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Waring's problem (previously asked about here) asks, for each integer $k \ge 2$, what is the smallest integer $g(k)$ such that any positive integer can be written as a sum of $g(k)$ kth powers. We have $g(2) = 4$, $g(3) = 9$, etc. A (harder) variant asks what the smallest integer $G(k)$ is such that all sufficiently large integers can be written as a sum of $G(k)$ kth powers.

I have two related questions:

  1. What is known if we relax the condition ``any positive integer'' and only require a positive-density subset? More precisely, we look for the smallest $g'(k)$ for which there is some $S \subset \mathbb{Z}_{>0}$ of positive density such that any $x \in S$ can be written as $g'(k)$ $k$th powers. Then we have $g'(2) = 3$, while $G(2) = 4$; and $g'(3) = 4$, while it is only known that $4 \le G(3) \le 7$. Is anything known about $g'(k)$ for k = 4,5, or larger?

  2. For fixed k, is there an efficient algorithm that, given n, writes n as a sum of $g(k)$ kth powers? What about decomposing n into the minimal number of kth powers for that n? (Here `efficient' means polynomial in log(n).)

Edit: Wikipedia says that ``In the absence of congruence restrictions, a density argument suggests that G(k) should equal k + 1.'' So perhaps this is the answer to (1)?

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    $\begingroup$ If I understand correctly (never guaranteed), then g'(2) = 3. See LeVeque, Topics in Number Theory, Volume II, Section 7-5, The integers representable as the sum of two squares. The count up to some x is asymptotically constant * x / sqrt( log x). Available Dover reprint. $\endgroup$
    – Will Jagy
    Commented Jan 22, 2010 at 1:39
  • $\begingroup$ Yes, you are right. I was recalling the result that any prime congruent to 1 mod 4 is the sum of two squares. But the primes aren't dense in Z. :) Post edited. Actually for my application densities like c*x/poly(log x) would be sufficient...but no need to complicate things. $\endgroup$
    – David Mandell Freeman
    Commented Jan 22, 2010 at 1:56
  • $\begingroup$ I'm starting to understand the question. How do you know g'(3) = 4 ? $\endgroup$
    – Will Jagy
    Commented Jan 22, 2010 at 2:28
  • $\begingroup$ g'(2) = 3 since a positive-density subset of the integers can be written as a sum of three squares. Specifically, all the integers which are not of the form $4^m (8k+7)$ can be written this way; that set has density 5/6. @Will: cx/sqrt(log x) is correct, and not enough; we need a constant times x. $\endgroup$
    – Michael Lugo
    Commented Jan 22, 2010 at 3:09
  • $\begingroup$ The result g'(3) = 4 is the main result of Davenport, H. "On Waring's Problem for Cubes." Acta Math. 71, 123-143, 1939 $\endgroup$
    – David Mandell Freeman
    Commented Jan 22, 2010 at 7:57

1 Answer 1

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Regarding question 1, it is easy to prove that $g'(k) \ge k$, and the expectation is that $g'(k)=k$ for all $k \ge 3$. But the equality is an open question even for $k=3$. See Deshouillers, Jean-Marc; Hennecart, François; Landreau, Bernard; Sums of powers: an arithmetic refinement to the probabilistic model of Erdös and Rényi. Acta Arith. 85 (1998), no. 1, 13-33, and other papers by these authors.

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