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PROCEEDINGS of the AMERICAN MATHEMATICAL SOCIETY Volume 96. Number

1. lanuary

1986

WARING'S PROBLEM FOR FINITE INTERVALS MELVYN B. NATHANSON Abstract. Let f(n, k. s) denote the cardinality of the smallest set A of nonnegative A-th powers such that every integer in [0. n] is a sum of s elements of A. and let ß(k, s) = limsup„JK log/(/í, k, s)/log n. Clearly, ß[k. s) 5= l/s. In this paper it is proved that/(», k, s) < oi1/1,~? nl(k, s). where g(k) is defined as in Waring's problem, and ß(k. s) - l/s as s -» oo.

Lagrange proved that every nonnegative integer is the sum of four squares. Wieferich proved that every nonnegative integer is the sum of nine cubes, and Linnik showed that every sufficiently large integer is the sum of seven cubes. For k > 2, let g(k) (resp. G(k)) denote the smallest number such that every (resp. every sufficiently large) nonnegative integer is the sum of g(k) (resp. G(k)) nonnegative

klb powers. Hilbert proved that g(k) and G(k) exist. In general, if A is a set of nonnegative s not necessarily distinct elements of A. nonnegative integers n, then A is a basis Define the counting function A(x) of

integers, let sA denote the set of all sums of If n e sA for all (resp. all sufficiently large) (resp. asymptotic basis) of order s. the set A by

A(x)=

E

I-

ue/t 1< a « x

\f A is a basis or asymptotic basis of order s, then A(x) > c0xl/s for some c0 > 0 and all x > xQ. Raikov [5] and Stöhr [6] constructed "thin" bases of order s such that A(x) < cxxl/s, and Cassels [1] constructed a basis A = {ak }^=0 of order s such that ak — aks + 0(ks_1) for some a > 0. The sets constructed by Cassels, Raikov, and Stöhr are highly artificial. It is more interesting to prove the existence of thin bases among the classical sequences of additive number theory. Some results are known for A:th powers and primes. Using a probabilistic method, Erdös and Nathanson [3] showed that for every e > 0 there exists a set A of squares such that (i) every nonnegative integer is the sum of four squares in A, (ii) if n ¥= 4'(8/c + 7), then n is the sum of three squares in A; and (iii) A(x) ~ c2x(1/3) + e. They also proved that for e > 0 there exists a set A of numbers of the form p and pq, where p and q are primes, such that (i) every sufficiently large even integer is the sum of two numbers in A, and (ii) A(x) - c3x(1/2) + e. Nathanson [4] proved that if e > 0 and 5 > s0(k), there exists a sequence A of klb powers such that (i) A is a basis of

orders, and (ii) A(x) ~ xOAXi-VO+t, Received by the editors February 11, 1985.

1980 Mathematics Subject Classification.Primary 10J06, 10L05, 10L02; Secondary 10J05. Key words and phrases. Waring's problem, additive bases, sums of k th powers. ©

1986

American

0002-9939/86

15

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M. B. NATHANSON

If A is a finite set of nonnegative integers, let card(^4) denote the cardinality of A. Let [a, b] denote the set of integers m such that a ^ m < b. The set A is a basis of order s for n if [0, n] e sA. Clearly, card(^) > c0nl/s if A is a basis of order í for n. Choi, Erdös, and Nathanson [2] constructed a finite set A of squares such that [0, n] ç 4,4 and card(^4) < (4/log2)«1/3log n. They conjectured that for every

e > 0 there exists a basis A of squares of order 4 for « such that card(^) < c/?(1/4)+ E. Zöllner [7] has recently proved this conjecture. It is not known if the exponent can be reduced from (1/4) + e to 1/4. Much less is understood about sums of /cth powers. Nathanson [4] proved that if 5 > g(k) and n > «(e), there exists a set A of Arth powers such that [0, n] e sA and card(/4) < (2 + e)«1/(/i + 1). For s > s0(k) + 1, the exponent can be reduced to y/(k + y), where y = 1 - (l/s) + e. Let f(n, k, s) denote the cardinality of the smallest set A of kth powers such that

[0,n]QsA.

Define on \ v \ogf(n,k,s) ß(k,s) = hm sup -r—-. n-»oo

o

Clearly, ß(k, s) > l/s. In this paper I prove that ß(k, s) < l/(s - g(k) + k) for all 5 > g(k), and so ß(k, s) — l/s. This answers a question posed in [4]. Lemma 1. Let k > 2 and n > 1. Le? X, p e R with 0 < X < ju < 1. TAev?¿Aereis a sei 5 of positive kth powers such that (i) card(fi)

< 2n^~x, and (ii) if m e [«A,2«M]

there is bk e B such that 0 < m - bk < nx + 2*+1«"d-(i/*)). Proof.

Let [x] denote the integer part of x. Define the set B as follows: B=

Let m e [«\2«'i]. r?i/*wA/*]

{ [ql/knx/k]k\qe[l,2n»-x]}.

Let q = [m/nx]. Then q e [l,2n>l~x] and bk e B, where b -

Qearly, bk < {¿'knx'k >m-nx-

- \)k > qnx - 2k(qnx)(k~1)/k 2k(2n»)1H1/k).

This proves Lemma 1.

Lemma 2. Let 2 < k < t and let i e [1, t - fcj. For n > n0(k, t) there is a set B¡ of positive kth powers such that (i) card(B A < 2n1/'and(ii) if m e [n{k + '~1)/',2n{k + ')/'], there is bk e B( such that 0 < m - bk < 2nik + ,-l)/'.

Proof. Apply Lemma 2 with X = (k + i - l)/t and p = (k + i)/t. Since a - X = \/t and n\ + 2*+i„m. Proof. Let t = s - g(k) + k and nx(k, s) = n0(k, /). Let B0 = [bk\ 0 < b < (2n)1/l) and let 5, be the set defined in Lemma 2 for i = 1,2,...,/ — A:. Let ^ = U'ZoBi. Then [m1/']* e 50 n 5, and so

card(^) < 2(/ - fc + \)nl/' = 2(j - g(ifc) + l)«1/«*-«*)**). Let /0 = [0,2/1*/'] and /, = [« 2 a«i/e > 0. Fors > s(k, e) and n > n2(k, s) there exists a set A of kth powers such that A is a basis of order s for n and card(.4) < n(l/s) + e.

Corollary

2. Let k ^ 2. Then ß(k, s) - l/s as s -» oo.

Proof. The Theorem implies that

l/s