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The easier Waring problem asks for the least number $v=v(k)$ such that every every integer is a sum of $v$ $k$'th powers with signs, i.e. every $n\in \mathbb{N}$ is of the form $$n=x_1^k\pm x_2^k\pm\dotsb\pm x_v^k.$$

The problem is ``easier'' because unlike the usual Waring problem (without the signs) the existence of $v(k)$ is easy --- the bound $v(k)\leq 2^{k-1}+\tfrac{1}{2}k!$ follows from the repeated differencing. Of course, the upper bounds on the usual Waring problem apply, and so fact $v(k)=O(k\log k)$.

All the lower bounds on $v(k)$ I have seen come from the congruence considerations. For example, $v(3)\geq 4$ because we need at least four terms modulo $9$. However, if we discard the congruential obstacles is there a non-trivial lower bound? To put bluntly my question is

Is there $k$ large enough so that the set $\{x_1^k\pm x_2^k\pm x_3^k\pm x_4^k\pm x_5^k\}$has zero density?

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shouldn't k=6 come easily? Set N large , then if x_5 is the largest term (at least N) then at most 32N^4 additional terms can be formed with it? Gerhard "Ask Me About System Design" Paseman, 2011.05.11 –  Gerhard Paseman May 11 '11 at 18:31
    
Upon reflection, it is not clear at all that there is such a k. Gerhard "Needed To Drink Some Coffee" Paseman, 2011.05.11 –  Gerhard Paseman May 11 '11 at 18:43
    
Boris, One would think so, because your set is contained in the difference set S of the set of things which are the sum of at most 5 positive kth powers, and there seems little reason to suspect that $S$ behaves so much unlike a random set with $n^{5/k}$ elements up to $n$. Proving it would be quite a different matter. Maybe some papers of Browning and Heath-Brown are relevant.... –  Ben Green May 11 '11 at 22:49
    
@Ben: I agree that the answer is probably "yes, there is such a $k$". However, I found myself unable to prove it, hence the question. –  Boris Bukh May 11 '11 at 23:12

2 Answers 2

up vote 14 down vote accepted

As far as I am aware, nothing unconditional is known. The difficulty is that one has no obvious constraint on the size of the variables, so that the $x_i$ could be arbitrarily large in terms of $n$ in a solution. The difficulty of ruling out solutions (when congruence conditions do not rule out solubility) is related to proving insolubility of generalised Fermat equations.

However, there is a conditional approach assuming the truth of the generalised ABC Conjecture (or, as Pomerance describes it, the Alphabet Conjecture). Let's stick with the general situation with $v$ summands. Generalised ABC asserts that in any solution of $a_1+\ldots +a_s=0$ in which there are no vanishing subsums, one has $\underset{1\le i\le s}{\max}|a_i|\ll_{s,\epsilon} \left( \prod_{p|a_1\ldots a_s}p\right)^{(s-1)(s-2)/2+\epsilon}.$ There is disagreement about the specific exponent, but this does not matter so much in the conclusion (see a 1986 paper of Brownawell and Masser for a function field version, and I discuss this in a 1994 paper on Quasi-diagonal behaviour). The vanishing subsums in your equation $\pm x_1^k+\ldots +\pm x_v^k=n$ make life easier (treat them separately, and more powerful estimates are possible), so for simplicity suppose that the representations all have no vanishing subsums. Then one obtains $|x_i|^k\ll |nx_1\ldots x_v|^{v(v-1)/2+\epsilon}$. Provided that $k>v^2(v-1)/2$, it follows that $\max |x_i|\ll |n|^{\alpha+\epsilon}$, where $\alpha=v(v-1)/(2k-v^2(v-1))$. OK ... so far so good. What we have shown thus far is that the variables in a representation are bounded by $|n|^{\alpha +\epsilon}$. The total number of variables available to represent the integers $n$ between $N/2$ and $N$ is consequently no larger than a quantity which is $\ll (N^{\alpha+\epsilon})^v$ (there were $v$ variables). Whenever $v(\alpha+\epsilon)<1$, therefore, the set of integers represented must have zero density. If I have not made any computational errors along the way, this leads us to the conclusion that whenever $k>v^2(v-1)$, then the density of integers represented in the easier Waring problem will be zero. (But remember that this is all conditional on the Generalised ABC Conjecture.) For the specific problem with $v=5$, it looks as if $k>100$ conjecturally does the trick (though smaller $k$ should surely also work).

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Trevor, welcome to MO! –  Gerry Myerson May 12 '11 at 11:35

You do get zero density for $x^k - y^k$ with, say, $x,y > 0,$ as $ x^k - y^k = (x-y) (x^{k-1} + \cdots + y^{k-1} ) $ and either $x=y$ or $| x - y| \geq 1,$ so the number of $(x,y)$ pairs with $0 < x^k - y^k \leq n$ is no larger than $n^{2/(k-1)}.$ As soon as $k \geq 4$ we get zero density. Meanwhile, $x^3 - y^3$ also gives zero density, but this relates to the set of numbers up to $n$ that are represented by the positive quadratic form $x^2 + x y + y^2$ and the count is constant times $\frac{n}{\sqrt {\log n}}.$

However, it is now suspected that $\pm x^3 \pm y^3 \pm z^3$ gives full density, that being 7/9. Also, by easy identities, all numbers are the mixed sum of five cubes, so $v(3) \leq 5$

I would switch to asking your question with three variables instead of five, just completely ignore congruences and focus on your zero density, which seems a clever idea to me...

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