M ath. Res. Lett. 15 (2008), no. 2, c International Press 2008 SUM-PRODUCT ESTIMATES VIA DIRECTED EXPANDERS. Van H. Vu. 1.

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M ath. Res. Lett. 15 (2008), no. 2, 375 388 c International Press 2008 SUM-PRODUCT ESTIMATES VIA DIRECTED EXPANDERS Van H. Vu Abstract. Let F q be a finite field of order q and P be a polynoial in F q[x 1, x 2 ]. For a set A F q, define P (A) := {P (x 1, x 2 ) x i A}. Using certain constructions of expanders, we characterize all polynoials P for which the following holds If A + A is sall (copared to A ), then P (A) is large. The case P = x 1 x 2 corresponds to the well-known su-product proble. 1. Introduction Let Z be a ring and A be a finite subset of Z. The su-product phenoenon, first investigated in [8], can be expressed as follows If A + A is sall, then A A is large. ( ) Here sall and large are with respect to A. Earlier works on the proble focused on the case Z is R or Z. In the last few years, starting with [2], the case when Z is a finite field or a odular ring has been studied extensively. This study leads to any iportant contributions in various areas of atheatics (see [4] for a partial survey). One of the ain applications of su-product estiates is new constructions of expanders (see, e.g., [3]). In this paper, we investigate the reversed direction and derive su-product estiates fro certain constructions of expanders. In fact, our arguents lead to ore general results, described below. Let F q be a finite field and P be a polynoial in F q [x 1, x 2 ]. For a set A F q, define P (A) := {P (x 1, x 2 ) x i A}. As a generalization of ( ) (which is the case P = x 1 x 2 ), it is tepting to conjecture the following stateent If A + A is sall, then P (A) is large. ( ) A short consideration reveals, however, that ( ) does not hold for soe classes of polynoials. For instance, if P is linear then both A + A and P (A) can be sall at the sae tie. Received by the editors May 4, 2007. The author is supported by NSF Career Grant 0635606. 375

376 VAN H. VU Exaple. Set P 1 := 2x 1 + 3x 2. Let A = {1,..., n} F q, where q is a prie and 1 n q/10. Then A + A = 2n 1 and P 1 (A) = 5n 4. More generally, if P has the for P := Q(L(x 1, x 2 )) where Q is a polynoial in one variable and L is a linear for, then both A and P (A) can be sall at the sae tie. Exaple. Set P 2 := (2x 1 + 3x 2 ) 2 5(2x 1 + 3x 2 ) + 3. Let A = {1,..., n} F q, where q is a prie and 1 n q/10. Then A + A = 2n 1 and P 2 (A) = 5n 4. In this case, Q = z 2 5z + 3 and L = P 1 = 2x 1 + 3x 2. Our ain result shows that P := Q(L(x 1, x 2 )) is the only (bad) case where the ore general phenoenon ( ) fails. Definition 1.1. A polynoial P F q [x 1, x 2 ] is degenerate if it is of the for Q(L(x 1, x 2 )) where Q is an one-variable polynoial and L is a linear for in x 1, x 2. The following refineent of ( ) holds If A + A is sall and P is non-degenerate, then P (A) is large. ( ) Theore 1.2. There is a positive constant δ such that the following holds. Let P be a non-degenerate polynoial of degree k in F q [x 1, x 2 ]. Then for any A F q ax{ A + A, P (A) } A in{δ( A 2 q k 4 q )1/4, δ( k A )1/3 }. Reark 1.3. The estiate in Theore 1.2 is non-trivial when k 2 q 1/2 A q/k. In the case when P has fixed degree, this eans q 1/2 A q. This assuption is necessary as if A is a subfield of size q or q 1/2 then A + A = A and P (A) is at ost A. Here and later on a b eans a = o(b). Reark 1.4. Since P = x 1 x 2 is clearly non-degenerate, we obtain the following suproduct estiate, reproving a result fro [10] ax{ A + A, A A } A in{δ( A 2 q )1/4, δ( q A )1/3 }. Our arguents can be extended to odular rings. Let be a large integer and Z be the ring consisting of residues od. Let γ() be the sallest prie divisor of and τ() be the nuber of divisors of. Define g() := n τ()τ(/n). Theore 1.5. There is a positive constant δ such that the following holds. Let A be a subset of Z. Then ax{ A + A, A A } A in{δ γ()1/4 A 1/2, δ( g() 1/2 1/2 A )1/3 }.

SUM-PRODUCT ESTIMATES VIA DIRECTED EXPANDERS 377 Reark 1.6. This theore is effective when is the product of few large pries. Our study was otivated by two papers [14] and [10]. In these papers, the authors used an arguent based on Kloosteran sus estiates to study Cayley graphs and the su-product proble, respectively. Our approach here relies on a cobination of a generalization of this arguent and the spectral ethod fro graph theory. 2. Erdős distinct distances proble The following question, asked by Erdős in the 1940 s [7], is aong the ost well known probles in discrete geoetry Question 2.1. What is the iniu nuber of distinct distances (in euclidean nor) deterined by n points on the plane? For a point set A, we denote by (A) the set of distinct distances in A. It is easy to show that (A) = Ω( A 1/2 ). To see this, consider an arbitrary point a A. If fro a there are A 1/2 different distances, then we are done. Otherwise, by the pigeon hole principle, there is a circle centered at a containing at least A 1/2 other points. Take a point a on this circle. Since two circles intersect in at ost 2 points, there are at least A 1/2 1 2 distinct distances fro a to the other points on the circle. It has been conjectured that (A) A 1 o(1) (the o(1) ter is necessary as shown by the square grid). This conjecture is still open. For the state of the art of this proble, we refer to [15, Chapter 6]. What happens if one replaces the euclidean distance by other distances? One can easily see that for the l 1 distance, the conjectured bound (A) A 1 o(1) fails, as the square grid deterines only A 1/2 distances. On the other hand, it sees reasonable to think that there is no essential difference between the l 2 and (say) the l 4 nors. In fact, in [9], it was shown that certain arguents used to handle the l 2 case can be used, with soe ore care, to handle a wide class of other distances. The finite field version of Erdős proble was first considered in [2], with the euclidean distance (see also [11] for ore recent developent). Here we extend this work for a general distance. Let P be a syetric polynoial in two variables. (By syetry, we ean that P is syetric around the origin, i.e., P (x, y) = P ( x, y).) Define the P -distance between two points x = (x 1, x 2 ) and y = (y 1, y 2 ) in the finite plane F 2 q as P (y 1 x 1, y 2 x 2 ). Let P (A) be the set of distinct P -distances in A. Theore 2.2. There is a positive constant δ such that the following holds. Let P be a syetric non-degenerate polynoial of degree k and A be a subset of the finite plane F 2 q, then P (A) δ in{ A k 2 q, q k }.

378 VAN H. VU Reark 2.3. The polynoial P = x p + y p, which corresponds to the l p nor, is non-degenerate for any positive integer p 2. Reark 2.4. Assue that k = O(1). For A q, the ter A q A 1/2, and so P (A) A 1/2. If A q, one cannot expect a bound better than A 1/2, as A can be a sub-plane. Reark 2.5. The proof also works for a non-syetric P. In this case, dist(x, y) and dist(y, x) ay be different. 3. Directed expanders and spectral gaps Let G be a d-regular graph on n vertices and A G be the adjacency atrix of G. The rows and coluns of A G are indexed by the vertices of G and the entry a ij = 1 if i is adjacent to j in G and zero otherwise. Let d = λ 1 (G) λ 2 (G) λ n (G) be the eigenvalues of A G. Define λ(g) := ax{ λ 2, λ n }. It is well known that if λ(g) is significantly less than d, then G behaves like a rando graphs (see, for exaple, [6] or [1]). In particular, for any two vertex sets B and C e(b, C) d n B C λ(g) B C. where e(b, C) is the nuber of edges with one end point in B and the other in C. We are going to develop a directed version of this stateent. Let G be a directed graph (digraph) on n points where the out-degree of each vertex is d. The adjacency atrix A G is defined as follows: a ij = 1 if there is a directed edge fro i to j and zero otherwise. Let d = λ 1 (G), λ 2 (G),..., λ n (G) be the eigenvalues of A G. (These nubers can be coplex so we cannot order the, but by Frobenius theore all λ i d.) Define λ(g) := ax i 2 λ i. An n by n atrix A is noral if A A = AA. We say that a digraph is noral if its adjacency atrix is a noral atrix. There is a siple way to test whether a digraph is noral. In a digraph G, let N + (x, y) be the set of vertices z such that both xz and yz are (directed) edges. Siilarly, let N (x, y) be the set of vertices z such that both zx and zy are (directed) edges. It is easy to see that G is noral if and only if

SUM-PRODUCT ESTIMATES VIA DIRECTED EXPANDERS 379 (1) N + (x, y) = N (x, y) for any two vertices x and y. Lea 3.1. Let G be a noral directed graph on n vertices with all out-degree equal d. Let d = λ 1 (G), λ 2 (G),..., λ n (G) be the eigenvalues of A G. Then for any two vertex sets B and C e(b, C) d n B C λ(g) B C. where e(b, C) is the nuber of (directed) edges fro B to C. Proof. The eigenvector of λ 1 = d is 1, the all-one vector. Let v i, 2 i n, be the eigenvectors of λ i. A well known fact fro linear algebra asserts that if A is noral then its eigenvectors for an orthogonal basis of K n (where K denotes the field of coplex nubers). It follows that any vector x orthogonal to 1 can be written as a linear cobination of these v i. By the definition of λ we have that for any such vector x A G x 2 =< A G x, A G x > λ 2 x 2. Fro here one can use the sae arguents as in the non-directed case to conclude the proof. We reproduce these arguents (fro [1]) for the reader s convenience. Let V := {1,..., n} be the vertex set of G. Set c := C /n and let x := (x 1,..., x n ) where x i := I i C c. It is clear that x is orthogonal to 1. Thus, < Ax, Ax > λ(g) 2 x 2. The right hand side is λ 2 G c(1 c)n λ2 G cn = λ(g)2 C. The left hand side is v V ( N C(v) cd) 2, where N C (v) is the set of v C such that vv is an directed edge. It follows that (2) v B( N C (v) cd) 2 v V ( N C (v) cd) 2 λ 2 C. On the other hand, by the triangle inequality

380 VAN H. VU (3) e(b, C) d n B C = e(b, C) cd B v B N C (v) cd. By Cauchy-Schwartz and (2), the right hand side of (3) is bounded fro above by B ( (N C (v) cd) 2 ) 1/2 λ B C, v B concluding the proof. Now we are ready to foralize our first ain lea: Lea 3.2. (Expander decoposition lea) Let K n be the coplete digraph on V := {1,..., n}. Assue that K n is decoposed in to k + 1 edge-disjoint digraphs H 0, H 1,..., H k such that For each i = 1,..., k, the out-degrees in H i are the sae and at ost d and λ(h i ) λ. The out-degrees in H 0 are at ost d. Let B and C be subsets of V and K be a subgraph of K n with L (directed) edges going fro B to C. Then K contains edges fro at least in{ L B d 2λ B C, (L B d )n } 2d B C different H i, i 1. Proof. By the previous lea, each H i, 1 i k has at ost d n B C + λ B C edges going fro B to C. Furtherore, H 0 has at ost B d edges going fro B to C. Thus the nuber of H i, i 1, having edges in K is at least ( d n B C + λ B C ) 1 (L d B ) in{ L d B 2λ B C, (L d B )n } 2d B C copleting the proof.

SUM-PRODUCT ESTIMATES VIA DIRECTED EXPANDERS 381 4. Directed Cayley graphs Let H be a finite (additive) abelian group and S be a subset of H. Define a directed graph G S as follows. The vertex set of G is H. There is a direct edge fro x to y if and only if y x S. It is clear that every vertex in G S has out-degree S. (In general H can be non-abelian, but in this paper we restrict ourselves to this case.) Let χ ξ, ξ H, be the (additive) characters of H. It is well known that for any ξ H, s S χ ξ(s) is an eigenvalue of G S, with respect the eigenvector (χ ξ (x)) x H. It is iportant to notice that the graph G S, for any S, is noral, by (1). Indeed, for any two vertex x and y N + (x, y) = N (x, y) = (x + S) (y + S). We are going to focus on the following two cases Special case 1. H = F 2 q, with F q being a finite field of q = p r eleents, p prie. Using e(α) to denote exp( 2πi p α), we have χ ξ (x) = exp( 2πi Trace ξ x) = e(trace ξ x), p where Trace z := z + z p + + z pr 1 and ξ x is the inner product of ξ and x. Special case 2. H = Z 2. In this case we use e(α) to denote exp( 2πi α). We have χ ξ (x) = exp( 2πi ξ x) = e(ξ x). Our second ain ingredient is the following theore, which is a corollary of [12, Theore 5.1.1]. (We would like to thank B. C. Ngo for pointing out this reference.) Theore 4.1. Let P be a polynoial of degree k in F q [x 1, x 2 ] which does not contain a linear factor. Let Root(P) be the set of roots of P in F 2 q. Then for any 0 y F 2 q, x Root(P) e(x y) = O(k 2 q 1/2 ). Given a polynoial P and an eleent a F q, we denote by G a the Cayley graph defined by the set Root(P a). As a corollary of the theore above, we have Corollary 4.2. Let P be a polynoial of degree k in F q [x 1, x 2 ] and a be an eleent of F q such that P a does not contain a linear factor. Then λ(g a ) = O(k 2 q 1/2 ).

382 VAN H. VU It is plausible that a ring analogue of Theore 4.1 can be derived (with F q replaced by Z ). However, the (algebraic) achinery involved is heavy. We shall give a direct proof for Corollary 4.2 in the special case when P is quadratic. Let Ω be the set of those quadratic polynoials which (after a proper changing of variables) can be written in the for A 1 x 2 + A 2 y 2 with A 1, A 2 Z, the set of eleents co-prie with. (For exaple, both Q = x 2 + y 2 and Q = 2xy = (x + y) 2 (x y) 2 belong to Ω.) Fix a Q in Ω and for each a Z define the Cayley graph G a as before. Theore 4.3. For any 0 a Z, λ(g a ) g() γ(). 1/2 The proof of this theore will appear in Section 6. 5. Proofs of Theores 1.2 and 1.5 To prove Theore 1.2, consider a set A F q and set B := A A F 2 q. Since our estiate is trivial if A = O(k 2 q 1/2 ), we assue that A k 2 q 1/2. For each a F q, consider the polynoial P a = P a and define a Cayley graph G a accordingly. The out-degree in this graph is O(q). We say that an eleent a is good if P a does not contain a linear factor and bad other wise. Lea 5.1. Let P be a polynoial of degree k in F q [x 1,..., x d ]. Assue that P cannot be written in the for P = Q(L), where Q a polynoial with one variable and L is a linear for of x 1,..., x d. Then there are at ost k 1 eleents a i such that the polynoial P a i contains a linear factor. Proof. Let a 1,..., a k be different eleents of F q such that there are linear fors L 1,..., L k and polynoials P 1,..., P k F q [x 1,..., x d ] such that P a i = L i P i. If L i and L j had a coon root x, then P (x) a i = P (x) a j = 0, a contradiction as a i a j. It follows that for any 1 i < j d, L i and L j do not have a coon root. But since the L i are linear fors, we can conclude that they are translates of the sae linear for L, i.e., L i = L b i, for soe b 1,..., b k F q. It now suffices to prove the following clai Lea 5.2. Let P be a polynoial in F q [x 1,..., x d ] of degree k. Assue that there is a non-zero linear for L, a sequence a 1,..., a k of (not necessarily distinct) eleents of F q and a set {b 1,..., b k } F q such that P (x) = a i whenever L(x) = b i. Then there is a polynoial Q in one variable such that P = Q(L).

SUM-PRODUCT ESTIMATES VIA DIRECTED EXPANDERS 383 Assue, without loss of generality, that the coefficient of x 1 in L is non-zero. We are going to induct on the degree of x 1 in P (which is at ost k). If this degree is 0 (in other words P does not depend on x 1 ), then P is a constant, since for any sequence x 2,..., x d, we can choose an x 1 such that L(x 1,..., x d ) = b 1, so P (x 1,..., x d ) = a 1. If the degree in concern is not zero, then we can write P = (L b 1 )P 1(x) + Q 1, where Q 1 does not contain x 1. By the above arguent, we can show that Q 1 = a 1. Furtherore, if L(x) = b i, 2 i k, then P 1 = (a i a 1 )/(b i b 1 ). Now apply the induction hypothesis on P 1, whose x 1 -degree is one less than that of P 1. If a is good, then λ(g a ) = O(k 2 q 1/2 ). Let the graph H 0 be the union of bad G a. By the above lea, the axiu out-degree of this graph is d = O(k 2 q). In F 2 q, define a directed graph K by drawing a directed edge fro (x, y) to (x, y ) if and only if either both x x and y y are in A or both x x and y y are in A. Consider the set C := (A + A) (A + A) F 2 q. Notice that in K any point fro B has at least A 2 edges going into C. Thus L, the nuber of directed edges fro B to C, is at least A 4. Since A k 2 q 1/2, we have L B d A 4 B d = A 4 A 2 O(k 2 q) = (1 o(1)) A 4. Applying the Expander Decoposition Lea and Corollary 4.2, we can conclude that the nuber of P a having edges fro B to C (which, by definition of B and C, is P (A) ), is at least ( A 4 Ω in{(1 o(1)) k 2 q 1/2 A A + A, (1 o(1)) A 4 q 2 ) kq A 2 A + A 2 }. fro which the desired estiate follows by Hölder inequality. The proof of Theore 1.5 (using Theore 4.3 instead of Corollary 4.2) is siilar and is left as an exercise. To prove Theore 2.2, consider a set A F 2 q where A kq. Let B = C = A and K be the coplete digraph on A. We can assue that A q. We have L = (1 + o(1)) A 2 and d = O(kq). Thus L d B = L d A = (1 + o(1))l = (1 + o(1)) A 2. By the Expander Decoposition Lea,

384 VAN H. VU The right hand side is ( A 2 P (A) = Ω in{(1 o(1)) k 2 q 1/2 A, (1 o(1)) A 2 q 2 ) kq A 2 }. copleting the proof. ( Ω in{ A k 2 q, q ) 1/2 k }, 6. Proof of Theore 4.3 We are going to follow an approach fro [14]. We need to use the following two classical estiates (see, for exaple, [13, page 19]) Theore 6.1. (Gauss su) Let be an positive odd integer. Then for any integer z co-prie to e(zy 2 ) =. y Z Theore 6.2. (Kloosteran su) Let be an positive odd integer. Then y Z e(ay + bȳ) τ()(a, b, ) 1/2, where (a, b, ) is the greatest coon divisor of a, b and, τ() is the nuber of divisors of, and ȳ is the inverse of y. Let p 1 < < p k be the prie divisors of and set Ω() := { i I p i I {1,..., k}, I }. Notice that g() satisfies the following recursive forula: g(1) := 0 and g() := τ() + d Ω() g(/d). Let S be the set of roots of Q a. We are going to use the notation G S instead of G a. We use induction on to show that λ(g S ) g() γ(). 1/2 The case = 1 is trivial, so fro now on we assue > 1. Cayley s graphs, the eigenvalues of G S are By properties of

SUM-PRODUCT ESTIMATES VIA DIRECTED EXPANDERS 385 λ ξ = s S e(ξ s), where ξ Z 2. For ξ = 0, we obtain the largest eigenvalue S, which is the degree of the graph. In what follows, we assue that ξ 0. Recall that s S if and only if Q(s) = a. We have (4) λ ξ = e( av)e(ξ x + vq(x)) = F (v) v Z x Z 2 v Z \{0} where F (v) := x Z 2 e( av)e(ξ x + vq(x)), taking into account the fact that F (0) = 0. For d = i I p i Ω(), let η(d) = I + 1. By the exclusion-inclusion forula, (5) F (v) = F (v) + F (v). v Z \0 v Z d Ω()( 1) η(d) d v Let us first bound S 0 := v Z F (v). We write x = (x 1, x 2 ) where x 1, x 2 Z. As Q is non-degenerate, by changing variables we can rewrite e(ξ x + vq(x)) as e(v(a 1 x 2 1 + A 2 x 2 2) + (B 1 x 1 + B 2 x 2 + C)) where B 1, B 2, C ay depend on ξ, but A 1, A 2 Z depends only on Q. We have (thanks to the fact that v, A 1, A 2, 2, 4 are all in Z ) v(a 1 x 2 1 + A 2 x 2 2) + (B 1 x 1 + B 2 x 2 + C) =va 1 (x 1 + B 1 2vA 1 ) 2 + va 2 (x 2 + B 2 2vA 2 ) 2 + (C B2 1 4vA 1 B2 2 4vA 2 ). It follows that (6) S 0 = v Z e(c)e( av ( B2 1 4A 1 + B2 2 4A 2 ) v) x 1,x 2 Z e(va 1 (x 1 + B 1 2vA 1 ) 2 + va 2 (x 2 + B 2 2vA 2 ) 2 ). Notice that

386 VAN H. VU (7) x 1,x 2 Z e(va 1 (x 1 + B 1 2vA 1 ) 2 + va 2 (x 2 + B 2 2vA 2 ) 2 ) = e(va 1 y 2 ) e(va 2 y 2 ). y Z y Z Set b := B2 1 4vA 1 + B2 2 4vA 2, we have ( ) (8) S 0 = e(c) e(va 1 y 2 ) e(va 2 y 2 ) y Z y Z v Z e( av b v). By Theores 6.1 and 6.2 and the fact that (a, b, ) γ() (since a 0), we have (9) S 0 τ()(a, b, ) 1/2 1/2 τ() γ(). 1/2 2 Now we bound the second ter in the right hand side of (5), using the induction hypothesis. Fix d Ω() and consider S d := F (v) = e(ξ x) e(v(q(x) a)). d v x Z 2 d v Write = d d, v = dv, where d := /d and v Z d. Each vector x in Z 2 has a unique decoposition x = x [1] + d x [2] where x [1] Z 2 d and x [2] Z 2 d. Finally, there is a Z d such that a a (od d ). Since Q(x) Q(x [1] ) (od d ), we have e(v(q(x) a)) = exp ( 2πi d v (Q(x [x1] a ) ). Therefore, (10) e(v(q(x) a)) = d v v Z d exp (2πi d v (Q(x [1] a ) which equals d if Q(x [1] ) a (od d ) and zero otherwise. It follows that

SUM-PRODUCT ESTIMATES VIA DIRECTED EXPANDERS 387 S d = d e(ξ x). x Z 2,Q(x[1] )=a (od d ) Next, we rewrite e(ξ x) as exp( 2πi (ξ x[1] + d ξ x [2] )). This way, we have (11) S d = d x [1] Z 2 d,q(x [1] )=a e(ξ x [1] ) x [2] Z 2 d exp( 2πi d ξ x 2). The su x [2] Z 2 exp( 2πi d ξ x 2) is d 2 if both coordinates of ξ are divisible by d and d zero other wise. Set ξ d = ξ/d, we have S d = d d 2 exp( 2πi ξ d x [1] ). d Q(x [1] )=a Notice that 2πi Q(x [1] )=a exp( d ξ d x [1] is a (non-trivial) eigenvalue of a Cayley s graph defined by Q on Z 2 d, where d = /d. Thus, by the induction hypothesis, This iplies exp( 2πi ξ d x [1] /d /d ) g(/d) g(/d) d γ(/d) 1/2 γ(). 1/2 Q(x [1] )=a 2 (12) S d g(/d) γ(). 1/2 By (5), (9), (12) and the triangle inequality λ ξ copleting the proof. 2 γ() 1/2 ( τ() + d Ω() 2 g(/d) ) = g() γ() 1/2 7. Open questions Our study leads to several questions:

388 VAN H. VU Proble 1. What happens if A q 1/2 in the case q is a prie? Proble 2. If q is not a prie and both A + A and P (A) is sall, can one prove that ost of A is contained in a subfield? Proble 3. Characterize all pairs P 1, P 2 of polynoials such that the following generalization of ( ) holds: If P 1 (A) is sall, then P 2 (A) is large. Proble 4. Assue that q is a prie. Let ɛ be a sall positive constant. A polynoial in F q [x 1, x 2 ] is generic if for any sufficiently large A F q, P (A) in{q, A 1+ɛ }. Can one characterize all generic polynoials? References [1] N. Alon, J. Spencer, The probabilistic ethod (Second edition), Wiley-Interscience, 2000. [2] J. Bourgain, N. Katz, and T. Tao, A su-product estiate in finite fields, and applications, Geo. Funct. Anal. 14 (2004), no. 1, 27 57. [3] J. Bourgain and A. Gaburd, New results on expanders, C. R. Math. Acad. Sci. Paris, 342 (2006), no. 10, 717 721. [4] J. Bourgain, More on the su-product phenoenon in prie fields and its applications, Int. J. Nuber Theory, 1 (2005), no. 1, 1 32. [5] P. Brass, W. Moser, and J. Pach, Research probles in discrete geoetry, Springer, New York, 2005. [6] F. Chung, R. Graha, and R. Wilson, Quasi-rando graphs, Cobinatorica 9 (1989), no. 4, 345 362. [7] P. Erdős, On sets of distances of n points, Aer. Math. Monthly, 53 (1946), 248 250. [8] P. Erdős and E. Szeerédi, On sus and products of integers, Studies in pure atheatics, 213 218, Birkhuser, Basel, 1983. [9] J. Garibaldi, Erdős Distance Proble for Convex Metrics, Ph. D. Thesis, UCLA 2004. [10] D. Hart, A. Iosevich, and J. Solyosi, Su-product estiates in finite fields via Kloosteran sus, Int. Math. Res. Not. IMRN 2007, no. 5, Art. ID rn007, 14 pp. [11] A. Iosevich and M. Rudnev, Erdős distance proble in vector spaces over finite fields, Trans. Aer. Math. Soc. 359 (2007), no. 12, 6127 6142. [12] N. Katz, Soes exponentielles, Asterisque 79, Socit Mathatique de France, Paris, 1980. [13] H. Iwaniec and E. Kowalski, Analytic nuber theory, Aerican Matheatical Society, 2004. [14] A. Medrano, P. Myers, H. M. Stark, and A. Terras, Finite analogues of Euclidean space, J. Coput. Appl. Math. 68 (1996), no. 1-2, 221 238. [15] T. Tao and V. Vu, Additive cobinatorics, Cabridge University Press, 2006. Departent of Matheatics, Rutgers, Piscataway, NJ 08854 E-ail address: vanvu@@ath.rutgers.edu