SUM-PRODUCT ESTIMATES APPLIED TO WARING S PROBLEM OVER FINITE FIELDS

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1 #A68 INTEGERS 11 (011) SUM-PRODUCT ESTIMATES APPLIED TO WARING S PROBLEM OVER FINITE FIELDS Todd Cochrane 1 Department of Mathematics, Kansas State University, Manhattan, Kansas cochrane@math.ksu.edu James Cipra Department of Mathematics, Kansas State University, Manhattan, Kansas cipra@math.ksu.edu Received: 5/13/11, Accepted: 10/31/11, Published: 11/7/11 Abstract Let A be the set of nonzero k-th powers in F q and γ (k, q) denote the minimal n such that na = F q. We use sum-product estimates for na and na na, following the method of Glibichuk and Konyagin to estimate γ (k, q). In particular, we obtain γ (k, q) 633(k) log 4/ log A for A > 1 provided that γ (k, q) exists. Dedicated to the memory of John Selfridge 1. Introduction Let F q be a finite field in q = p f elements and k be a positive integer. The smallest s such that the equation x k 1 + x k + + x k s = a (1) is solvable for all a F q (should such an s exist) is called Waring s number for F q, denoted γ(k, q). Similarly, the smallest s such that ±x k 1 ± x k ± ± x k s = a, () is solvable for all a F q is denoted δ(k, p). If d = (k, q 1) then clearly γ(d, q) = γ(k, q) and so we may assume k (q 1). If A is the multiplicative subgroup of k-th powers in F q then we write γ(a, q) = γ(k, q), δ(a, q) = δ(k, q). 1 Current address: Dept. of Mathematics, Georgia Institute of Technology, Atlanta, Georgia

2 INTEGERS: 11 (011) Also, we let γ (A, q), δ (A, q) denote the minimal s such that every element of F q is the sum (± sum ) of exactly s nonzero k-th powers, that is, (1), () resp. are solvable with all x i 0. It is well-known that γ(a, q), δ(a, q), γ (A, q), δ (A, q) exist if and only if A contains a set of f linearly independent points over F p ; see Lemma 6. For any subsets S, T of F q and positive integer n, let S + T = {s + t : s S, t T }, S T = {s t : s S, t T }, ns = S+S+ +S (n-times), ST = {st : s S, t T }, S n = SS S (n-times). Note that (ns)t n(st ). We let nst denote the latter, n(st ). Also, for any a F q we let as = {as : s S}. If A is a multiplicative subgroup of F q then γ (A, q) (if it exists) is the minimal s such that sa = F q, while γ(a, q) is the minimal s such that sa 0 = F q, where A 0 = A {0}. It is well-known that γ(k, q) k with equality if k = q 1 or (q 1)/. This was first observed by Cauchy [5] for the case q = p. Our first result is the analogue for γ (k, q). The proof makes use of Kneser s lower bound for A + B. Theorem 1. If A is the multiplicative subgroup of k-th powers in F q, A > and γ (A, q) exists, then γ (A, q) k + 1. When A = (that is, q is an odd prime and A = {±1}) then γ (A, q) = k. For A > it was established by Tietäväinen [0], for odd q, and by Winterhof [], [3, Lemma 1], for even q, that γ(a, q) [k/] + 1. It is an open question whether the same improvement holds for γ (A, q). For the case of prime fields Heilbronn [17] formulated two conjectures, which in the more general setting of F q can be stated as follow: 1. If A > then γ(a, q) k.. For any ɛ > 0 there exists a constant c(ɛ) such that if A > c(ɛ) then γ(a, q) ɛ k ɛ. The second conjecture was proven by Konyagin [18] for prime fields. Cipra, Cochrane and Pinner [8] established the first conjecture for prime fields, and the explicit bound γ(a, p) 83 k was obtained in [9]. Cipra [6, Theorem 4] proved the first conjecture for the general finite field F q, obtaining { 16 k + 1, for q = p. γ(a, q) 10 k + 1, for q = p f (3), f 3, whenever γ(a, q) is defined.

3 INTEGERS: 11 (011) 3 Next, let A = A F p, so that A = ( A, p 1). Cipra [6], sharpening the work of Winterhof [], established the bound (k + 1) 1/f 1 γ(k, q) 8f A, (4) whenever γ(k, q) exists. He also established the bound γ(k, q) fk log 4 f log A log log(k), (5) which resolved the second Heilbronn conjecture provided A f is sufficiently large. For prime fields Glibichuk and Konyagin [15] used methods of additive combinatorics to obtain γ (A, p) 400 k log 4 log A, (6) for any multiplicative subgroup A with A > 1. Cochrane and Pinner [9, Corollary 7.1] obtained a similar bound for q = p, and Glibichuk [13] established the same type of bound for q = p. The main result of this paper is a generalization of (6) to arbitrary F q, thus resolving the second Heilbronn conjecture for any finite field. Theorem. If A is a multiplicative subgroup of F q for which γ (A, q) is defined and A > 1, then with k = (q 1)/ A, we have γ (A, q) 633(k) log 4 log A. After submitting this work, the author s learned that Glibichuk [14, Corollary 1] recently proved a similar result, albeit with weaker constants. In particular, if A = p ɛ, then our result gives γ (A, q) 4 1/ɛ, whereas his result gives γ (A, q) 6 1/ɛ. For δ (A, q) we establish the stronger bound, Theorem 3. If A is a multiplicative subgroup of F q for which δ (A, q) is defined and A > 1, then with k = (q 1)/ A, we have δ (A, q) (40/3)k log 4 log A. As noted in Theorem 9, if q is even or A is even then γ (A, q) = δ (A, q) and thus the stronger bound in Theorem 3 applies to γ (A, q) as well. Further relations between δ(a, q) and γ(a, q) are given in Theorem 9. The exponent on k in the theorem improves on (5) when A > ( A, p 1) f and on (4) when A > 4 f. For small A ( A = O(1) as p ) one can obtain a stronger result by employing the lattice method of Bovey. In this manner we prove, Theorem 4. For any positive integer t there is a constant c 1 (t) such that if A is a multiplicative subgroup of F q with A = t, and such that γ(a, q) is defined, then γ(a, q) c 1 (t)k 1/φ(t).

4 INTEGERS: 11 (011) 4 The constant c 1 (t), estimated in [4] for prime fields, depends on the size of the coefficients of the cyclotomic polynomial of order t. Corollary 5. For any positive integer l there is a constant c(l) such that if A is a multiplicative subgroup of F q of order t such that φ(t) l and γ(a, q) exists, then γ(a, q) c(l)k 1/l. Proof. Suppose that φ(t) l. Put c = max t 4 l c 1 (t), c(l) = max{c, 1/l 633}, with c 1 (t) as defined in Theorem 4. If t > 4 l then, by Theorem, γ(k, q) 633 1/l k 1/l c(l)k 1/l. If t 4 l then, by Theorem 4, γ(k, q) ck 1/φ(t) c(l)k 1/l.. Preliminary Lemmas The first lemma gives equivalent conditions for the existence of γ(k, q). It is wellknown, and follows from the fact that the set of all sums of k-th powers is a multiplicatively closed set and therefore a subfield of F q ; see Tornheim [1, Lemma 1], or Bhaskaran [1]. Lemma 6. Let A be the set of nonzero k-th powers in F q. equivalent. The following are (i) γ(k, q) exists, that is, every element of F q is a sum of k-th powers. (ii) A is not contained in any proper subfield of F q ; that is, A contains a set of f linearly independent points over F p. (iii) A does not divide p j 1 for any j f, j < f, that is, pf 1 p j 1 for any j f, j < f. does not divide k It is also not hard to show that γ (A, q) exists if and only if A > 1 and γ(a, q) exists. An important tool needed throughout this paper is Rusza s triangle inequality (see, e.g., Nathanson [19, Lemma 7.4]), for any S, T F q, and its corollary S + T S 1/ T T 1/, (7) 1 ns S n 1 S S 1 1 n 1 S S 1 1 n, (8) for any positive integer n. The first inequality in (8) follows by induction on n, and the second from the trivial bound S S S. The following is a key lemma for showing that a sum-product set fills up F q. Lemma 7. Let A, B be subsets of F q and m 3 be a positive integer.

5 INTEGERS: 11 (011) 5 (a) If B A 1 m ( ) /m > q 1 B q then mab = Fq. (b) If B A q then 8AB = F q. (c) If B A > q and either A or B is symmetric (A = A) or antisymmetric (A A = ) then 8AB = F q. A slightly weaker form of part (a) was proven by Bourgain [, Lemma 1] for q = p and m = 3 and by Cochrane and Pinner [9, Lemma.1] for q = p and general m. A similar proof works for F q and is provided in Section 7. In the earlier versions of this ( statement, an extra hypothesis, 0 A, was included, and the factor 1 B q ) /m was excluded. Part (b) is due to Glibichuk and Konyagin [15, Lemma.1] for prime fields and to Glibichuk and Rudnev [16] for general F q. Part (c) is due to Glibichuk [1] for prime fields and to Glibichuk and Rudnev [16] as well as Cipra [7] for general F q. In particular, if A is a multiplicative subgroup, then applying (c) with B = A we see that γ (A, q) 8 provided that A > q. In the cases where A = 3, 4 or 6 one can actually evaluate γ(a, q). This was done in [8] for the case of prime moduli. Theorem 8. Let A is a multiplicative subgroup of F q of order 3,4 or 6 for which γ(a, q) exists. Then q = p or p. If q = p then γ(a, q) = p 1. If q = p then a + b 1, if A = 3, γ(a, p) = c 1, if A = 4, 3 a b, if A = 6 where, if A = 3 or 6, then a, b are the unique positive integers with a > b and a +b +ab = p, while if A = 4 then c, d are the unique positive integers with c > d and c + d = p. In particular, for A = 3, 4 or 6 we have 3k γ(a, q) k, if A = 3, k 1 γ(a, q) k 1, if A = 4, k 1 γ(a, q) 3 6k, if A = 6. Proof. Since A = 3, 4 or 6, every element of A is of degree 1 or over F p and therefore A F p. Thus, in order for γ(k, q) to exist we must have q = p or p. The case q = p is just Theorem of [8] and the case q = is trivial, so we shall assume q = p with p an odd prime and that A is not contained in F p. Case i: A = 3. Say A = {1, T, T } where T F p F p satisfies T + T + 1 = 0. In particular, p (mod 3). We claim that γ(a, q) = p 1 and consequently,

6 INTEGERS: 11 (011) 6 since 3k = p 1, γ(k, q) = 3k Let w = x + yt denote a typical element of F q where 0 x, y p 1 and let γ(w) denote the minimal number of elements of A required to represent w. First note that γ(0) = 3 since 1 + T + T = 0 so we assume that w 0. Suppose that x y. If x + y < p then trivially γ(w) < p. If x y < x then we write w = (y x)t + (p x)t and get γ(w) p+y x < p. If y x and y > 3p then we write w = (x y+p) 1+(p y)t and get γ(w) (x y + p) p 3 y < p. If y x and y < 3p, then x + y 3 y < p. A similar argument holds for x y. Finally, one can check that γ( 1 3 (p + 1) + 3 (p + 1)T ) = p 1. Case ii: A = 4. Say A = {±1, ±T }, with T = 1, T F p F p. In particular, p 3 (mod 4). Any element of F q may be written x+yt with x, y p 1, and so γ(a, q) p 1. Also, it is plain that γ( p 1 + p 1 T ) = p 1. Thus, γ(a, q) = p 1. Case iii: A = 6. Say A = {±1, ±T, ±T } with T T + 1 = 0. As in case ii, any element of F q may be written x+yt with x, y p 1, and so γ(a, q) p 1. Also, with just a little work one again sees that γ( p 1 + p 1 T ) = p 1. Thus, γ(a, q) = p 1. The precise relationship between γ(k, q) and δ(k, q) is an important unresolved problem. It is not known whether γ(k, q) Cδ(k, q) for some constant C. Bovey [4, Lemma ] established γ(k, p) (log p + 1)δ(k, p) for prime moduli, and improvements were given in [8]. Here we prove the analogue of [8, Theorem 4.1] for general finite fields. Theorem 9. Let A be the set of nonzero k-th powers in F q with k (q 1), such that γ(k, q) is defined. Then, (a) γ(k, q) 3 log ( 3 log q log A ) δ(k, q). (b) γ(k, q) 3 log (4δ(k, q)) δ(k, q) (c) γ(k, q) log (log (q)) δ(k, q). (d) γ(k, q) (p min 1)δ(k, q), where p min is the minimal prime divisor of A. (e) If q is even or A is even then δ(k, q) = γ(k, q). If A is odd and p is odd, then δ(k, q) = γ( k, q). Proof. a) Put A 0 = A {0}, δ = δ(k, q). Since δa 0 δa 0 = F q we obtain from (8), (observing that this inequality is strict for S > 1), jδa 0 > δa 0 δa 0 1 1/j = q 1 1/j, (9) ( ) for any positive integer j. Hence if j log 3 log q log A we have jδa 0 A 1 3 q, and so by Lemma 7 (a) with m = 3, 3(jδA 0 )A = F q, that is, 3jδA 0 = F q.

7 INTEGERS: 11 (011) 7 b) This follows from part (a) and the trivial bound ( A +1) δ q, when A 11. Indeed, in this case, log q log( A + 1) δ < 4 log A log A 3 δ. For A < 11, the result follow from part (d) of this theorem, since p min 7 for such A. c) We repeat the proof given by Cipra [6]. If j log (log (q)) then q 1/j and so by (9), jδa > q/. Thus, jδa = F q. (Here we have used the fact that if S is a subset of a finite group G with S > G / then S + S = G.) d) Let l be the minimal prime divisor of A. Then A has a subgroup G of order l and x G x = 0 so that 1 is a sum of l 1 elements of A. e) If q is even then 1 = 1, and so trivially δ(k, q) = γ(k, q). If A is even then 1 is a k-th power, and so again γ(k, q) = δ(k, q). If A is odd then k must be even (for p ) and A ( A) is the set of k/-th powers. 3. Proof of Theorem 1 Let k (q 1), A be the set of nonzero k-th powers in F q and A 0 = A {0}. Before addressing Theorem 1, which is concerned with representing elements as sums of nonzero k-th powers, we start by reviewing the proof of Cauchy s theorem, γ(k, q) k, which allows for some terms to be zero. For any positive integer n, na 0 = {0} Ax 1 Ax l, for some distinct cosets Ax i of A, 1 i l. If na 0 F q then (n+1)a 0 contains na 0 and, assuming that γ(k, q) exists, must be strictly larger. Therefore, (n + 1)A 0 na 0 + A. By induction we get a Cauchy-Davenport type inequality, na 0 min{q, 1 + n A }, (10) for n 1, and in view of the equality k A = q 1, deduce that γ(k, q) k whenever γ(k, q) exists. To estimate γ (k, q) we have to work a little harder since (n + 1)A doesn t contain na in general, so it is not immediate that it has larger cardinality. However, we are able to recover the following analogue of (10), and Theorem 1 is an immediate consequence. Lemma 10. If A is a multiplicative subgroup of F q containing f linearly independent points over F p and A >, then for any positive integer n, na min{q, n A }. Proof. Let A be a multiplicative subgroup of F q containing f linearly independent points over F p. We first show that if B is any subset of F q such that AB B

8 INTEGERS: 11 (011) 8 then either A + B = F q or A + B A + B 1. This follows from Kneser s inequality (see [19, Theorem 4.1]): A + B A + B stab(a + B), where stab(a + B) = {x F q : A + B + x = A + B}, an additive subgroup of F q, that is, an F p subspace of F q. We need only establish that if stab(a + B) {0} then A+B = F q. Suppose x is a nonzero element of stab(a+b). Then A+B+x = A+B. Since AB B and AA = A we get A+B+Ax A+B. Thus Ax stab(a+b), but Ax contains f linearly independent points over F p. Thus stab(a+b) is of dimension f over F p, and so stab(a + B) = F q. Plainly, we must also have A + B = F q since for any point c A + B, c + stab(a + B) A + B. Now let n be any positive integer and B = na. Then AB B and so either (n+1)a = F q or (n+1)a na + A 1. The proof now follows by induction on n. Suppose na n A for a given n and that (n + 1)A F q. Then (n + 1)A (n + 1) A 1, but (n + 1)A is a union of cosets of A together (possibly) with 0. If A >, this forces (n + 1)A to be a union of at least (n + 1) cosets of A together (possibly) with 0. Thus (n + 1)A (n + 1) A. When A = ; that is, q = p and A = {±1}, then na = n + 1 for n < p. Thus γ (A, q) = p 1 = k. 4. Estimating δ (A, q) and Proof of Theorem 3 Following the method of Glibichuk and Konyagin [15], for any subsets X, Y of F q 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 a generalization of a lemma of Glibichuk and Konyagin [15, Lemma 3.] to finite fields. Lemma 11. Let q = p f, X, Y F q and a 1, a,..., a f F q be a set of f linearly independent points over F p. If X X Y Y F q then for some a i we have XY XY + a i Y a i Y X Y. Proof. Let S = X X Y Y. Assume S F q and that a 1,..., a f are linearly independent values in F q. We claim that for some a i, S +a i S, for otherwise S +k 1 a 1 +k a + + k f a f S for all nonnegative integers k 1,..., k f, implying that S = F q. Say x 1 x y 1 y + a i / S, for some x 1, x X, y 1, y Y. Then the mapping from X Y into XY XY + a i Y a i Y given by (x, y) (y 1 y )x + (x 1 x + a i y 1 a i y )y, is one-to-one and the lemma follows.

9 INTEGERS: 11 (011) 9 Applying the lemma to a multiplicative subgroup A of F q containing a set a 1,..., a f of linearly independent points, we immediately obtain, Lemma 1. Let A be a multiplicative subgroup of F q containing f linearly independent points over F p and X be any subset of F q such that AX X and X X A A F q. Then X X + A A X A. We also need the following elementary result. Lemma 13. Let A be a multiplicative subgroup of F q and X, Y be subsets of F q such that AX X, AY Y. If X X Y Y q A then X X Y Y F q. 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. Thus X X X X ( Y Y 1) Y Y. A Since the right-hand side is less than q by assumption, the result follow. For l N, let n 1 = 1 and n l = 5 4 4l 1 3, for l, so that n = 3, n 3 = 13, n 4 = 53, n 5 = 13 and n l+1 = 4n l + 1, for l. (11) Put A 1 = A and, for l, A l = (n l A n l A) so that, for l, A l 1 A l 1 + A A = A l. (1) Lemma 14. Let A be a multiplicative subgroup of F q containing f linearly independent points over F p. Then for l 1, (a) If A l 1 A l 1 A A < q A then A l A l. (b) In all cases, A l min{ A l, q/ A }. One can compare the above result with Lemma 5. of [15] where it is shown for F p that A l 3 8 min{ A l, p 1 }. Proof. The proof of (a) is by induction on l, the statement being trivial for l = 1. For l > 1, put X = A l 1, Y = A. If A l 1 A l 1 A A < q A then by Lemma 13, X X Y Y F q. Also, by (1) we have X X + A A = A l. Thus by Lemma 1, A l A l 1 A and so by the induction assumption, A l A l. If A l 1 A l 1 A A q A then since A A A we have A l 1 A l 1 q/ A. Since A l = n l A n l A n l 1 A n l 1 A = A l 1 A l 1, we obtain A l A l 1 A l 1 q A / A A q/ A.

10 INTEGERS: 11 (011) 10 Lemma 15. Let A be a multiplicative subgroup of F q containing f linearly independent points over F p. Set l = log(q 1)/ log A. Then δ (A, q) 16n l. Proof. For such l we have l + 1 > log(q 1)/ log A and so A l+1 q. Thus, by Lemma 14 (b), A l A min{ A l+1, q} = q. Since ( A, q) = 1 we must in fact have A l A > q. Since A is symmetric ( 1 A) or antisymmetric ( 1 / A), it follows from Lemma 7 (c) that 8A l A = F q, that is, 8n l A 8n l A = F q. Proof of Theorem 3. With l as in Lemma 15 we have using k A = (q 1), log(q 1) l = 1 + log k log A log A. Thus by Lemma 15, δ (A, q) 16n l l log k/ log A, thereby finishing the proof. 5. Estimating γ (A, q) and Proof of Theorem Let A be a multiplicative subgroup of F q containing f linearly independent points. We start by obtaining growth estimates for na. If A = 1 then q = p and na = 1 for any n. If A = then q = p, A = ±1 and na = min{p, n + 1}. Next we note that { A 3/ if A A < q A, 4A q 1/ A 3/8 (13), otherwise. Indeed, by (7) and Lemma 14 (a) we have 4A A 1/ 3A 3A 1/ = A 1/ A 1/ A 3/, if A A < q A. Otherwise, A A (q A ) 1/. In particular, A > (q A ) 1/, and so A q 1/3. Thus, by (8), 4A A A 15/16 (q A ) 15/3 q 15/3 A 3/3 A 1/3 q 1/ A 3/8. For l N set m l = l l 3 + 9, so that m 1 = 1, m = 4, m 3 = 17, m 4 = 70, m 5 = 83 and m l = m l 1 + n l for l, with n l as defined in the previous section. Lemma 16. Let A be a multiplicative subgroup of F q containing f linearly independent points over F p. Then for l 1 we have m l A max A l 1+ 1 { ( A A A l 1, if l = 1, or l and A l 1 A l 1 A A < q A, ) } 3/4 q 3/4, A 1/4 q 1/, otherwise. (14)

11 INTEGERS: 11 (011) 11 Proof. The result is trivial when l = 1. Assume that the theorem holds for l 1 with l. Suppose that A l 1 A l 1 A A < q A. In particular, if l 3 then A l A l A A < q A. Then using m l = n l + m l 1, inequality (7), the induction assumption and Lemma 14, we have m l A m l 1 A 1/ n l A n l A 1/ A (l + 1 l ) 1 A l/ = A l 1+ 1 l 1. Suppose next that A l 1 A l 1 A A q A. (15) Then, since m l > n l > 4n l 1, we have by inequality (8) that m l A (n l 1 A) n l 1 A n l 1 A 3/4 = A l 1 A l 1 3/4 ( ) 3/4 q A. A A To prove the second inequality, m l A A 1/4 q 1/, under the assumption of (15), more work is required. A stronger result was established (13) for l =, so we assume l 3. First observe that by Lemma 1, with X = A A, if A A A A < q A, (so that by Lemma 13, (X X)/(A A) F q ), then and so by (15), 5A 5A = (A A)A (A A)A + A A A A A, 5A 5A n l 1 A n l 1 A = 5A 5A A l 1 A l 1 A A A A l 1 A l 1 q A. Since n l 1 > 5 for l 3, it follows that n l 1 A n l 1 A > 5A 5A 1/ n l 1 A n l 1 A 1/ q 1/ A. (16) Also, since for any set B, B B B, using (15), and so n l 1 A A n l 1 A n l 1 A A A = A l 1 A l 1 A A q A, n l 1 A A q. (17) Thus since m l > n l > 4n l 1 we obtain from (7), (16) and (17), m l A 4n l 1 A n l 1 A 1/ n l 1 A n l 1 A 1/ n l 1 A 1/ q 1/4 A 1/ = ( n l 1 A 1/ A 1/4 )q 1/4 A 1/4 q 1/4 q 1/4 A 1/4 = q 1/ A 1/4. There remains the case A A A A q A. In this case A A q 1/ A 1/ and A A q. The latter implies A (q/ A ) 1/. Thus, by (7), 4A A 1/ A A 1/ q 1/4 A 1/4 q 1/4 A 1/4 = q 1/, and m l A 6A 4A 1/ A A 1/ q 1/4 q 1/4 A 1/4 = q 1/ A 1/4, which finishes the proof.

12 INTEGERS: 11 (011) 1 Lemma 17. Let A be a multiplicative subgroup of F q containing f linearly independent points over F p. Then for l 1 we have m l A min { A l 1+ 1 l 1, } q. (18) Proof. If l = 1 or A = 1 the statement is trivial. For A =, m l A min{m l + 1, q} = min{ l l , q} min{l, q}. For A 4 the result follows from Lemma 16, since q 1/ A 1/4 q in this case. For A = 3 and l = the result follows from (13) in a similar manner. Suppose that A = 3 and l 3. Then A = {1, α, α } = {1, α, 1 α}, where α is a primitive cube root of 1, and A A = {0, ±(1 α), ±(+α), ±(α+1)}, whence A A = 7. Then, by Lemma 16, m l A q 3/4 A 3/4 A A 3/4 (3/7) 3/4 q 3/4 q, provided that q 4(7/3) 3 = , and so the result follows from Lemma 16 when q 51. We are left with testing the prime powers less than 50. If q is a prime then we can use the Cauchy-Davenport inequality to get m k A 17A min{q, 35} q. The prime powers remaining with 3 q 1 are 4,16,5,49. We can t have q = 16 or 49 since 3 ( 1) and 3 (7 1), implying that A does not contain a set of f linearly independent points. For q = 4, A = F 4, A = F 4 and trivially A q. For q = 5 one can check that 3A = Applying Lemma 7 (b) with A = B = m l A we immediately obtain, Lemma 18. Let A be a multiplicative subgroup of F q containing f linearly independent points over F p. If A (q) 1 (l 1+ 1 l 1 ) 1, then 8m l A = F q. In particular, 8A = F q for A > q 1/, 18A = F q for A > 1.6 q 1/3, 31A = F q for A > 1.17 q /9, 3900A = F q for A > 1.1 q 4/5, 64071A = F q for A > 1.09 q 8/65.

13 INTEGERS: 11 (011) 13 In comparison, for F p Cochrane and Pinner [9] obtained 8A = F p for A > p 1/, 3A = F p for A > 3.91 p 1/3, 39A = F p for A >.78 p 1/4, 888A = F p for A > 3.19 p 1/5, 1800A = F p for A >.8 p 1/6, 56448A = F p for A >.43 p 1/7, 8488A = F p for A > 1.91 p 1/8. We cannot do quite as well for F q because we do not have a lower bound on A. For F p it was shown in [9, Theorem 5.] that A min{ 1 4 A 3/, p }. We do not know if such a bound holds in F q. Proof of Theorem. An integer l satisfies the hypothesis of Lemma 18 provided that l + 1 log q + 1. (19) l 1 log A We claim that the value log((q 1)) l = log A + 1, suffices. To see this, first observe that for this choice of l, l < log (4(q 1)), that is, q 1 > l, and so using the inequality we have log(q) log((q 1)) = log(q/(q 1)) < q q 1 1 = 1 q 1, log(q) log((q 1)) 1 < log A log A (q 1) log A < 1 l 1 log A < 1 l 1. Thus (19) is satisfied, and so by Lemma 18, γ (A, q) 8m l. Since m l l, we have γ (A, q) 8(5/18) log (q 1) 4 +1 log (q 1) log A < log A log 4/ log A = ((q 1)) = ( A k) log 4 log A 633(k) log 4 log A, completing the proof of Theorem. We cannot do quite as well for F q as for F p because we do not have a good lower bound on A. For F p we were able to use the bound [9, Theorem 5.], A min{ 1 4 A 3/, p }. We do not know if such a bound holds in F q.

14 INTEGERS: 11 (011) Proof of Theorem 4 For small A we use the method of Bovey [4] to bound δ(a, q) and γ(a, q). We start by generalizing [4, Lemma 3]. For any n-tuple u = (u 1,..., u n ) R n let u 1 = n i=1 u i. Lemma 19. Let F q be any finite field and u 1, u,..., u n F q. Let T : Z n F q be the linear function T (x 1,..., x n ) = n i=1 x iu i. Suppose that v 1,..., v n Z n are linearly independent vectors over R with T (v i ) = 0, 1 i n. Then for any value a in the range of T there exists a vector u Z n with T (u) = a and u 1 1 n i=1 v i 1. Proof. Let w Z n with T (w) = a. Write w = n i=1 y iv i for some y i R, 1 i n. Say y i = x i + ɛ i for some x i Z and ɛ i R with ɛ i 1/, 1 i n. Put u = n i=1 ɛ iv i = w n i=1 x iv i. Then u Z n, T (u) = a and u 1 1 n i=1 v i 1. Proof of Theorem 4. By Theorem 9 (d), γ(k, q) (t 1)δ(k, q), where t = A, and so it suffices to prove the statement of Theorem 4 for δ(k, q). Put r = φ(t). Let R be a primitive t-th root of unity in F q, 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 and the set of nonzero k-th powers in F q is just {1, R, R,..., R t 1 }. Let f : Z r Z[ω] be given by f(x 1, x,..., x r ) = x 1 + x ω + + x r ω r 1. Then f is a one-to-one Z-module homomorphism. Let T : Z r F q be the linear map T (x 1,..., x r ) = r i=1 x ir i 1 and L be the lattice of points satisfying T (x 1,..., x r ) = 0. Since the set of k-th powers 1, R,..., R r 1 spans all of F q we have V ol(l) = q. Thus by Minkowski s fundamental theorem there is a nonzero vector v 1 = (a 1, a,..., a r ) in L with a i q 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 points in L and so, by Lemma 19, for any a F q there is an r-tuple of integers u = (u 1,..., u r ) such that u 1 + u R + u 3 R + + u r R r 1 = a, and r i=1 u i 1 r i=1 v i 1. Thus δ(k, q) 1 r i=1 v i 1. Now plainly v i 1 t q 1/r, (indeed, as shown in [4], v i 1 r(a(t) + 1) r p n/r, where A(t) is the maximal absolute value of the coefficients of Φ t (x)). Thus δ(k, q) t q 1/r. 7. Proof of Lemma 7(a) Let m 3 be a positive integer, A, B F q, A = A {0}, a F q and N denote the number of m-tuples (x 1,..., x m, y 1,... y m ) with x 1, x A, x 3,..., x m A,

15 INTEGERS: 11 (011) 15 y i B, 1 i m, and x 1 y x m y m = a. Let ψ denote the additive character on F q, ψ(z) = e πit r(z)/p. Then qn = A A m B m + ψ(λ(x 1 y x m y m a)) λ 0 x 1,x A x 3,..,x m A y i B = A A m B m + Error, (0) with Error = ψ( λa) ψ(λxy) λ 0 x A y B m ψ(λxy). x A y B Now, by the Cauchy-Schwarz inequality, ψ(λxy) x A y B ψ(λxy) B 1/ ψ(λxy) y B x A y B x A B 1/ ψ(λxy) 1/ = B 1/ (q A ) 1/. y Fq x A Also, ψ(λ(xy)) λ F q x A y B = x 1,x A y 1,y B λ F q ψ(λ(x 1 y 1 x y )) 1/ = q {(x 1, x, y 1, y ) : x 1, x A, y 1, y B, x 1 y 1 = x y } q A B, the latter following since 0 A, so x 1 y 1 = x y can be written y 1 = x 1 1 x y. Thus, Error A m B m q m ψ(λ(xy)) λ 0 x A y B = A m B m q m ψ(λ(xy)) λ F q x A y B ( A m B m q m q A B A B ) ( = A m 1 A B m q m 1 B q ), A B and we see that the main term in (0) exceeds the error term provided that ( ) A m 1 B m m > q 1 B q.

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