (n) 1/2 ) 5/2 A.

11 INTEGERS: 12 (2012) 11 Proof. For any c C, let Nf c (A,, D) denote the nuber of triples (a, b, d) A D such that ab = f(c, d). y the pigeon-hole principle and Theore 19, there exists c 0 C such that D N c0 f (A,, D) A 4n1/2 3/4 D 7(r + s + n + (n) 1/2 ) 5/2 A. esides, for any fixed a, b and c 0, f(c 0, d) ab is a polynoial of degree on d. Therefore, the nuber of d such that ab = f(c 0, d) is at ost. The corollary follows. We following lea follows fro Corollary 20 in a siilar way that Lea 14 follows fro Corollary 9. Lea 21. Suppose that A,, C, D F with A, 4n1/2 3/4 D k. D Then there are a 1,..., a k A, b 1,..., b k such that a i b j f(c, D) for all 1 i, j k. Let c = c(k) = 1 4(k+2) and 1. Then A,,, D 1 c. It follows that 4n 1/2 3/4 k D 3/4 c/2+kc 1 c A,. D Theore 6 now follows fro Lea Another proble In [1], Csikvári, Sárközy and Gyarati proposed soe further related probles. One of these probles is the following (Proble 4 in [1]): Is it true that for all ε > 0, there is a k 0 = k 0 (ε) such that if k N, k > k 0, p > p 0 = p 0 (ε, k) and A, F with then in{ A, } > ε, a 1 + a 2 = b 1... b k, a 1, a 2 A, b 1,..., b k (12) can be solved? In this section, we give a negative answer for this uestion by proving the following theore.

12 INTEGERS: 12 (2012) 12 Theore 22. For all ε < 1/2, k N, there exists two sets A, F with such that E. (12) cannot be solved. A, > ε Proof. Let ν be a generator of F and t = ε + 1. We choose = {1, ν,..., ν t }. Then > ε and k = {b 1... b k : b i } = {1, ν,..., ν kt }. Now we choose eleents of A inductively. Let T 0 = /2 = {b/2 : b }, A 0 = {a 0 } with a 0 / T 0. Suppose that we have T i and A i = {a 0,..., a i }. We then construct T i+1 and A i+1 as follows: T i+1 = T i ( k a i ) {a i }, A i+1 = A i {a i+1 }, for soe a i+1 / T i+1. It is easy to check that under this construction, (A i + A i ) k = for all i. Since T i+1 T i + k + 1 T i + tk + 1, we can continue the process until i(tk + 1) <. Therefore, we can choose a set A, such that A ( 1)/(kt + 1) ε and (A + A) k =. This copletes the proof of the theore. If F is not a prie field, we can do slightly better. Suppose that = p 2 for soe prie power p. We construct the Paley su graph P + with the vertex set F, and two vertices a, b are adjacent if and only if a + b is a suare residue. It is well known that the axial cliue of P + has size p. Since P + is self-syetric, the axial independent set of P + also has size p. Therefore, we can find a set A with A = 1/2 such that a + a is suare non-residue for all a, a F. Let be the set of all suare residues, then = /2 and E. (12) is not solvable in A,. References [1] P. Csikvári, A. Sárközy and K. Gyarati, Density and Rasey type results on algebraic euations with restricted solution sets, Cobinatorica, to appear. [2] M. Z. Garaev, The su-product estiate for large subsets of prie fields, Proc. Aer. Math. Soc. 136 (2008), [3] M. Z. Garaev and V. Garcia, The euation x 1 x 2 = x 3 x 4 + λ in fields of prie order and applications, J. Nuber Theory 128(9) (2008), [4] K. Gyarati and A. Sárközy, Euations in finite fields with restricted solution sets, II (algebraic euations), Acta Math. Hungar. 119 (2008), [5] A. Sárközy, On sus and products of residues odulo p, Acta. Arith. 118 (2005), [6] A. Sárközy, On products and shifted products of residues odulo p, Integers 8(2) (2008), A9. [7] I. E. Shparlinski, On the solvability of bilinear euations in finite fields, Glasgow Math J 50 (2008),