The Density of Numbers Represented by Diagonal Forms of Large Degree

# The Density of Numbers Represented by Diagonal Forms of Large Degree

Brandon Hanson Pennsylvania State University
University Park, PA
and  Asif Zaman University of Toronto
Toronto, ON
###### Abstract.

Let be a fixed positive integer and be arbitrary. We show that, on average over , the density of numbers represented by the degree diagonal form

 a1xk1+⋯+asxks

decays rapidly with respect to .

## 1. Introduction

The classical version of Waring’s problem asks whether every positive integer can be written as a sum of at most positive integers, each of which is a ’th power. In other words, is there an integer (which depends on ) such that for each we have a solution to the equation

 (1) n=xk1+⋯+xks

in non-negative integers ? The least value of which is admissible is usually referred to as , and Waring’s problem is thus the assertion that for any . Waring’s problem has a long history; for a nice exposition see [8].

The “easier” version of Waring’s problem, a name attributed to Wright [10], asks whether there is a solution to the equation

 (2) n=xk1±⋯±xks.

The least for which this equation is soluble for each is usually referred to as , and establishing that is a fairly simple argument, which can be found in [6]. Clearly any upper bound for in the usual Waring problem extends to a bound for as well. However, the freedom to use negative summands may make considerably smaller.

One can verify that . Indeed, in order for to be written as a sum of ’th powers, we only have ’s at our disposal. For these reasons, one usually considers instead , which is the least such that (1) is soluble for all sufficiently large. Here, the bound is still quite simple. To represent each in the range , the variables can be no larger than . Thus the vector is a lattice point in the box , and there are at most such lattice points. To represent all integers in the desired range, we must therefore have . The introduction of negative summands causes this argument to fail completely, because one is no longer counting lattice points in a bounded region. This motivates the following question, which was asked in [2]:

###### Problem.

For sufficiently large, is it true that the set of integers of the form

 n=xk1±xk2±⋯±xks

has asymptotic density zero?

A result of Wooley (see [2] and [9]) asserts that, for , the set of integers of the form

 n=xk1±⋯±xks

has density zero and in fact more is true – one can obtain fairly good decay rates in the proportion of integers up to which can be represented. However, Wooley’s result is conditional on a generalized version of the -conjecture and, as far as the authors are aware, there seems to be little known unconditionally for large values of , say . We prove a result in this direction which is much weaker, but unconditional. We will not be able to prove that the set of integers represented has zero density, but we will establish bounds on the asymptotic density of these integers. These bounds will, on average, decay quite rapidly with respect to .

In fact, we will establish something a bit more general in that we will allow for arbitrary integer coefficients, not just ’s and ’s. Let be fixed and let be arbitrary. We consider the form

 (3) F{a},k({x})=a1xk1+⋯+asxks

and the set

 R({a},k)={n:n=F{a},k({x}) for % some {x}∈Zs}

of numbers which this form represents.

We shall estimate the average asymptotic (upper)-density

 (4) δk=limsupN→∞|R({a},k)∩[1,N]|N

as a function of . This number implicitly depends on a, but the results we shall prove about are uniform over a. For , the following theorem establishes that the value of is large on average.

###### Theorem 1.

Let be fixed and be arbitrary. Let be sufficiently large depending at most on and define as in (4). Then

 (5) 1X∑1≤k

We will use the convention that when . Thus, we expect that the quantity on the lefthand side in (5) is infinite for all sufficiently large depending only on and a. Perhaps it is instructive to compare Theorem 1 to a conditional result. Let denote the number of primes satisfying and let denote Euler’s totient function.

###### Proposition 2.

Let be fixed and be arbitrary. Let be sufficiently large depending at most on and define as in (4). If

 (6) π(X;k,1)=1φ(k)Li(X)+Oε(X1/2+εk1/2)

for any and then

 (7) log(1/δk)≫k1s−1logk.

Assumption (6) is one of the strongest widely-believed conjectures regarding the distribution of primes in arithmetic progressions and Theorem 1 unconditionally obtains the corresponding average result for . Note that the special case is addressed by classical work of Mahler [5] which implies for . For further details on the case , see for example [1] and [7].

## 2. Local densities and a conditional result

The method of proof for Theorem 1 and Proposition 2 is to bound the density by considering local constraints. For instance, a very simple first observation is that for prime

 δp−1≤2sp.

This is just the trivial observation that the set of powers modulo consist of the residue classes and modulo . Thus there are at most admissible values of the diagonal form modulo . Our aim is then to improve this estimate for a given , and subsequently obtain good density estimates for the average exponent . To this end, define

 δk(p)=1p∣∣{F{a},k({z})modp:{z}∈Fsp}∣∣.

The Chinese Remainder Theorem then gives:

###### Lemma 3.

For any integer ,

 δk≤∏pδk(p).

We will combine this with a simple development of the idea we used to bound . Let denote the greatest common divisor of two integers and .

###### Lemma 4.

Let and let be a prime. Then

 δk(p)≤αk,p

where

 αk,p=1p(p−1(k,p−1)+1)s.
###### Proof.

By the structure theorem for cyclic groups, the set non-zero powers modulo forms a subgroup of the unit group of size . Adding in the class modulo , there are values of modulo . Thus, the proportion of admissible residue classes modulo is at most . ∎

Using these two lemmas, we can establish Proposition 2.

###### Proof of Proposition 2.

Suppose is a prime satisfying . With defined as in Lemma 4, observe that if and only if

 p−1k+1

Since , we have that whenever

 p<14k1/(1−1/s)=14k1+1/(s−1).

Let be sufficiently large and set . Thus, by Lemmas 3 and 4,

 log(1/δk)≥∑plog(1/δk(p)) ≥∑p≡1 (mod k)p

Since , the above is

 ≥∑p≡1 (mod k)p

after fixing to be sufficiently large, depending at most on . Then, using assumption (6) to bound the sum on the right in the above inequality, it follows that

 log(1/δk)≫k1+1/(s−1)φ(k)logk≫k1/(s−1)logk,

after bounding trivially by . This proves the proposition. ∎

## 3. Global density is small on average

This section is dedicated to proving Theorem 1 for which we require one additional lemma. For integers and , let

 (8) ψ(X;q,a)=∑n

where equals if is a power of a prime and equals otherwise.

###### Lemma 5.

Let be arbitrary. For ,

 (9) ∑1≤m

 (10) ∑1≤m
###### Proof.

We divide the sum in (9) dyadically. For , note . Hence, by the Bombieri-Vinogradov theorem [3, Theorem 17.1], we have that

 ∑M≤m<2Mψ(mX;m,1) ≥∑M≤m<2Mψ(MX;m,1) =(∑M≤m<2MMXφ(m))+O(MX(logMX)−2) =ζ(2)ζ(3)log2ζ(6)MX+O(XlogM+MX(logMX)−2).

In the last step, we applied the classical fact [4] that, for ,

 ∑n≤x1φ(n)=ζ(2)ζ(3)ζ(6)(logx+γ−∑plogpp2−p+1)+O(x−1logx).

Summing the prior estimate over with and recalling yields desired result. To prove (10), we proceed similarly. For and , we may apply the Brun-Titchmarsh inequality [3, Theorem 6.6] to and deduce that

 ∑M≤m<2M∫mX(m+1)sψ(t;m,1)t(logt)2dt ≪∑M≤m<2M∫mX(m+1)s1φ(m)(logt)2dt ≪∑M≤m<2M1φ(m)∫2MX(M+1)s1(logt)2dt ≪∑M≤m<2MMXφ(m)log2(MX) ≪MXlog2X.

The desired bound follows by dyadically summing this estimate. ∎

###### Proof of Theorem 1.

By Lemma 3, observe that

 ∑1≤k

where

 S1=∑1≤k

It suffices to show . By Lemma 4,

 δk(p)≤min{αk,p,1}

for each prime . Let , so for some integer with . Thus, each pair in the above sum corresponds to a unique triple of positive integers and such that and . Moreover,

 αk,p<1⟺p>(m+1)s.

Collecting these observations, we deduce that

 S1 ≥∑m≥1∑m∣p−1p>(m+1)s∑1≤d≤mXp−1(d,p−1)=1(logp−slog(m+1)).

Whenever , the inner sum over contains . Thus, the above is

 ≥∑m≥1∑p≡1 (mod m)(m+1)s

By positivity, we may restrict the outer sum to for some parameter satisfying

 (11) (Y+1)s

Recalling (8), it follows by partial summation that

 (12) S1 ≥∑1≤m

Set where is a parameter which will be specified. For , we have that

 (13) 1−slog(m+1)log(mX)=1−s1+logXlogm+O(1mlogX).

If , then from (13) we see that

 1−slog(m+1)log(mX)≥ηs+O(X−1/(2s−1))≥ηs+O(1logX).

Otherwise, for , we similarly have that

 1−slog(m+1)log(mX)≥12+O(1logX)≥ηs+O(1logX).

Substituting these bounds into (12) and noting , we deduce that

 S1≥(∑1≤m

Since for , Lemma 5 therefore implies that

 S1≥ζ(2)ζ(3)log2ζ(6)⋅ηs⋅XY+O(XYlogX).

Note the implied constant is independent of and depends only on . Choose where is a fixed sufficiently large constant depending only on . Thus, satisfies (11) and it follows that . ∎

## References

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