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WARING'S PROBLEM WITH POLYNOMIAL SUMMANDS

Published online by Cambridge University Press:  01 June 2000

KEVIN FORD
Affiliation:
Department of Mathematics, University of South Carolina, Columbia, SC 29208, USA; [email protected]
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Abstract

Let f(x) be an integer-valued polynomial with no fixed integer divisor [ges ] 2, that is, for no integer d [ges ] 2 does d[mid ]f(x) for all integers x. One generalization of the famous Waring problem is to determine whether, for large enough s, the equation

formula here

is solvable in positive integers x1, …, xs for sufficiently large integers n. The existence of such s for every f was established by Kamke [5] in 1921. Subsequent authors (Pillai, Hua [2–4], Vinogradov, Načaev [7], and others) have studied the problem of bounding G(f), the least s for which (1.1) is solvable for all large n.

Questions of local solubility of (1.1), that is, solubility of the congruence

formula here

play a more important and complicated role in this problem than in the classical Waring problem. Let Γ0(f) denote the least number s so that (1.2) is solvable for every pair n, q. It is well known that Γ0(xk) [les ] 4k for every k, but Hua [4] found that for every k, the polynomial

formula here

satisfies Γ0(fk) [ges ] 2k − 1 (take s = 2k−2, q = 2k and n = (−1)k in (1.2)). Clearly G(f) [ges ] Γ0(f), but one can say more by restricting the values of n under consideration, as has been done by several authors in the case f(x) = x4 (for example [1, 6]).

The singular series

formula here

where e(z) = eiz, encapsulates the local solubility information. In particular, [Sfr ]s, f(n) [ges ] 0 for every n and [Sfr ]s, f(n) > 0 if and only if (1.2) is soluble for every q. Define G(f) to be the least number s so that for every δ >0 and every n>n0(δ) with [Sfr ]s, f(n) [ges ] δ, (1.1) is soluble. The reason for taking [Sfr ]s, f(n) [ges ] δ instead of [Sfr ]s, f(n) > 0 is that we wish to exclude from consideration certain n lying in sparse sequences for which (1.1) is insoluble but [Sfr ]s, f(n) > 0. For example, taking f(x) = x4, s = 15 and nj = 79·16j (j = 0, 1, …), it can be shown that (1.1) is not soluble for n = nj, that [Sfr ]s, f(nj) > 0 for all j, and that [Sfr ]s, f(nj) → 0 as j → ∞. It is known that G(x4) = 16 (see [1]) and that G(x4) [ges ] 11 almost holds (see [6]).

Type
Research Article
Copyright
The London Mathematical Society 2000

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