Monatsh Math (2011) 162:385–407 DOI 10.1007/s00605-009-0171-6

The asymptotic formula in Waring’s problem for one square and seventeen fifth powers Jörg Brüdern · Koichi Kawada

Received: 15 August 2009 / Accepted: 3 November 2009 / Published online: 21 November 2009 © Springer-Verlag 2009

Abstract Let Rk,s (n) denote the number of solutions of the equation n = x 2 + y1k + y2k + · · · + ysk in natural numbers x, y1 , . . . , ys . By a straightforward application of the circle method, an asymptotic formula for Rk,s (n) is obtained when k ≥ 3 and s ≥ 2k−1 + 2. When k ≥ 6, work of Heath-Brown and Boklan is applied to establish the asymptotic formula under the milder constraint s ≥ 7 · 2k−4 + 3. Although the principal conclusions provided by Heath-Brown and Boklan are not available for smaller values of k, some of the underlying ideas are still applicable for k = 5, and the main objective of this article is to establish an asymptotic formula for R5,17 (n) by this strategy. Keywords Waring’s problem · Asymptotic formula · Sums of one square and powers Mathematics Subject Classification (2000)

11P05 · 11P55

1 Introduction In this paper we shall be concerned with the diophantine equation n = x 2 + y1k + y2k + · · · + ysk

(1)

Communicated by J. Schoißengeier. J. Brüdern Institut für Algebra und Zahlentheorie, Universität Stuttgart, 70511 Stuttgart, Germany K. Kawada (B) Department of Mathematics, Faculty of Education, Iwate University, Morioka 020-8550, Japan e-mail: [email protected]

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where s and k are natural numbers, k ≥ 3. This family of equations belongs to the small stock of variants of Waring’s problem that have been studied by various writers since the early days of the Hardy–Littlewood method. A purely heuristical application of that method, based on a major arc analysis only, suggests that the number Rk,s (n) of solutions to (1) in natural numbers x, y1 , . . . , ys satisfies the asymptotic relation Rk,s (n) =

( 23 )(1 + k1 )s ( 21

+

s k)

s

1

Sk,s (n)n k − 2 (1 + o(1))

(2)

as n tends to infinity, provided only that s > 21 k. Here the singular series is defined by Sk,s (n) =

∞ 

q −s−1

q=1

⎛ ⎞   2   k s  q q q   ax ⎝ ay ⎠ an . e e e − q q q

a=1 x=1 (a,q)=1

(3)

y=1

A first analysis of the problem was made by Stanley [11] in 1930. Following the pattern laid down by Hardy and Littlewood [4,5] in their classic series Partitio Numerorum, she established the asymptotic formula (2) for s ≥ s1 (k) where  s1 (3) = 7, s1 (4) = 14, s1 (5) = 28, s1 (k) =

2k−2

 1 k − 1 + O(k) (k > 5). 2 (4)

Later, Sinnadurai [10] verified (2) for R3,6 (n), and Hooley [9] gave a different proof for this result. When k ≥ 4, however, the authors are not aware of any improvements of Stanley’s bounds (4) recorded in the literature. Yet, since the 1920s, the theory of Waring’s problem has experienced waves of innovation, resulting in significantly smaller lower bounds for s for which (2) can be demonstrated. The current state of the art is that when k ≥ 3 and s ≥ 2k−1 + 2, then for any ε > 0, one has Rk,s (n) =

( 23 )(1 + k1 )s ( 21 + ks )

s 1 1 1−k

s 1 Sk,s (n)n k − 2 + O n k − 2 − k 2 +ε .

(5)

Moreover, there exists a function s0 (k) satisfying s0 (k) ≤

1 2 k log k + O(k 2 log log k), 2

and such that (5) holds whenever s ≥ s0 (k). Certainly well-known to experts in the field, we shall not dwell on a proof of (5) for long, but give a brief sketch in Sect. 3 below. Note that when k = 3 we may take s = 6, so the formula for R3,6 (n) that was already established by Sinnadurai and Hooley, is included in (5). A mild benefit is the explicit exponent in the error term in (5). Connaisseurs will notice that the saving in (5) over the main term is exactly the saving provided by Weyl’s inequality. In such an

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instance, the method only just fails with one kth power removed from the representation problem. Thus, one would hope to handle the case s = 2k−1 + 1 by a further refinement of the basic method. When k = 3, this is a much wanted result. Vaughan [12] obtained R3,5 (n)  n 7/6 , but the corresponding asymptotic formula remains a desideratum. Likewise, when k = 4, we have not been able to establish (1) for s = 9. Leaving aside the case k = 5 for a while, work of Heath-Brown [6] and Boklan [1] suggests that further improvements for k ≥ 6 can be made, and this is indeed the case. Theorem 1 Let k ≥ 6 and s ≥ 7 · 2k−4 + 3. Then, one has Rk,s (n) =

( 23 )(1 + k1 )s ( 21

+

s k)

s

1

Sk,s (n)n k − 2 (1 + O((log n)−1 )).

The proof of this result depends crucially on Boklan’s minor arc estimate, improving Heath-Brown’s pivotal argument that led to a new form of Hua’s lemma when k ≥ 6. Although the principal conclusions of Heath-Brown [6] and Boklan [1] are not available for smaller values of k, some of the internal mechanics still apply when k = 5. The main objective of the current communication is to show how this observation can be made precise for the problem at hand. Thus, when k = 5, we are able to establish the same conclusion as in Theorem 1, with an improved error term. Theorem 2 There is a positive number δ such that R5,17 (n) =

( 23 )( 65 )17 39 ( 10 )

S5,17 (n)n 29/10 + O(n 29/10−δ ).

The proofs of Theorems 1 and 2 depart from a use of Weyl’s inequality for the square and an application of Hua’s lemma for the kth powers. For this initial step to be successful, one has to use a Farey dissection in which the major arcs are so wide and numerous that they deny a standard√treatment by Poisson summation. To strengthen our rhetoric, we choose 1 ≤ Q ≤ 21 n, and denote by M(Q) the union of the pairwise disjoint intervals |qα − a| ≤ Q/n with 1 ≤ a ≤ q ≤ Q and (a, q) = 1. Then, our major arcs will be M(Q) with Q of approximate size n 2/k or larger. For comparison, classically major arcs would be defined with Q = n 1/k . Hence, we are forced to prune substantially, and we shall apply two different strategies to achieve this. When k ≥ 6, Boklan’s estimates will be coupled with a divisor sum technique that derives from work of Wolke [15]. The latter procedure might be applicable elsewhere, and we therefore highlight the outcome of our analysis as Lemmata 5 and 6 below. Alternatively, one may apply some features of Heath-Brown’s major arc treatment, and we follow this path when k = 5. Ultimately, the proof of Theorem 2 then depends on the following mean value theorem that might be of some independent interest. Theorem 3 Let P = [n 1/5 ]. There exists δ > 0 such that one has  M(P 2 )\M(P)

30  5 e(αx ) dα  P 25−δ . x≤P

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This estimate should be compared with Theorem 2 of Heath-Brown [6] and the work of Boklan [1]. Both authors estimate the 78 2k th moment of a kth power Weyl sum when k ≥ 6. Heath-Brown obtains an essentially best possible upper bound, and Boklan was able to use ideas of Vaughan [13] to make extra savings while integrating over suitable minor arcs only. Note that 28 = 78 25 , but for k = 5 the methods of Heath-Brown and Boklan seem to fail when integrating over a unit interval. Yet, within the set M(P 2 ), most of the ideas are still applicable, and we shall prove Theorem 3 by a suitable adjustment of Heath-Brown’s argument. One would hope that one could prove the cognate upper bound  M(P 2 )

28  5 23+ε e(αx ) dα  P x≤P

(6)

along the same lines. This would constitute a cleaner analogue of Heath-Brown’s Theorem 2 when k = 5, but for applications our Theorem 3 performs almost as well as (6) would do. We briefly compare our results with the situation in Waring’s problem. The asymptotic formula for the number of representations by sums of s kth powers of natural numbers has been verified for s ≥ s2 (k) where, in particular, s2 (5) = 32 (Vaughan [13]), s2 (k) = 78 2k for k ≥ 6 (Boklan [1]) and s2 (k) = k 2 log k(1 + o(1)) (Ford [3]). Thus, with the presence of an extra square, we can nearly halve the number of kth powers, at least when k is large. We shall see in Sect. 3 why this is possible. Also, for smaller k, our improvements are proportional to those of Heath-Brown, in that the number of kth powers involved gets reduced, approximately, by a factor 78 . Notation We use the abbreviation e(α) = exp(2πiα), and apply the “ε-convention”: whenever the letter ε appears in a statement, it is asserted that the statement is true for any positive real number ε. Implicit constants in Landau or Vinogradov symbols may depend on the actual value of ε. 2 Warming up Let n be a natural number, and write Pl = n 1/l . The Weyl sums fl (α) =



e(αx l )

x≤Pl

occur in the integral representation 1 Rk,s (n) =

f 2 (α) f k (α)s e(−αn)dα, 0

123

(7)

The asymptotic formula in Waring’s problem

389

which in turn is an immediate consequence of orthogonality. We split this integral into major and minor arcs. For the major arcs, we choose Mk = M(Pk ). Their contribution to the integral in (7) is described in Lemma 1, the proof of which is standard. Lemma 1 Let k ≥ 3 and s ≥ 2k. Then, the singular series Sk,s (n) as defined by (3) converges absolutely, and one has 1  Sk,s (n)  1. Moreover, one has  f 2 (α) f k (α)s e(−αn)dα =

( 23 )(1 + k1 )s

Mk

( 21

+

s k)

s−1 1

s 1 Sk,s (n)n k − 2 + O n k − 2 .

Proof By Theorem 4.2 of Vaughan [14], the series Sk,s (n) converges absolutely and uniformly with respect to n. This establishes the upper bound for Sk,s (n), and for the lower bound, one may apply traditional methods as described in Sections 2.6 and 4.5 of Vaughan [14]. We omit the routine details. The major arc integral can be evaluated by closely following the proof of Theorem 4.4 of Vaughan [14], so that we may again spare the reader the details. Lemma 2 Let k ≥ 3. Then 1 | f 2 (α) f k (α)2

k−1 +1

1

|dα  n k (2

k−1 +1)− 1 +ε 2

.

0

We sketch a proof of Lemma 2 in the next section. It is now straightforward to establish a minor arcs estimate that complements the conclusion of Lemma 1. When Q ≥ 1, let m(Q) denote the complement of M(Q) in [1/Q, 1 + 1/Q], and in particular, write mk = m(Pk ). Then, the estimate provided by Lemma 2 may be combined with Weyl’s inequality [14, Lemma 2.4], to infer that whenever s ≥ 2k−1 + 2, then 

s

1

1 1−k +ε

| f 2 (α) f k (α)s |dα  n k − 2 − k 2

.

mk

By (7) and Lemma 1, the conclusion described in (5) now follows. Likewise, we may verify the clause following (5), concerning large values of k. Instead of invoking Lemma 2, we simply appeal to the following estimate. Lemma 3 There exists a number c such that whenever k ≥ 3 and t ≥ 21 k 2 log k + ck 2 log log k, then 1

t

1

| f 2 (α) f k (α)t |dα  n k − 2 +ε .

0

Proof The proof depends heavily on the work of Ford [3]. By Theorem 1 of that paper, one readily confirms that

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1

1

2

| f k (α)|8k dα  n 8k− 2 ,

(8)

0

and we then deduce the estimate 1

2

| f 2 (α)2 f k (α)8k |dα  n 8k+ε .

(9)

0

To see this, note that the left-hand side equals the number of solutions of k k k x12 − x22 = y1k + · · · + y4k 2 − y4k 2 +1 − · · · − y8k 2

(10)

subject to x j ≤ P2 , y j ≤ Pk . First choose y1 , . . . , y8k 2 such that the right-hand side of (10) is non-zero. There are O(n 8k ) choices for the y j , and then by x12 − x22 = (x1 − x2 )(x1 + x2 ) and a standard divisor estimate, the Eq. (10) leaves only O(n ε ) choices for x1 , x2 . Next choose y1 , . . . , y8k 2 such that the right-hand side of (10) is 1 zero. By (8) and orthogonality, there are O(n 8k− 2 ) choices for the y j , and (10) implies 1 x1 = x2 in this case, leaving O(n 2 ) choices for x1 . This proves (9). Another reference to Theorem 1 of Ford [3] in conjunction with Theorem 5.1 of Vaughan [14] also provides a constant c1 such that whenever h ≥ k 2 log k + c1 k 2 log log k one has 1

h

| f k (α)|h dα  n k −1 .

(11)

0

Now use (9), (11) and Schwarz’s inequality to complete the proof of Lemma 3.

3 The easy goal Before we begin our approach to Lemma 2, we introduce some notation for frequent use later, and √ recall some standard √ estimates. In the interest of brevity, we now write N = M( 21 n) and n = m( 21 n). Then, for α ∈ N there is a unique triple q, a, β √ √ with 1 ≤ a ≤ q ≤ 21 n, (a, q) = 1 and q|β| ≤ 1/(2 n) such that α = a/q + β, and we may define functions fl∗ and l on N by l = fl − fl∗ and

fl∗ (α)

=q

−1

Pl Sl (q, a)

e(βt l )dt 0

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The asymptotic formula in Waring’s problem

391

where  l q  ax . Sl (q, a) = e q

(12)

x=1

Then, by Theorems 4.1 and 4.2 of Vaughan [14] and a partial integration, whenever α ∈ N, one has 1

l (α)  q ε (q + nq|β|) 2 ,

1

fl∗ (α)  Pl (q + nq|β|)− l .

(13)

In particular, we see that for α ∈ N and k ≥ 3, one has the bounds 1

2 (α)  P22

+ε

f 2∗ (α)k (α)  P21+ε .

,

(14)

We also require Weyl’s inequality in the form 1

sup | f 2 (α)|  P22

+ε

α∈n

,

(15)

which is immediate from Lemma 2.4 of Vaughan [14], as well as a special case of Hua’s Lemma [14, Lemma 2.5] which asserts that 1 | f k (α)|2

k−1

dα  Pk2

k−1 −k+1+ε

.

(16)

0

The analogous bound 

| f k∗ (α)|2

k−1

dα  Pk2

k−1 −k+1

(17)

N

follows from the second inequality in (13), by straightforward estimations (in fact, without much effort, one may improve (17) by a factor Pk , but we do not need this here). When k = 3, the bound (16) alone will not suffice. Instead, we again appeal to Hua’s Lemma [14, Lemma 2.5] and then use Hölder’s inequality to deduce that 1 0

⎛ 1 ⎞ 34 ⎛ 1 ⎞ 41   11 +ε | f 3 (α)|5 dα ≤ ⎝ | f 3 (α)|4 dα ⎠ ⎝ | f 3 (α)|8 dα ⎠  P3 4 . 0

(18)

0

We are ready to derive Lemma 2. It will be convenient to write K = 2k−1 . When k ≥ 4, a trivial estimate f k (α)  Pk applied to one factor suffices to infer from (15)

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and (16) that 

1

| f 2 (α) f k (α) K +1 |dα  P22

+ε

PkK −k+2+ε  P2 PkK +1 n ε−1 .

(19)

n

By (18), in the case k = 3, the upper bound O(n 7/6+ε ) for the integral in (19) follows in the same way, and is consistent with the last expression in (19). For the contribution from the major arcs, we begin with f 2 = f 2∗ + 2 and then use the inequality 

| f 2 f kK +1 |dα

N

1 ≤ sup |2 | α∈N

| f k | K +1 dα +

0



| f 2∗ f kK +1 |dα.

(20)

N

The first term on the right-hand side here may be estimated through (14) and (16) or (18) as O(P2 PkK +1 n ε−1 ), as in the previous argument. For the second integral, we use binomial expansion to deduce that | f k | K +1 ≤ | f k∗ | K +1 + | f k − f k∗ |

K 

| f k f k∗ K − j |. j

j=0

Now multiply with | f 2∗ | and integrate over N. Then 

| f 2∗ f kK +1 |dα 

N



| f 2∗ || f k∗ | K +1 dα + sup | f 2∗ k |

N

α∈N

K  

| f k | j | f k∗ | K − j dα.

j=0 N

Here, by (13), the first integral on the right is readily estimated by O(P2 PkK +1 n ε−1 ). For the second term, the same upper bound follows from (14) and the inequalities  | f k | j | f k∗ | K − j dα  PkK −k+1+ε N

for any 0 ≤ j ≤ K ; the latter are immediate from (16) and (17) via another appeal to Hölder’s inequality. Collecting together, we see that the integral on the left-hand side of (20) is O(P2 PkK +1 n ε−1 ). When combined with (19), this establishes Lemma 2. The reader familiar with modern forms of the circle method will realise that up to this point only routine arguments have been applied. It is therefore perhaps justified to consider the results described around (5) as part of the folklore. 4 A divisor sum and its applications In this section, we estimate a divisor sum associated with symmetric sums of kth powers by a method of Wolke [15]. The principal result appears as Lemma 5 below. This is

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The asymptotic formula in Waring’s problem

393

then used within a technique known as Ramanujan sum pruning to control redundant logarithms in certain mean value bounds for Weyl sums. The latter estimates, formulated in some generality in Lemma 6, may serve as amplifiers in pruning routines, and should prove useful elsewhere. We begin with a special case of Satz 1 of Wolke [15]. Here and later, the number of positive divisors of m is denoted by τ (m). Lemma 4 Let C1 , . . . , C6 denote positive absolute constants. Suppose that {an } is a sequence of natural numbers satisfying an  n C1 . Put A(x, d) =



an and R(x, d) = A(x, d) −

n≤x d|an

(d) x, d

where (d) is a multiplicative function satisfying 0 ≤ ( pl ) ≤ C2 l C3 and ( p) < p,

(21)

for all primes p and natural numbers l. Suppose further that |R(x, d)| ≤ C4 x 1−C5 , for all natural numbers d ≤ x C6 . Then one has

   τ (an )  x exp

( p)/ p . n≤x

p≤x

Proof On putting f = τ in the statement of Satz 1 of Wolke [15], the lemma follows immediately. Lemma 5 Let k and h be fixed integers greater than 1, and let ψ(m; P) denote the number of solutions of the equation m=

h  k (x kj − x h+ j ), j=1

in natural numbers x j ≤ P (1 ≤ j ≤ 2h). Then one has 

ψ(m; P)τ (m)  P 2h log P.

m≥1

Proof Let ψ ∗ (m; P) denote the number of those solutions counted by ψ(m; P) where the additional coprimality condition (x1 , x2 , . . . , x2h ) = 1 is satisfied, and consider the sum  ψ ∗ (m; P)τ (m). (P) = m≥1

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J. Brüdern, K. Kawada

We shall apply Lemma 4 to the latter sum, and show that one has (P)  P 2h log P.

(22)

Elementary transformations then suffice to deduce the conclusion of Lemma 5. To see this, write d = (x1 , x2 , . . . , x2h ), and arrange the solutions counted by ψ(m; P) according to the value of d. Then ψ(m; P) =



ψ

∗



d≤P d k |m

m P ; dk d

 ,

and therefore we have 

ψ(m; P)τ (m) =

m≥1



ψ

∗



m P ; dk d

d≤P m≥1 d k |m

 τ (m) ≤



 τ (d )

d≤P

k

P d

 .

With the inequality (22) in hand, we may then infer the bound 

ψ(m; P)τ (m) 

m≥1



 τ (d k )

d≤P

P d

2h

 log

2P d

  P 2h log P,

as required. To derive (22) from Lemma 4, we first examine the sum A(d) =



ψ ∗ (m; P).

m≥1 m≡0 (mod d)

Note that A(d) equals the number of solutions of the congruence h 

k (x kj − x h+ j) ≡ 0

(mod d)

j=1

subject to x j ≤ P (1 ≤ j ≤ 2h) and (x1 , . . . , x2h ) = 1. Now let λ(a1 , . . . , a2h ) denote the number of x1 , . . . , x2h with 1 ≤ x j ≤ P satisfying the congruences x j ≡ a j (mod d) for 1 ≤ j ≤ 2h and the coprimality constraint (x1 , x2 , . . . , x2h ) = 1. Then one has A(d) =

 a1 ,...,a2h

123

λ(a1 , . . . , a2h ),

(23)

The asymptotic formula in Waring’s problem

395

where the summation is over a1 , . . . , a2h satisfying h  k (a kj − ah+ j ) ≡ 0 (mod d), 1 ≤ a j ≤ d, (a1 , . . . , a2h , d) = 1.

(24)

j=1

By the familiar property of the Möbius function μ(l), we have 

λ(a1 , . . . , a2h ) =



···

x1 ≤P x1 ≡a1 (mod d)



μ(l).

x2h ≤P l|(x1 ,...,x2h ) x2h ≡a2h (mod d)

Here, in view of the coprimality condition in (24), each l appearing in the innermost sum is necessarily coprime to d. It follows that ⎛ ⎜ 2h ⎜  ⎜ ⎜ λ(a1 , . . . , a2h ) = μ(l) ⎜ l≤P j=1 ⎜ ⎝ (l,d)=1

⎞ 



 =

P d

x j ≤P x j ≡a j (mod d) x j ≡0 (mod l)

⎟ ⎟ ⎟ 1⎟ ⎟ ⎟ ⎠

⎛ ⎞

2h    P 2h−1 μ(l) +O⎝ + 1 ⎠. l 2h dl l≤P (l,d)=1

(25)

l≤P

It is now easy to see that whenever h ≥ 2, the sum over l appearing in the first term on the right-hand side here is equal to  p d

(1 − l

−2h

)+O



 l

−2h

l>P

=

1  (1 − p −2h )−1 + O(P 1−2h ). ζ (2h) p|d

Here ζ (s) denotes the Riemann zeta function. We write r (d) for the number of a1 , . . . , a2h satisfying (24), and then define

(d) = r (d)d 1−2h

 (1 − p −2h )−1 . p|d

By (23) and (25), we infer that the identity A(d) =

(d) P 2h + R(d) d ζ (2h)

(26)

holds with some R(d) satisfying R(d)  r (d)(P/d)2h−1 + r (d)P.

(27)

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J. Brüdern, K. Kawada

Next, we analyse the functions r (d) and (d). By the Chinese remainder theorem, the function r (d) is multiplicative, and so is then also (d). Recalling the exponential sum Sk (q, a) introduced in (12), it follows from orthogonality that, for any prime p, one has p  −1 |Sk ( p, a)|2h − 1. r ( p) = p a=1

But Sk ( p, a)  holds whenever p  a (see Lemma 4.3 of Vaughan [14]), and by (12) one has Sk ( p, p) = p. This yields the formula r ( p) = p 2h−1 + O( p h ), and consequently, one also has p 1/2

( p) = 1 + O( p 1−h ).

(28)

Define the integer θ = θ (k, p) by the conditions p θ |k and p θ  k. Then put γ = γ (k, p) = θ (k, p) + (k, p, 2). The significance of γ is that for any a with p  a and any l ≥ γ , the numbers of solutions of the congruences x k ≡ a (mod pl ) and x k ≡ a (mod p γ ) are the same. This much follows from the elementary theory of kth power residues, and we may deduce that r ( pl ) = p (2h−1)(l−γ )r ( p γ ). Expressed in terms of , this yields the identity

( pl ) = p (1−2h)l (1 − p −2h )−1r ( pl ) = p (1−2h)γ (1 − p −2h )−1r ( p γ ) = ( p γ ), valid whenever l ≥ γ . When p  k, we have γ = 1, and so, by (28) we see that

( pl )  1 holds for all l ≥ 1. When p|k, on the other hand, both p and γ are O(1), and we have ( pl )  1 for l ≥ γ . This holds for l < γ as well, for trivial reasons. Moreover, we have, obviously, 0 ≤ r ( p) ≤ p 2h − 2. We conclude that 0 ≤ ( pl )  1 and ( p) < p

(29)

hold for all primes p and all natural numbers l. Also, by multiplicativity, we derive from (29) that one has (d)  d ε . Consequently, r (d)  d 2h−1+ε , and by (27) it follows that R(d)  P 2h−1+ε

(30)

holds uniformly in the range d ≤ P 1/2 . We are ready to apply Lemma 4 in which we take {an } to be the sequence of natural numbers m with multiplicity ψ ∗ (m; P). In view of (29), (26) and (30) in particular, we may indeed apply Lemma 4 to bound the sum (P). Then, recalling Mertens’ theorem together with (28), we have ⎛

⎞  ( p) ⎠  P 2h log P, (P)  P 2h exp ⎝ p 2h p≤P

as required at (22). This completes the proof of Lemma 5.

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397

Before we move on to applications, we remark that there is an alternative, perhaps more natural route to the estimate given in Lemma 5. One may try to choose {an } to be the sequence of natural numbers m with multiplicity ψ(m; P) in Lemma 4, but this strategy gets stuck with the requirement on the upper bound for ( pl ) in (21). However, an inspection of the proof of Satz 1 of Wolke [15] reveals that the assertion of Lemma 4 is still justified if the constraint (21) is replaced by the weaker requirement that 0 ≤ ( p) < min{ p, C2 } and 0 ≤ ( pl ) ≤ C2 pl(1/2−C3 ) hold for all primes p and integers l ≥ 2. This adjusted form of Lemma 4 would be applicable in the new context, and one would reach the desired conclusion with the above choice of the sequence {an } quite directly. We have preferred the proof given above because this only depends on more readily accessible references. Lemma 6 Let k and h denote integers√greater than 1. Suppose that the real number X satisfies the inequalities 1 ≤ X ≤ 21 n. Then one has 

| f 2∗ (α) f k (α)h |2 dα

1  X (log n)

M(X )

| f k (α)|2h dα + Pk2h (log n)2 . 0

Proof As in the proof of Lemma 2 of Brüdern [2], one finds that X



| f 2∗ (α) f k (α)h |2 dα ≤ n

 q≤X

M(X )

  2h q qn  a −1 + β dβ. q (1 + n|β|)−1 f k q a=1 X − qn

In the notation introduced in Lemma 5, we have  | f k (α)|2h = ψ(m; Pk )e(mα). m

Hence, by orthogonality and the obvious fact that ψ(m; Pk ) = ψ(−m; Pk ), one confirms that  2h q   fk a + β = q q



m≡0 (mod q)

a=1



ψ(m; Pk )e(−mβ)  q

ψ(m; Pk ).

m≥0 m≡0 (mod q)

We now infer 

X

| f 2∗ (α) f k (α)h |2 dα  n

M(X )

qn ψ(m; Pk ) (1 + nβ)−1 dβ.





q≤X

m≥0 m≡0 (mod q)

0

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J. Brüdern, K. Kawada

The last integral here is simply O(n −1 log n), and we also have 



q≤X

m≥0 m≡0 (mod q)

ψ(m; Pk ) ≤ X ψ(0; Pk ) +



ψ(m; Pk )

m≥1



1.

q|m

Hence we have 

⎛ | f 2∗ (α) f k (α)h |2 dα

 (log n) ⎝ X ψ(0; Pk ) +



⎞ ψ(m; Pk )τ (m)⎠ ,

m≥1

M(X )

By orthogonality, 1 ψ(0; Pk ) =

| f k (α)|2h dα, 0

and the lemma now follows from Lemma 5. We shall apply Lemma 6 in the next section with h = 2 and h = 3, for k ≥ 5. It is convenient, therefore, to record here the following bounds. Lemma 7 When 2 < j < k, one has 1

1 | f k (α)| dα  4

Pk2

0

2j−j

j

| f k (α)|2 dα  Pk

and

1

(log n)ε− 2 j ( j−1) .

0

Proof The former inequality is due to Hooley [7,8], and the latter is the improvement of Hua’s inequality established as Theorem B of Vaughan [13]. We deduce from Lemma 7 that for k ≥ 4 one has 1

⎛ | f k (α)|6 dα  ⎝

0

1 0

⎞ 21 ⎛ 1 ⎞ 21  7/2 | f k (α)|4 dα ⎠ ⎝ | f k (α)|8 dα ⎠  Pk (log n)ε−3/2 . 0

Equipped with the former inequality of Lemma√7 and the estimate just confirmed, Lemma 6 then asserts that whenever 1 ≤ X ≤ 21 n, one has 

| f 2∗ (α) f k (α)2 |2 dα  X Pk2 (log n) + Pk4 (log n)2

(31)

M(X )

and  M(X )

123

| f 2∗ (α) f k (α)3 |2 dα  X Pk

7/2

+ Pk6 (log n)2 .

(32)

The asymptotic formula in Waring’s problem

399

5 Proofs of Theorems 1 and 2 We launch the proofs of Theorems 1 and 2, and therefore we assume that k ≥ 5 throughout this section, but otherwise retain the notation used in Sects. 2 and 3. Our proof of Theorem 1 relies on the following estimate contained in Boklan [1]. Lemma 8 Let χ (k) = 1 or 0, according as k is even or odd, and define 1 3 η(k) = 2(k − 3)3/2 − k 2 − k + 13 + χ (k)(τ (k) − 1). 2 2 Then, for k ≥ 6, one has  k−3 k−3 | f k (α)|7·2 dα  Pk7·2 −k (log Pk )η(k)+ε . mk

Proof This is a crude form of Theorem 2 of Boklan [1]. In fact, the desired estimate is confirmed by combining the bounds (6.6), (8.4), (8.5) and the inequality preceding to (10.3) of [1], on noting that our minor arcs mk are slightly narrower than those of Boklan [1], U\M in the notation of [1]. We are now ready to establish Theorem 1. A similar argument will reduce the proof of Theorem 2 to that of Theorem 3. It is convenient for us to put J = 7 · 2k−4 + 3, and write  (33) ϒk = | f 2 (α) f k (α) J | dα. mk

Then, in view of Lemma 1 and (7), the proofs of Theorems 1 and 2 will be complete once the bounds ϒ5  P2−1 P517−δ , ϒk  P2−1 PkJ (log n)−1 (k ≥ 6)

(34)

have been demonstrated with some positive number δ. The initial steps of the estimation can be performed for any k ≥ 6, we shall describe an alternative route for k = 5 later. When k ≥ 6, one has 47 2k−2 < J < 2k−1 . Therefore, on interpolating between the cases j = k − 2 and j = k − 1 of [14, Lemma 2.5] by means of Hölder’s inequality, one finds that there is a positive number η for which the bound 1

J −k+ 54 −η

| f k (α)| J dα  Pk

(35)

0

holds. We now dissect the set mk into L = M(Pk )\Mk and m(Pk ). Let ϒk∗ and ϒk† denote the contributions to the integral (33) that arise from integrating over the 5/2 sets L and m(Pk ), respectively. Then 5/2

5/2

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J. Brüdern, K. Kawada

ϒk = ϒk∗ + ϒk† .

(36) −5/4

5/2

By (13) and (15) one finds that whenever α ∈ m(Pk ), then f 2 (α)  P2 Pk Hence, by (35), we infer that J −k−η

ϒk†  P2 Pk

J −η

 P2−1 Pk

.

(37)

For the estimation of ϒk∗ , we first apply (13) to see that f 2 (α)  | f 2∗ (α)| + Pk Hence, by a crude use of (35), we deduce that whenver k ≥ 6, one has

5/4+ε

J −k+ 52

ϒk∗  Pk

 +

.

| f 2∗ (α) f k (α) J | dα.

.

(38)

L

A straightforward treatment of the integral on the right-hand side of (38) is possible when k ≥ 7. In this situation, we first apply Schwarz’s inequality, and then use (32) and Lemma 8 to observe that 

⎛

⎞1

2 ⎛ ⎞1 2  ⎟ ⎝ | f k |7·2k−3 dα ⎠ | f 2∗ |2 | f k |6 dα ⎟ ⎠



⎜ | f 2∗ (α) f k (α) J | dα ≤ ⎜ ⎝

mk

5/2

L

M(Pk ) 1

 (Pk6 (log n)2 ) 2 (Pk7·2

k−3 −k

1

(log n)η(k)+ε ) 2

 P2−1 PkJ (log n)1+η(k)/2+ε . Since η(k) < −5 for k ≥ 7, the desired estimate (34) follows by (36), (37) and (38) for these k. When k = 6, however, this plain argument fails as η(6) > −0.608. Although Boklan’s estimates in Lemma 8 could be improved so that the latter treatment would provide a bound adequate to our ultimate purpose for k = 6 as well (see the footnote on p.330 of Boklan [1], for example), a more direct and simpler alternative is more efficient in the present context. All what is now required is a successful estimation of the integral on the right-hand side of (38). We initiate the treatment by splitting L into the sets 5/2

P = M(P62 )\M(P6 ), Q = M(P6 )\M(P62 ). Recall that when k = 6 then J = 31. We begin with the set P. For α ∈ m6 , Weyl’s 1−η inequality gives f 6 (α)  P6 , for some η > 0. Hence, by Schwarz’s inequality,  P

123

⎛ 1−η ⎜ | f 2∗ f 631 | dα  P6 ⎜ ⎝



M(P62 )

⎞1

2 ⎛ ⎞1 2  ⎟ ⎝ | f 6 |56 dα ⎠ . | f 2∗ f 62 |2 dα ⎟ ⎠

m6

The asymptotic formula in Waring’s problem

401

Consequently, by (31) and Lemma 8 with k = 6, we conclude that 

| f 2∗ f 631 | dα  P2−1 P6

31−η

log n.

(39)

P 5/2

Now suppose that P62 ≤ Q ≤ P6 , and consider a slice M(2Q)\M(Q). For any α in this set, one finds from (13) that | f 2∗ (α)|2  n Q −1 . Consequently, by (31), 

| f 2∗ f 6 |4 dα  n Q −1 (Q P62 log n + P64 (log n)2 ).

M(2Q)\M(Q)

We take Q = 2l P62 and sum over l. Then, recalling the definition of Q, it follows that 

| f 2∗ (α) f 6 (α)|4 dα  n P62 (log n)2 .

(40)

Q

Finally, by Hölder’s inequality, one has  Q

⎛ | f 2∗ f 631 | dα  ⎝



⎞1 ⎛ 4

| f 2∗ f 6 |4 dα ⎠ ⎝

Q



⎞1 ⎛ 4

| f 6 |56 dα ⎠ ⎝

m6

1

⎞ 21 | f 6 |32 dα ⎠ .

0

Three integrals on the right-hand side are estimated, respectively, by (40), Lemma 8 with k = 6, and Lemma 7 with k = 6 and j = 5, and consequently the integral on the left-hand side is readily seen to be bounded by O(P2−1 P631 (log n)−4 ). In view of (39) and (38), this yields also an admissible upper bound for ϒ6∗ . Hence, by (36) and (37), we now have ϒ6  P2−1 P631 (log n)−1 , as required in (34). This completes the proof of Theorem 1. When k = 5, an even simpler approach suffices. Here J = 17 = 24 + 1, and Hua’s lemma in its historical form (16) is appropriate. Moreover, we note that Weyl’s 15/16+ε , uniformly for α ∈ m5 . For α ∈ m(P52 ), we also inequality gives f 5 (α)  P5 deduce from (13) and (15) that f 2 (α)  P2 P5−1 . On combining these remarks with (16), it follows that 

12−η

| f 2 (α) f 5 (α)17 | dα  P2 P5

(41)

m(P52 )

is a valid upper bound whenever η < 1/16. Moreover, on writing P = M(P52 )\M(P5 ), then by (14), (16) and a trivial estimate for an excessive f 5 , we

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J. Brüdern, K. Kawada

have 

 | f 2 f 517 | dα



P514+ε

P

+

| f 2∗ f 517 | dα.

(42)

P

By Schwarz’s inequality, 

⎛ ⎜ | f 2∗ f 517 | dα ≤ ⎜ ⎝



⎞1 ⎛ ⎞1 2 2  ⎟ ⎜ ⎟ 30 | f 2∗ f 52 |2 dα ⎟ ⎠ ⎝ | f 5 | dα ⎠ .

M(P52 )

P

P

By (31), the first integral on the right-hand side is bounded by O(P54+ε ), and Theorem 3 bounds the second by O(P525−δ ). On combining these estimates with (42), it follows that  12− 1 δ+ε | f 2 (α) f 5 (α)17 | dα  P2 P5 2 . P

This inequality together with (41) yields a bound for ϒ5 as required in (34). This completes the proof of Theorem 2. 6 Proof of Theorem 3 In this section we prove Thorem 3. Throughout this section we write P = [P5 ]. As is already mentioned in the introduction, this part heavily relies on the ideas of HeathBrown [6], and for easy reference we apply the notation from that paper. In particular, we put I(α) = [m P −3 , (m + 1)P −3 ) where m is the integer with m ≤ P 3 α < m + 1. Now define  3 T (α) = max e(yn + zn) , I,y,z n∈I

where I runs over subintervals of [1, P], and y and z run over real numbers with y ∈ I(α) and z ∈ [0, 1], respectively. Finally, put T (α) =

2 80P 

T (αh),

h=1

for short. Then, applying Weyl’s differencing twice, one finds that (see Lemma 1 of Heath-Brown [6]) | f 5 (α)|4  P 3 + P 1+ε T (α).

123

(43)

The asymptotic formula in Waring’s problem

403

Let σ be a real number with 0 < σ < 10−2 . Let D be the set of α ∈ M(P 2 )\M(P) for which one has T (α) ≥ P 8/3+σ , and write d for the complement of D in M(P 2 )\M(P), so that M(P 2 )\M(P) = D ∪ d.

(44)

Whenever α ∈ d, the bound (43) yields | f 5 (α)|4  P 11/3+σ +ε , and on combining this with (16), we infer that 1

 | f 5 (α)| dα  sup | f 5 (α)| 30

α∈d

d

| f 5 (α)|16 dα

14 0

 (P 11/3+σ +ε )7/2 · P 12+ε  P 25−1/6+4σ .

(45)

To estimate the corresponding contribution from D, we follow Heath-Brown [6] and cover D by a union of intervals of the form I (q, Z ) =



{α ∈ [0, 1]; Z − P −5 ≤ α − a/q ≤ 2Z }.

(a,q)=1

Here β denotes the distance of β to the nearest integer. Then, since D ⊂ M(P 2 )\M(P), we find that  | f 5 (α)| dα ≤ 30

 Q

D



 | f 5 (α)|30 dα,

Z Q
where Q and Z run over powers of 2 in the ranges 1 < 2Q ≤ P 2 and max{(2Q P 4 )−1 , P −5 } ≤ Z ≤ (Q P 3 )−1 .

(46)

We now use (43) to estimate | f 5 (α)|24 and note that the first term on the right-hand side of (43) is irrelevant whenever α ∈ D. It follows that there are values of Q and Z satisfying (46) for which one has  | f 5 (α)| dα  P 30

D

6+ε



 sup

Q
T (α)

| f 5 (α)|6 dα.

6

(47)

I (q,Z )

The quantity on the right-hand side of (47) may be bounded by the methods of Heath-Brown [6]. When (46) holds, the proof of Lemma 7 of [6] remains valid when k = 5, and yields the bound 

sup

Q
T (α)6  P 14+ε (Q Z )−1 .

(48)

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J. Brüdern, K. Kawada

Further progress now depends on a minor refinement of Lemma 8 of [6], which we state as our next lemma. Lemma 9 Suppose that Q and Z satisfy (46), and that Q < q ≤ 2Q. Then one has  | f 5 (α)|6 dα  q Z P 5−1/8+ε + P 1−1/16+ε . I (q,Z )

Proof For d|q and a real number R, we define FR (α) =





e(αx 5 ) and FR,d (α) =

R
e(αx 5 ).

R
Since we may express f 5 (α) as the sum of O(log P) sums FR (α) with 1 < 2R ≤ P we have   6 6 | f 5 (α)| dα  (log P) sup |FR (α)|6 dα. (49) 0
I (q,Z )

I (q,Z )

We write I R for the integral on the right-hand side. On writing 

FR (α) =

FR,d (α) +

d|q d≤P ξ



FR,d (α) = FR∗ (α) + FR∗∗ (α),

d|q d>P ξ

say, where ξ is a positive number to be determined later, we have  IR 

|FR (α)

2

FR∗ (α)4 |dα

I (q,Z )





|FR (α)2 FR∗∗ (α)4 |dα

+ I (q,Z )

⎛

1/3 ⎜ |FR (α)2 FR∗ (α)4 |dα + I R ⎝

 I (q,Z )

⎞2/3



⎟ |FR∗∗ (α)|6 dα ⎠

.

I (q,Z )

It follows that  IR 

|FR (α) I (q,Z )

2

FR∗ (α)4 |dα

 +

|FR∗∗ (α)|6 dα.

(50)

I (q,Z )

We now concentrate on the first integral on the right-hand side of (50) for a while. First, for α ∈ I (q, Z ), there is an a coprime to q such that Z − P −5 ≤ |α −a/q| ≤ 2Z , and also we may take coprime integers q  and a  such that |q  α − a  | ≤ (4P 4 )−1 and

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The asymptotic formula in Waring’s problem

405

q  ≤ 4P 4 . Since (46) and Q < q ≤ 2Q hold, if q  < P/4 then we have   a − a ≤ 2Z + 1 ≤ 2 + 1 ≤ 8q P + q < 1 , q q 4q  P 4 q P3 4q  P 4 4qq  P 4 qq  P 2 so we must have q = q  and a = a  . But we then see that Q < q = q  < P/4 and 1 α − a ≥ Z − P −5 ≥ 1 , ≥ 4q  P 4 q 4Q P 4 which implies q = q  ≤ Q, a contradiction. Thus for α ∈ I (q, Z ), we must have q  ≥ P/4, and Weyl’s inequality now yields the estimate FR (α)  R 1+ε (q 

−1

+ R −1 + q  R −5 )1/16  R 1+ε (P 4 R −5 )1/16 .

We now deduce that   |FR (α)2 FR∗ (α)4 |dα  P (1+ε)/2 R 11/8 I (q,Z )

I (q,Z )

 P 1/2+ε R 11/8



|FR∗ (α)|4 dα  |FR,d (α)|4 dα.

(51)

d|q I (q,Z ) d≤P ξ

Following the proof of (4.2) of Heath-Brown [6], we find that  |FR,d (α)| dα ≤ 4

q 2Z 

|FR,d (α)|4 dα  q Z N ,

(52)

a=1−2Z

I (q,Z )

where N denotes the number of solutions of x15 + x25 ≡ x35 + x45 (mod q), |x15 + x25 − x35 − x45 | ≤ Z −1 in integers x j satisfying R < x j ≤ 2R and (x j , q) = d. Applying the argument that leads an upper bound for the quantity corresponding to N on p. 228 of [6], we derive the inequality N  (R/d)3 P ε (1 + q0−1 d −1 Z −1 R −4 ), where q0 = q/(q, d 5 ). Therefore, making use of the trivial bound (q, d 5 ) ≤ d 5 , we deduce from (52) that  |FR,d (α)|4 dα  P ε (q Z R 3 d −3 + d R −1 ). (53) I (q,Z )

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Substituting the last estimate into (51), we see that  |FR (α)2 FR∗ (α)4 |dα  q Z P 1/2+ε R 35/8 + P 1/2+ξ +ε R 3/8 .

(54)

I (q,Z )

Turning to the second integral on the right-hand side of (50), we first observe that    ∗∗ 6 ε |FR (α)| dα  P |FR,d (α)|6 dα. (55) d|q I (q,Z ) d>P ξ

I (q,Z )

Next, note that that the estimate (53) is still valid in the current situation, and use the trivial bound |FR,d (α)|  R/d to derive that  I (q,Z )

|FR,d (α)|6 dα  (R/d)2 P ε (q Z R 3 d −3 + d R −1 )  P ε (q Z R 5 d −5 + Rd −1 ),

whence by (55), 

|FR∗∗ (α)|6 dα  q Z R 5 P ε−5ξ + R P ε−ξ .

I (q,Z )

Now take ξ = 1/16 and combine the last inequality with (54) to conclude that uniformly for 0 < R ≤ P, one has I R  q Z P 5−1/8+ε + P 1−1/16+ε , and the lemma is thus immediate in view of (49). Now equipped with (48) and Lemma 9 the proof of Theorem 3 is swiftly completed. In fact, applying Lemma 9 within (47), and then appealing to (48), we find that   | f 5 (α)|30 dα  P 6+(ε−1)/16 sup T (α)6 (Q Z P 5 + P) D

Q
 P 20−1/16+ε (P 5 + (Q Z )−1 P)  P 25−1/16+ε , because (46) implies Q Z  P −4 . Theorem 3 follows from (44), (45) and this last estimate. References 1. Boklan, K.D.: The asymptotic formula in Waring’s problem. Mathematika 41, 329–347 (1994) 2. Brüdern, J.: A problem in additive number theory. Math. Proc. Cambridge Philos. Soc. 103, 27–33 (1988)

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3. Ford, K.B.: New estimates for mean values of Weyl sums. Int. Math. Res. Notices 155–171 (1995) 4. Hardy, G.H., Littlewood, J.E.: Some problems of “Partitio Numerorum”. I: a new solution to Waring’s problem. Göttinger Nachrichten 33–54 (1920) 5. Hardy, G.H., Littlewood, J.E.: Some problems of “Partitio Numerorum”. VI: further researches in Waring’s problem. Math. Z. 23, 1–37 (1925) 6. Heath-Brown, D.R.: Weyl’s inequality, Hua’s inequality, and Waring’s problem. J. Lond. Math. Soc. 38(2), 216–230 (1988) 7. Hooley, C.: On another sieve method and the numbers that are a sum of two hth powers. Proc. Lond. Math. Soc. 43(3), 73–109 (1981) 8. Hooley, C.: On another sieve method and the numbers that are a sum of two hth powers. II. J. Reine Angew. Math. 475, 55–75 (1996) 9. Hooley, C.: On a new approach to various problems of Waring’s type. In: Recent Progress in Analytic Number Theory (Durham, 1979), vol. 1, pp. 127–191. Academic Press, London (1981) 10. Sinnadurai, J.St.-C.L.: Representation of integers as sums of six cubes and one square. Q. J. Math. Oxf. Ser. 16(2), 289–296 (1965) 11. Stanley, G.K.: The representation of a number as the sum of one square and a number of k-th powers. Proc. Lond. Math. Soc. 31(2), 512–553 (1930) 12. Vaughan, R.C.: On Waring’s problem: one square and five cubes. Q. J. Math. Oxf. Ser. 37(2), 117– 127 (1986) 13. Vaughan, R.C.: On Waring’s problem for smaller exponents. II. Mathematika 33, 6–22 (1986) 14. Vaughan, R.C.: The Hardy–Littlewood method. In: Cambridge Tracts in Mathematics, vol. 125, 2nd edn. Cambridge University Press, Cambridge (1997) 15. Wolke, D.: Multiplikative Funktionen auf schnell wachsenden Folgen. J. Reine Angew. Math. 251, 54–67 (1971)

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