SUM-PRODUCT ESTIMATES APPLIED TO WARING S PROBLEM MOD P

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1 SUM-PRODUCT ESTIMATES APPLIED TO WARING S PROBLEM MOD P TODD COCHRANE AND CHRISTOPHER PINNER Abstract. Let γ(k, p) denote Waring s number (mod p) and δ(k, p) denote the ± Waring s number (mod p). We use sum-product estimates for na and na na, following the method of Glibichuk and Konyagin, to estimate γ(k, p) and δ(k, p). In particular, we obtain explicit numerical constants in the Heilbronn upper bounds: γ(k, p) 8 k 1/, δ(k, p) 0 k 1/ for any positive k not divisible by (p 1)/. 1. Preliminaries Let p be a prime and k a positive integer. The smallest s such that the congruence (1.1) x k 1 + x k + + x k s a (mod p) is solvable for all integers a is called Waring s number (mod p), denoted γ(k, p). Similarly, the smallest s such that (1.) ±x k 1 ± x k + ± x k s a (mod p), is solvable for all a is denoted δ(k, p). If d = (k, p 1) then clearly γ(d, p) = γ(k, p) and so we assume henceforth that k p 1. If A is the multiplicative subgroup of k-th powers in Z p then we write γ(a, p) = γ(k, p), δ(a, p) = δ(k, p). Cauchy [4] established the uniform bound γ(k, p) k with equality if k = p 1 or (p 1)/, and many improvements to this bound have been made since then; see [6] for references. Heilbronn [11] made the following conjectures: Let t = A = (p 1)/k. I: For any ε > 0, γ(k, p) ɛ k ε for t > c ε. II: For t >, γ(k, p) k 1/. The first conjecture was proved by Konyagin [1] and the second by Cipra and the authors [6]. For t =, 4, 6 it was shown [6] that (1.) k 1 γ(k, p) k, and thus the exponent 1/ is sharp. Indeed, the exact value of γ(k, p) was determined for these three cases. The purpose of this paper is to show how sum-product estimates can be used to obtain explicit constants in the Heilbronn upper bounds. Theorem 1.1. For t > we have the uniform upper bound γ(k, p) 8 k 1/. The proof of the theorem (section 9) uses the sum-product method of Glibichuk and Konyagin [9] for t 4 (sections 6,7) and the lattice method of Bovey [] for t < 4 (section 8). An explicit version of the first Heilbronn conjecture is given in Corollary 7.1. Date: September 6, Mathematics Subject Classification. 11P05;11T99. Key words and phrases. Waring s problem, sum-product sets. 1

2 TODD COCHRANE AND CHRISTOPHER PINNER For delta we obtain δ(k, p) 0 k 1/ ; Corollary 10.. We also explore the relationship between γ(k, p) and δ(k, p) (section 4) proving in particular, γ(k, p) log (log (p)) δ(k, p). Bovey [] proved the weaker bound γ(k, p) δ(k, p) log p. We leave open the following Question 1. Does there exist a constant C such that γ(k, p) C δ(k, p)? For any subsets S, T of Z p let. Sum-Product Estimates. S + T = {s + t : s S, t T }, ST = {st : s S, t T }, S T = {s t : s S, t T }, ns = S + S + + S (n times). Note that (ns)t n(st ). We let nst denote the latter, n(st ). If A is a multiplicative subgroup of Z p then for any l, A l = A, na l = na and (na)(ma) nma. The basic strategy for bounding Waring s number is to first obtain good lower bounds for na and then apply the following lemma to sets of the form na, ma to obtain all of Z p. Lemma.1. Let A, B be subsets of Z p and m a positive integer. a) If 0 / A and B A 1 m > p then mab = Zp. b) If B A p then 8AB = Z p. Part (a) was proven by Bourgain [1, Lemma 1] for the case m =. We prove the general case in section. Part (b) is due to Glibichuk and Konyagin [9, Lemma.1]. It follows from (b) that if na p (for a multiplicative group A) then γ(a, p) 8n. We shall make frequent use of the Cauchy-Davenport inequality, for any S, T Z p, and its corollary S + T min{ S + T 1, p}, ns min{n( S 1) + 1, p}. Another key tool we need is Rusza s triangle inequality, (see eg. Nathanson [15, Lemma 7.4]). (.1) S + T S 1/ T T 1/, for any S, T Z p, and its corollary 1 (.) ns S n 1 S S 1 1 n 1 S S 1 1 for any positive integer n. In section 5 we obtain lower bounds for A A and A + A using the method of Stepanov. Next we obtain lower bounds for na na (section 6), followed by lower bounds for na (section 7).. Proof of Lemma.1(a) Let a Z p and N denote the number of m-tuples (x 1,..., x m, y 1,... y m) Z m p x 1y x my m = a. We first note that e p(λ(xy)) = e p(λ(x 1y 1 x y )) x A y B x 1,x A y 1,y B λ F p λ Z p = p {(x 1, x, y 1, y ) : x 1, x A, y 1, y B, x 1y 1 = x y } p A B, n, with

3 WARING S PROBLEM the last inequality following from the assumption that 0 / A (and thus x 1y 1 = x y implies y 1 = x 1 1 xy.) Now, pn = A m B m + (.1) e p(λ(x 1y x my m a)) λ 0 x i A y i B = A m B m + ( m (.) e p( λa) e p(λxy)). λ 0 By the Cauchy-Schwarz inequality, for λ 0, x A,y B x A y B e p(λxy) ( e p(λxy) B 1/ e p(λxy) y B x A y B x A B 1/ e p(λxy) 1/ = B 1/ (p A ) 1/, y Fp x A and so by the note above, ( m e p( λa) e p(λxy)) λ 0 x A y B m ( A B p) )1/ e p(λ(xy)) λ Z p A m +1 B m p m. We conclude from (.) that N is positive provided that yielding the result of the theorem. A m B m > A m +1 B m p m, x A y B 4. Relations between γ(k, p) and δ(k, p) Theorem 4.1. Let A ( be the set ) of nonzero k-th powers in Z p with k (p 1), k p 1. a) γ(k, p) log log p δ(k, p). log A b) γ(k, p) (log (δ(k, p)) + 4) δ(k, p). c) γ(k, p) log (log (p)) δ(k, p). d) γ(k, p) (p min 1)δ(k, p), where p min is the minimal prime divisor of A. e) If A is even then δ(k, p) = γ(k, p). If A is odd, then δ(k, p) = γ( k, p). Proof of Theorem 4.1. a) Put A 0 = A {0}, δ = δ(k, p). Since δa 0 δa 0 = Z p we obtain from (.) (4.1) jδa 0 δa 0 δa 0 1 1/j = p 1 1/j ( ) for any positive integer j. Hence if j > log log p we have jδa log A 0 A 1 > p, and by Lemma.1(a), (jδa 0)A = Z p, that is, jδa 0 = Z p. b) This follows from part (a) and the trivial bound ( A + 1) δ p, when A. c) If j log (log (p)) then p 1/j and so by (4.1) jδa p/, and thus jδa = Z p. d) Let q be the minimal prime divisor of A. Then A has a subgroup G of order q and x G x = 0 so that 1 is a sum of q 1 elements of A. e) If A is even then 1 is a k-th power, and so γ(k, p) = δ(k, p). If A is odd then k must be even (for p ) and A ( A) is the set of k/-th powers.

4 4 TODD COCHRANE AND CHRISTOPHER PINNER 5. Lower Bounds for A + A and A A We give two estimates for A + A and A A with A a multiplicative subgroup of Z p, the first effective when A p / and the second when A < p /. Throughout this section A ± A will denote either one of these two sets. Theorem 5.1. If A is a multiplicative subgroup of Z p then ) 1 A ± A p (1 + p. A In particular A ± A p if A p/. Proof. Let N denote the number of solutions of the congruence x 1 ± x y 1 ± y (mod p) with x 1, x, y 1, y A, and N a the number of solutions of x 1 ±x a (mod p), x 1, x A, for a Z p. By the Cauchy-Schwarz inequality A = a Na A ± A 1/ N 1/. The lower bound for A ± A then follows from the estimate of Hua and Vandiver [1] and Weil [16], N A 4 p + A p. Theorem 5.. (a) Let A be a multiplicative subgroup of Z p and σ be a positive integer. If 4σ(4σ ) A, then A ± A (σ + 1) A. p 4σ (b) In particular, if A is a multiplicative subgroup of Z p with A < p /, we have A ± A 1 4 A ( A ) > 1 4 A /. The theorem is a refinement of a special case of Bourgain, Glibichuck and Konyagin [, Lemma 7] which gives A A 1 9 A / for A < p 1/. The case σ = 1 is comparable to what one obtains from the Cauchy-Davenport Theorem, A min{p, A }, for any multiplicative subgroup A. If 0 is included there is the stronger result A 0 min{p, A +1}, for any multiplicative subgroup A with A, where A 0 = A {0}; see [15, Theorem.8]. It is plain that the exponent / in the lower bound of the theorem cannot be improved if we allow A to approach p / in size, but we are lead to ask Question. For A < p 1/ can the exponent / in the theorem be improved? Question. For A p / do we have A + A Z p, that is, γ(a, p)? (Note, 0 may not be in A + A even when A = p 1.) It is known that γ(a, p) for A > p/4. Proof. We use the Stepanov method as developed by Heath-Brown and Konyagin [10]. Let A be a multiplicative subgroup of Z p with t = A and σ be a positive integer. Suppose that 4σ(4σ ) A p. We proceed with a proof by contradiction. Assume that 4σ A ± A < (σ + 1)t. Write A ± A as a union of disjoint cosets of A in Z p, A ± A = Ax 1 Ax Ax s {0}, where the {0} is omitted if 0 / A + A. In particular, (5.1) A ± A = st + 1 or st, and so s σ. For any coset Ax j let N j = {x A : x ± 1 Ax j} = {(x, y) A A : x ± y = x j}. Now for any x A, x 1, x ± 1 Ax j for some j and so s (5.) N j = t 1 or t. j=1

5 WARING S PROBLEM 5 The next lemma is extracted from the proof of [14, Lemma.]. Lemma 5.1. Let a, b, d be positive integers such that sad + 1 sd(d 1) < ab, ab t, tb p. Then s a 1 + t(b 1) N j. d j=1 Proof. The lower case a, b, d in the lemma correspond to the upper case A, B, D in [14]. In equation (.11) of [14] we actually have sad + 1 sd(d 1) < ab by summing over k in the preceding line of their proof. while We apply the lemma with a = 4s, b = 4s, d = 8s 5. Then sad + 1 sd(d 1) = 64s 64s + 15s, ab = 64s 64s + 16s, so the first hypothesis holds. Next, ab = 4s(4s ) 4σ(4σ ) t. Finally, since t we have tb p (4σ ) = p. Thus, by the lemma, p 4σ s N j j=1 4σ 4s 1 + t(4s ) 8s 5 = t 1 t + 6 1s 8s 5 < t 1, the latter inequality following from 1s 6 4s(4s ) t. This contradicts the inequality in (5.). For part (b) simply choose σ = [ 1 ( t )] and observe that t < p / implies 4 t p 4σ. 6. Lower Bounds for na na, Part I We follow the method of Glibichuk and Konyagin [9], which builds upon ideas in []. For any subsets X, Y of Z p let { } X X Y Y = x1 x : x 1, x X, y 1, y Y, y 1 y. y 1 y The key lemma is Lemma 6.1. [9, Lemma.] For X, Y Z p with Y > 1 and X X Y Y XY XY + Y Y X Y. Zp we have Proof. If X X Y Y Zp then there exist x1, x X, y1, y Y such that x 1 x y 1 y + 1 / X X. Y Y But then the mapping from X Y into XY XY + Y Y given by is clearly one-to-one and the lemma follows. We also use the elementary (x, y) (y 1 y )x + (x 1 x + y 1 y )y, Lemma 6.. Let A be a multiplicative subgroup of Z p and X, Y be subsets of Z p such that AX X, AY Y. Then X X X X ( Y Y 1) Y Y. A Proof. If c = (x 1 x )/(y 1 y ) for some x 1, x X, y 1 y Y, then c = (ax 1 ax )/(ay 1 ay ) for any a A.

6 6 TODD COCHRANE AND CHRISTOPHER PINNER For k N let a k = 4k 1, b k = 4k + 8, 6 so that a 1 = 1, a = 5, a = 1, a 4 = 85, b 1 =, b = 4, b = 1, b 4 = 44, and for k 1, (6.1) a k+1 = 4a k + 1, b k+1 = 8a k Put A k = (a k A a k A), B k = (b k A b k A). Theorem 6.1. Let A be a multiplicative subgroup of Z p. a) For k 1, b) For k, A k A A A k 1 if k = 1 or A k 1 A k 1 A A < p A. B k A A A k if A k A k A A < p A. Proof of Theorem 6.1. a) The statement is trivial for k = 1. For k > 1, put X = A k 1, Y = A. The hypothesis A k 1 A k 1 A A < p A implies, by Lemma 6., that Zp. Noting that by relation (6.1) X X Y Y XY XY + Y Y = A k 1 A k 1 + A A = (4a k 1 + 1)A (4a k+1 + 1)A = A k, we obtain A k A k 1 A by Lemma 6.1. The theorem now follows by induction on k. b) Put X = A k, Y = A A. Under the assumption of the theorem (X X)/(Y Y ) Z p. Now, by relation (6.1) and so by Lemma 6.1 XY XY + Y Y (8a k + 4)A (8a k + 4)A = B k, Part (b) follows from the bound in part (a). B k A k A A. Theorem 6.. Let A be a multiplicative subgroup of Z p and λ be a positive real number such that A A λ A /. a) For k 1, A k min{ 1/ p /, λ A k+ 1 }. b) For k, B k min{ /7 p 4/7, λ A k }. Proof. a) The result is immediate for k = 1 or A = 1, so we assume k and A. If A k 1 A k 1 A A < p A the inequality follows from Theorem 6.1. Otherwise, A k 1 A k 1 p A / A A. Then A k = a k A a k A 4a k 1 A 4a k 1 A A k 1 A k 1 p A A A p A A 1/. Also by the Cauchy-Davenport relation A k A = 5A 5A A A (for A > 1). Thus A k (p / A A )( A A ) = p and the result follows. b)we may assume A > 1. If A k A k A A < p A the result follows from Theorem 6.1. Assume that A k A k A A p A. Then B k = b k A b k A 8a k 1 A 8a k a A a k A a k A A k A k p A A A p A A /4. Also B k 1A 1A > A A and so B k 7 (p 4 / A A ) A A.

7 WARING S PROBLEM 7 Thus with λ = 1 (as given by Lemma 5.) we have for any multiplicative subgroup A 4 of Z p, A A min{ 1 4 A /, p/} A A min{ A, p / } 5A 5A min{ 1 4 A 5/, 1/ p / } 1A 1A min{ 1 16 A, /7 p 4/7 } 1A 1A min{ 1 4 A 7/, 1/ p / } 44A 44A min{ 1 16 A 4, /7 p 4/7 } 85A 85A min{ 1 4 A 9/, 1/ p / } The bound for A A is from Theorems 5.1 and 5.. The bound for A A follows from Lemma 6.1 when A A < p A and from the Cauchy-Davenport inequality otherwise. Further lower bounds on na na are given in section 10. For k N, put 7. Lower bounds for na m k = 1 4k+1 + k 1, n k = 4k+1 + k 14, so that m 1 =, m = 19, m = 84, n 1 = 7, n = 40, n = 169. Theorem 7.1. Suppose that A is a multiplicative subgroup of Z p and λ is a positive real number such that A λ A / and A A λ A /. Then for any k N, where α k = λ k, β k = λ k. a) m k A min{ p, α k A k+ 1 }, b) n k A min{ p, β k A k+1 }, Observing that A = Z p when A > p / (see eg. [7]) and that by Theorem 5. we can take λ = 1/4 for A < p /, we obtain in particular that for any multiplicative subgroup A of Z p, A min{.5 A /, p/} 4A min{ A /, p 5/8 } 7A min{.5 A, p} 19A min{.15 A 5/, p} 40A min{.177 A, p} 84A min{.106 A 7/, p} 169A min{.16 A 4, p}. The estimate for 4A comes from 4A A 1/ A A 1/ A / for A A < p A, 4A A A 15/16 (p A ) 15/ p 5/8, otherwise. In comparison [9, Lemma 5.] has 1A 8 A 1/7 for A p 1, 5A 8 A 0/7 for A p 1, 1A 8 A 7/7 for A 4 < p 1, etc.. Proof of Theorem 7.1. The inequalities m k A 1 A k+ 1 and nk A 1 A k+1 follow immediately from the Cauchy-Davenport estimates of m k A and n k A for A < 5 and so we assume A 5.

8 8 TODD COCHRANE AND CHRISTOPHER PINNER We prove parts (a) and (b) simultaneously by induction on k. First note that the validity of part (a) for k implies the validity of part (b) for k. If m k A p then trivially n k A p. Otherwise m k A α k A k+ 1. Then since nk = m k + a k+1 we have by Rusza s inequality (.1) n k A m k A 1/ a k+1 A a k+1 A 1/ m k A 1/ A k+1 1/. If A k A k A A < p A then by Theorem 6.1 and the bound in part (a), n k A λ k A k A A 1/ A k/ β k A k+1. If A k A k A A p A then in particular a k A a k A = A k A k p 1/ A 1/ and a k A A p. Thus n k A (a k A) a k A 1/4 a k A a k A /4 a k A 1/4 p /8 A /8 = ( a k A A ) 1/8 A 1/4 p /8 A 1/4 p 1/ p. For k = 1 m 1A = A and so the inequality in (a) is trivial. Suppose the theorem is true for k 1. Note that for k, m k = n k 1 + b k+1 and so by inequality (.1) (7.1) m k A n k 1 A 1/ b k+1 A b k+1 A 1/ = n k 1 A 1/ B k+1 1/. If A k 1 A k 1 A A < p A then by Theorem 6.1(b) and the induction assumption we have m k A λ 4 k 1 A k k A A A λ 8 4 k +1 A k+ 1 = αk A k+ 1. If A k 1 A k 1 A A p A then in particular a k 1 A a k 1 A p 1/ A 1/ and a k 1 A A p. Thus m k A 4(a k 1 A) a k 1 A 1/8 a k 1 A a k 1 A 7/8 a k 1 A 1/8 p 7/16 A 7/16 ( a k 1 A A ) 1/16 A 1/4 p 7/16 A 1/4 p 1/ p. ( Theorem 7.. Put γ k = α k ) 1/(k+1), ( δk = subgroup of Z p and k N. a) If A γ k p 1/(k+1), then 8m ka = Z p. b) If A δ k p 1/(k+) then 8n ka = Z p. β k ) 1/(k+). Let A be a multiplicative Proof of Theorem. Under the given hypotheses, it follows from Theorem 7.1 that m k A p and nk A p and so by the result of Glibichuk and Konyagin, Lemma.1 (b) the theorem follows. Letting λ = 1/4 we obtain the following for any multiplicative subgroup A of Z p: 8A = Z p for A > p 1/ A = Z p for A >.18p 1/ 9A = Z p for A >.8p 1/4 888A = Z p for A >.64p 1/5 1800A = Z p for A > p 1/ A = Z p for A >.11p 1/7 8488A = Z p for A > 1.7p 1/8

9 WARING S PROBLEM 9 The result for 8A is due to Glibichuk [8, Corollary 4]. Note that m k k+1 and n k k+1 for any k 1. Define c 1 = c = 1 and c l = { γ l 1 δ l if l is odd Then we obtain from Theorem 7. that for l, if l 4 is even (7.) A c l p 1/l = 57 4 l A = Z p. Corollary 7.1. For any prime p, l and multiplicative subgroup A of Z p with c l p 1/l A < c l 1 p 1/(l 1), we have γ(a, p) 14.5 p ln( A /c l 1). Proof. A c l p 1/l and so l A = Z p. We also have (l 1) ln( A /c l 1 ) ln p. Thus γ(a, p) ln p ln( A /c l 1) 14.5p ln( A /c l 1). 8. Bovey s method for small A. For small A we use a method of Bovey to bound δ(k, p) and γ(k, p). Let t = A so that tk = (p 1) and put r = φ(t). Let R be a primitive t-th root of one (mod p), that is, a generator of the cyclic group A, and Φ t(x) be the t-th cyclotomic polynomial over Q of degree r and ω be a primitive t-th root of unity over Q. In particular, Φ t(r) 0 (mod p). Let f : Z r Z[ω] be given by f(x 1, x,..., x r) = x 1 + x ω + + x rω r 1. Then f is a one-to-one Z-module homomorphism. Consider the linear congruence (8.1) x 1 + Rx + R x + + R r 1 x r 0 (mod p). By the box principle, we know there is a nonzero solution of (8.1) in integers v 1 = (a 1, a,..., a r) with a i [p 1/r ] (p 1) 1/r, 1 i r. For i r set v i = f 1 (ω i 1 f(v 1)). Then v 1,..., v r form a set of linearly independent solutions of (8.1) and by [, Lemma ] δ(k, p) 1 t v i 1, where (x 1, x,..., x t) 1 system i=1 = t i=1 xi. To determine the latter sum we start with the a 1 + a ω a rω r 1 a 1ω + a ω a rω r a 1ω + a ω a rω r+1 a 1ω + a ω a rω r+ a 1ω 4 + a ω a rω r a 1ω r 1 + a ω r a rω r and then reduce the higher powers of ω to powers less than r using Φ t or any other relation that is convenient. Note that for 0 i r 1, ω i occurs i + 1 times in the array, while for r i r, ω i occurs r 1 i times. If ω i can be expressed as a sum/difference of w i powers of ω less than r then we will call w i the weight of ω i in the above system. We see that ( r ) δ(k, p) 1 r i + w i(r 1 i) (p 1) 1/r. i=1 i=r.

10 10 TODD COCHRANE AND CHRISTOPHER PINNER In passing from δ(k, p) to γ(k, p) we use the relation of Theorem 4.1 (d), (8.) γ(k, p) (p min 1)δ(k, p). where p min is the minimal prime divisor of t. To illustrate the method we consider a few special cases. Case 1: Suppose t is a prime power q α so that r = q α q α 1. Then ω r+qα 1 = 1 and ω r = q i=0 ωqα 1i. It follows that w i = q 1 for i = r,..., r + q α 1 1 and that w i = 1 for i = r + q α 1,..., r. Thus (8.) δ(k, p) 1 r+q r α 1 1 r i + (r 1 i)(q 1) + (r 1 i) (p 1) 1/r i=1 i=r i=r+q α 1 and so δ(k, p) 1 4 qα 1 ( q α 1 (4q 11q + 8) (q ) ) (p 1) 1/r < t + 1 r k 1/r, γ(k, p) 1 4 (q 1)qα 1 ( q α 1 (4q 11q + 8) (q ) ) (p 1) 1/r < t + 1 r k 1/r. In particular, for t = α, we have and for prime t = q δ(k, p) t 8 (p 1)1/r, (8.4) δ(k, p) (t t +.5)(p 1) 1/(t 1), γ(k, p) (t 1)(t t +.5)(p 1) 1/(t 1) Case : Suppose t = q where q is a prime, so that r = q 1 and we have ω q = 1, ω q 1 = 1 + ω + ω q. We obtain r v i 1 ( t i=1 t + 5)(p 1)/(t ) δ(k, p) (.5t 1.5t +.5)(p 1) /(t ), γ(k, p) (.5t 1.5t +.5)(p 1) /(t ) Case : t = 1, r = 1. We have ω 1 = ω 11 ω 9 + ω 8 ω 6 + ω 4 ω + ω 1, and ω 14 = ω 7 1. Thus ω 1 = ω 11 ω 10 + ω 8 ω 7 ω 6 + ω 5 ω + ω 1 giving it a weight of 9. ω 14 to ω 18 each have weight, ω 19 weight 9, ω 0 weight 8, ω 1 and ω each of weight 1. Altogether we get r v i 1 ( (9+ +5) (+1))p 1/1 = 89p 1/1, i=1 and δ(k, p) 194.5p 1/1 γ(k, p) 89p 1/1. In a similar manner we obtain the following table of upper bounds for δ(k, p) and γ(k, p). The values for t =, 4 and 6 were determined in [6]. The p s appearing in the

11 WARING S PROBLEM 11 table may be replaced by (p 1). t δ(k, p) γ(k, p) t δ(k, p) γ(k, p) (p 1)/ (p 1)/ p 1/1 89p 1/1 k k p 1/ p 1/10 4 k 1 k p 1/ 10175p 1/ 5 1.5p 1/4 50p 1/4 4 4p 1/8 4p 1/8 6k 6k 5 7.5p 1/ p 1/ p 1/6 18p 1/ p 1/1 1.5p 1/1 8 8p 1/4 8p 1/ p 1/18 441p 1/18 9 4p 1/6 48p 1/ p 1/1 14.5p 1/ p 1/4 1.5p 1/ p 1/8 118p 1/ p 1/10 905p 1/ p 1/8 74p 1/ p 1/4 10.5p 1/ p 1/0 6115p 1/ p 1/1 1590p 1/1 18p 1/16 18p 1/ p 1/6 0.5p 1/6 58.5p 1/0 1167p 1/ p 1/8 148p 1/ p 1/ p 1/16 16 p 1/8 p 1/8 5 1p 1/4 49p 1/ p 1/16 848p 1/ p 1/1 97.5p 1/1 18 4p 1/6 4p 1/ p 1/6 4578p 1/ p 1/ p 1/ p 1/ p 1/ p 1/8 51.5p 1/8 9. Proof of Theorem 1.1 Let t = A >. As noted in (1.) for t =, 4, γ(k, p) k and so we may assume t 5. The inequality γ(k, p) [k/] + 1 of S. Chowla, Mann and Strauss [5], implies the theorem for k 7551 and so we assume k > The first step is to prove the theorem for t < 4 using the table from the previous section. Suppose t is a prime. Then by (8.4), γ(k, p) (t 1)(t t +.5)t 1/(t 1) k 1/(t 1) 8 k 1/, provided that k > 10 6, t < 4. For k < 10 6, p < < 5 and so by Theorem 4.1 (c), γ(k, p) 10δ(k, p). Thus we get the improved (for t > 10) upper bound γ(k, p) 10(t t +.5)t 1/(t 1) k 1/(t 1). With the aid of a calculator one can check that the latter quantity is less than 8k 1/ for t 1 and k 755. For nonprime values of t < 4, we turn to the table in the previous section. We note that if γ(k, p) C(p 1) 1/r then γ(k, p) 8 k 1/ provided that k > (C/8) r/(r ) t /(r ). Using the values of C in the table one checks that the statement is valid for k > Finally, suppose that t 4 and that k > If t > c 6p 1/6 we have by Theorem 7., γ(k, p) 1800 < 8 k 1/. Next, assume t < c 6p 1/6 = p 1/6. Say c l p 1/l t < c l 1 p 1/(l 1) for some l 7. Then by Corollary 7.1, and noting that.10 > c 7 > c 6 > c 8 > c 9... we have γ(k, p) 14.5 p ln(t/c 7) 14.5 (4 + 1/755) 14.5 (t + 1/k) ln(4/c 7) k ln(t/c 7) k ln(t/c 7 ) ln(4/c 7) 8 k.499.

12 1 TODD COCHRANE AND CHRISTOPHER PINNER 10. Lower Bounds for na na, Part II The lower bounds on A k and B k established in section 6 were sufficient for yielding good upper bounds on γ(k, p). One can achieve slightly better upper bounds on δ(k, p) by using the following variant of Theorem 6.1. Theorem For { any multiplicative subgroup A of Z p, 1 a) A A min{ A, p + 1} for any A A for A p 1/ { 8 b) For k 1, A k min{ A A A k 1, p+1 } A A A k 1 for A < p k+ 1 { min{ 16 c) For k, B k A A A k, p+1 }, A A A k for A < p k+4 1 The theorem follows from a couple lemmas of Glibichuk and Konyagin. Lemma [9, Corollary.5] For X, Y Z p with Y > 1, XY XY + Y Y > X Y (p 1) X Y + p 1. (Although their lemma is stated with a nonstrict inequality, the proof makes it clear that it is strict.) The following lemma is the same as Glibichuk and Konyagin [9, Lemma 5.1] applied to a slightly different set A k. Lemma 10.. Suppose A is a multiplicative subgroup of Z p with A 5. For any k and real number U with 0 U A A A k 1 we have U A k U 5 4 p 1. Proof. The proof is by induction on k, the case k = 1 being trivial. We use Lemma 10.1 with X = A k 1, Y = A. Noting that as above, we obtain XY XY + Y Y = A k 1 A k 1 + A A = A k, (10.1) A k A k 1 A (p 1) A k 1 A + p 1, and the proof proceeds identically as in [9]. Proof of Theorem a) Put X = Y = A in 10.1 to get A A A (p 1) A +p 1. A p 1 then A A 1 A, while if A > p 1 then A A > 1 (p 1). If A < p then A A A A A < p and so Lemma 6.1 gives A A A. b) Put U = min{ A A A k 1, p 1 }. Then by Lemma 10., A k min{ A 8 A A k 1, p 1 }, provided that A 5. For A = 1,,, 4 the inequality follows from the Cauchy-Davenport bound A k min{p, a k ( A 1) + 1} and A A = 1,, 7, 9 for A = 1,,, 4 respectively. The second inequality is proven by induction on k, the case k = 1 being trivial. Suppose the statement is true for k 1 and let A k+ < p. Put X = A k 1, Y = A. If X X < A A A k 1, in which case X X A A / A < p, we can apply Theorem 6.1 to get the result. If X X A A A k 1 then since A k X X the result is immediate. c) Suppose k. If A = 1 the bound is trivial, so assume A > 1. Put X = A k, Y = A A. Then XY XY + Y Y = 8 4k 1 A 8 4k 1 A + 4A 4A = B k, If

13 WARING S PROBLEM 1 and so by Lemma 10.1 { Ak A A B k min, p + 1 } and the result follows from the bound in part (b). Suppose now that A k+4 < p. If A k A k < A A A k, in which case X X Y Y / A < p, then the result follows from Theorem 6.1. Otherwise B k A k A k A A A k. Corollary Let A be a multiplicative subgroup of Z p with A > 1 and λ < 1 be a positive real with A A λ A /. a) For k 1 if A ( 8 λ )/(k+1) p /(k+1) then a k A a k A = Z p. b) For k, if A > ( 8 ) 1/k p 1/k then b λ k A b k A = Z p. Proof. a) As seen in (10.1) { Ak 1 A A k > min, p } Under the given hypotheses we have by Theorem 10.1 { } 1 A A k 1 min 8 λ A k+ p + 1, 8 A for A > 5. Thus A k > p and A k = Z p. If A =, or 4 the corollary follows readily from the Cauchy-Davenport inequality, p, a k A a k A min{p, 4a k ( A 1) + 1}. For A = 5 the conditions require k. Using the bound for δ(a, p) from the table in Section 8 (t = 5), we get δ(a, p) 1.5p 1/4 1.5(λ/8) 1/4 5 k k/4 (4k 1) = a k. b) Under the given hypothesis 16 λ A k > p and so by Theorem 10.1, B k > p. Thus B k = Z p. Thus we obtain (with λ =.5) 4A 4A = Z p for A > p 10A 10A = Z p for A >.58p /5 16A 16A = Z p for A >.18p 1/ 4A 4A = Z p for A > 1.97p /7 88A 88A = Z p for A >.56p 1/4 170A 170A = Z p for A > 1.70p /9 44A 44A = Z p for A >.1p 1/5 68A 68A = Z p for A > 1.54p /11 168A 168A = Z p for A > 1.87p 1/6 The result for 4A 4A is due to Glibichuk [8]. The result for 16A 16A is obtained from A A.5 A / p for A >.18p 1/, and thus 8(A A)(A A) = Z p. Put (10.) d l = (8/(λ)) 1/l for l = /, 5/, 7/,.... Applying Corollary 10.1 (a) with k = l 1 we see that if A d lp 1/l then δ(a, p) 4a k = 4 (4k 1) = 4l 4. We deduce

14 14 TODD COCHRANE AND CHRISTOPHER PINNER Corollary 10.. For any prime p and multiplicative group A with d l p 1/l A d l 1 p 1/(l 1) for some half integer l 5/, we have δ(a, p) 8 p ln( A /d l 1). Corollary 10.. For t > we have the uniform upper bound, δ(k, p) 0k 1/. Proof. For k 1595 the result follows from δ(k, p) [k/] + 1 0k 1/. Suppose that k > We note that δ(k, p) C(p 1) 1/r implies δ(k, p) 0k 1/ provided that k > (C/0) r/(r ) t /(r ). Using the values of C given in the table in section 6 we see that the latter holds for t < 9, k > Next, if t > 1.70p /9 then δ(k, p) 40 0k 1/ for k > Otherwise, d l p 1/l t d l 1 p 1/l 1 for some half integer l 11/. If 9 t < 50 then we can take λ =.8 in the definition of d l (10.) since A A t , A / t / 50/ and get d 9/ = Thus by Corollary 10. δ(k, p) 8 (9 + 1 ) ln(9/1.650) k ln(9/1.650) 14k Finally, if t 50 then we take λ =.5, d 9/ = 1.70, and get δ(k, p) 8 ( ) ln(50/1.70) k References ln(50/1.70) 14k.41. [1] J. Bourgain, Mordell s exponential sum estimate revistited, J. Amer. Math. Soc. 18, no. (005), [] J. Bourgain, A.A. Glibichuk and S.V. Konyagin, Estimates for the number of sums and products and for exponential sums in fields of prime order, J. London Math. Soc. () 7 (006), [] J. D. Bovey, A new upper bound for Waring s problem (mod p), Acta Arith. (1977), [4] A. Cauchy, Recherches sur les nombres, J. Ecole Polytechnique 9 (181), [5] S. Chowla, H. B. Mann and E. G. Straus, Some applications of the Cauchy-Davenport theorem, Norske Vid. Selsk. Forh. Trondheim (1959), [6] J.A. Cipra, T. Cochrane, C. Pinner, Heilbronn s conjecture on Waring s Number (mod p), J. Number Theory 15, no., (007), [7] T. Cochrane and C. Pinner, Small values for Waring s number modulo p, preprint. [8] A.A. Glibichuk, Combinational properties of sets of residues modulo a prime and the Erdös- Graham problem, Mat. Zametki 79, no., (006), Translated in Math. Notes 79, no., (006), [9] A.A. Glibichuk and S.V. Konyagin, Additive properties of product sets in fields of prime order, Additive combinatorics, 79 86, CRM Proc. Lecture Notes, 4, Amer. Math. Soc., Providence, RI, 007. [10] D. R. Heath-Brown and S. Konyagin, New bounds for Gauss sums derived from kth powers, and for Heilbronn s exponential sum, Quart. J. Math. 51 (000), 1-5. [11] H. Heilbronn, Lecture Notes on Additive Number Theory mod p, California Institute of Technology (1964). [1] L.K. Hua and H.S. Vandiver, Characters over certain types of rings with applications to the theory of equations in a finite field, Proc. Nat. Acad. Sci. U.S.A. 5 (1949), [1] S. V. Konyagin, On estimates of Gaussian sums and Waring s problem for a prime modulus, Trudy Mat. Inst. Steklov 198 (199), ; translation in Proc. Steklov Inst. Math. 1994, [14] S. V. Konyagin and I. E. Shparlinski, Character sums with exponential functions and their applications, Cambridge Univ. Press, Cambridge, [15] M.B. Nathanson, Additive Number Theory, Inverse Problems and the Geometry of Sumsets, Graduate Texts in Mathematics 165, Springer, New York, 1996.

15 WARING S PROBLEM 15 [16] A. Weil, Number of solutions of equations in finite fields, Bull. AMS 55 (1949), Department of Mathematics, Kansas State University, Manhattan, KS address: cochrane@math.ksu.edu Department of Mathematics, Kansas State University, Manhattan, KS address: pinner@math.ksu.edu