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LXXXI.3 (1997)

Diagonal cubic equations

by

Hongze Li (Jinan)

1. Introduction. Let λ

j

be positive integers. We shall be concerned with solutions of the diophantine equation

(1.1) λ

1

x

31

+ λ

2

x

32

+ . . . + λ

7

x

37

= 0

which are small in the following sense. Let Λ = λ

1

λ

2

. . . λ

7

. Then we seek for solutions to (1.1) which satisfy

(1.2) 0 <

X

7 i=1

λ

i

|x

i

|

3

 Λ

g

with g as small as possible.

A first positive answer to the problem was found by R. C. Baker [1]. He established the solubility of (1.1) and (1.2) whenever g = 61. In this paper we prove the following theorem.

Theorem. Let g = 14. Then there are solutions to (1.1), (1.2).

The key for improvement is the mean value estimate for a class of ex- ponential sums presented in [2]. We state the special case we require as a lemma. Let X be a large parameter, and write

(1.3) θ

1

= 15/113, θ

2

= 14/113, η = 84/113.

We note that θ

1

+ θ

2

+ η = 1. Put

(1.4) M

1

= X

θ1

, M

2

= X

θ2

, Y = X

η

.

As usual we write e(α) = exp(2πiα). Now define the exponential sums

(1.5) g(α, P ) = X

P <X≤2P

e(αx

3

),

1991 Mathematics Subject Classification: Primary 11P55.

This work was supported by the National Natural Science Foundation of China.

[199]

(2)

(1.6) G(α, X) = X

M1<p1≤2M1

X

M2<p2≤2M2

g(α(p

1

p

2

)

3

, Y ), where, here and throughout, p, p

i

denote prime numbers.

Lemma 1. In the notation introduced above, for any ε > 0 we have

1

\

0

|G(α, X)|

6

dα  X

3+2θ1

. This is the case l = 3 of the Theorem in Br¨ udern [2].

2. Preliminaries. Now define (2.1) χ

3

(X)

= {p

1

p

2

y : X

θ1

< p

1

≤ 2X

θ1

, X

θ2

< p

2

≤ 2X

θ2

, X

η

< y ≤ 2X

η

} and

(2.2) χ

4

(X) = {py : X

θ

< p ≤ 2X

θ

, y ∈ χ

3

(X

1−θ

)}

in which θ = 233/1815.

Given positive integers µ

1

, µ

2

, we write S(µ

1

, µ

2

, B, E) for the number of solutions of the equation

(2.3) µ

1

x

31

+ µ

2

p

3

(y

13

+ y

23

) = µ

1

x

32

+ µ

2

p

3

(y

33

+ y

34

) in integers x

1

, x

2

, y

1

, y

2

, y

3

, y

4

and primes p satisfying

B < x

1

, x

2

≤ 2B, p - µ

1

x

1

x

2

, y

i

∈ χ

3

(E

1−θ

) (1 ≤ i ≤ 4), (2.4)

E

θ

< p ≤ 2E

θ

, p ≡ 2 (mod 3).

(2.5)

Lemma 2. Let B ≥ 1, E ≥ 1. Let µ

1

, µ

2

be positive integers, and (2.6) µ

2

E

3

 µ

1

B

3

 µ

2

E

3

, µ

1

 µ

2

, E

θ

≤ B

1/7

. Then

(2.7) S(µ

1

, µ

2

, B, E)  B

1+ε

E

2−θ

.

P r o o f. By the argument of Vaughan [7], p. 125, we have S(µ

1

, µ

2

, B, E)

 E

ε

S

0

, where S

0

is the number of solutions to (2.3) subject to (2.4), (2.5) and x

1

≡ x

2

(mod p

3

). The solutions with x

1

= x

2

contribute to S

0

at most BE

θ

E

2(1−θ)+ε

= BE

2−θ+ε

. For the solutions with x

1

6= x

2

we again follow through the argument in Vaughan [7] and find that there are at most 2S

1

remaining solutions where S

1

equals the number of solutions to

µ

1

h(3x

2

+ h

2

p

6

) = 4µ

2

(y

13

+ y

23

− y

33

− y

43

),

(3)

where x ≤ 4B, E

θ

< p ≤ 2E

θ

, 0 < h ≤ H, y

i

∈ χ

3

(E

1−θ

), and H = CBE

−3θ

for some constant C. On writing

Ω(α) = X

Eθ<p≤2Eθ

X

h≤H

X

x≤4B

e(αh(3x

2

+ h

2

p

6

))

we have, by H¨older’s inequality, S

1

=

1

\

0

Ω(µ

1

α)|G(4µ

2

α, E

1−θ

)|

4



1

\

0

|Ω(µ

1

α)|

3



1/3



1

\

0

|G(α, E

1−θ

)|

6



2/3

.

The argument of Vaughan [8], p. 39 shows that for B

1/8

< E

θ

≤ B

1/7

we have

1

\

0

|Ω(α)|

3

dα  H

2

(E

θ

B)

3/2+ε

. Since Ω(α) has period 1, we have

1

\

0

|Ω(µ

1

α)|

3

dα = 1 µ

1

µ

\

1

0

|Ω(α)|

3

dα =

1

\

0

|Ω(α)|

3

and by Lemma 1 we have

1

\

0

|G(α, E

1−θ

)|

6

dα  (E

1−θ

)

3+2θ1

. The lemma follows easily.

3. Outline of the method. For the proof of the Theorem, as in Baker [1] it suffices to prove the Theorem in the special case when λ

1

, . . . , λ

7

are cube-free, and no prime divides more than four of the λ

i

. We may suppose that

(3.1) λ

5

≥ λ

2

≥ λ

3

≥ λ

4

≥ λ

6

≥ λ

7

≥ λ

1

. Let

(3.2) N = C

0

Λ

g

, P = N

1/3

, P

j

= P λ

−1/3j

,

where C

0

is sufficiently large. Let δ denote a sufficiently small positive ab- solute constant, and let ε = δ

2

.

Let R(λ

1

, . . . , λ

7

) denote the number of solutions of the equation (3.3) λ

1

x

3

− λ

2

p

3

y

3

+ λ

3

p

3

p

331

p

332

p

333

z

33

+ λ

4

p

3

p

341

p

342

p

343

z

43

+ λ

5

u

3

+ λ

6

p

361

p

362

p

363

v

63

+ λ

7

p

371

p

372

p

373

v

37

= 0

(4)

in integers x, y, z

3

, z

4

, u, v

6

, v

7

and primes p, p

ij

satisfying (3.4) P

1

< x ≤ 2P

1

, W < y ≤ 2W, R

i

< z

i

≤ 2R

i

,

U < u ≤ 2U, V

i

< v

i

≤ 2V

i

,

(3.5) p - x, p

ij

- y (3 ≤ i ≤ 4, 1 ≤ j ≤ 3), p

ij

- u (6 ≤ i ≤ 7, 1 ≤ j ≤ 3), (3.6) Y < p ≤ 2Y, p ≡ 2 (mod 3), p - Λ,

(3.7) Z

ij

< p

ij

≤ 2Z

ij

, p

ij

≡ 2 (mod 3), (i = 3, 4, 6, 7, 1 ≤ j ≤ 3), p

ij

- Λ.

Here

(3.8) Y = C

1

P

11/7

, W = P

2

Y

−1

, R

i

= 1

200 (P

i

Y

−1

)

1176/1815

(i = 3, 4),

(3.9) U = 1

200 P

5

, V

i

= 1

200 P

i1176/1815

(i = 6, 7), (3.10) Z

i1

= 2

−i

(P

i

Y

−1

)

233/1815

, Z

i2

= 2

−i

(P

i

Y

−1

)

210/1815

,

Z

i3

= 2

−i

(P

i

Y

−1

)

196/1815

(i = 3, 4), (3.11) Z

i1

= 2

−i

P

i233/1815

, Z

i2

= 2

−i

P

i210/1815

,

Z

i3

= 2

−i

P

i196/1815

(i = 6, 7), where C

1

is sufficiently large.

The inequalities

(3.12) N  λ

1

P

13

, λ

2

Y

3

W

3

, λ

3

Y

3

Z

313

Z

323

Z

333

R

33

, λ

4

Y

3

Z

413

Z

423

Z

433

R

34

, λ

5

U

3

, λ

6

Z

613

Z

623

Z

633

V

63

, λ

7

Z

713

Z

723

Z

733

V

73

 N are easily verified.

Let (a

1

, a

2

) or more generally (a

1

, . . . , a

n

) denote greatest common divi- sor, while [a

1

, . . . , a

n

] denotes least common multiple. We write

(3.13) f

d

(X, α) = X

X<x≤2X (x,d)=1

e(αx

3

)

and use the notation f in place of f

1

.

Let p denote the ordered set p, p

31

, p

32

, p

33

, p

41

, p

42

, p

43

, p

61

, p

62

, p

63

, p

71

, p

72

, p

73

and let

(3.14) A =

Y

4 i=3

Y

3 j=1

p

ij

, B = Y

7 i=6

Y

3 j=1

p

ij

,

(3.15) F (p, α) = f

p

(P

1

, λ

1

α)f

A

(W, λ

2

p

3

α)f (R

3

, λ

3

p

3

p

331

p

332

p

333

α)

× f (R

4

, λ

4

p

3

p

341

p

342

p

343

α)

× f

B

(U, λ

5

α)f (V

6

, λ

6

p

361

p

362

p

363

α)f (V

7

, λ

7

p

371

p

372

p

373

α),

(5)

(3.16) F (α) = X

p (3.6),(3.7)

F (p; α).

The summation here is over ordered sets p satisfying (3.6), (3.7); we often use this type of notation below.

Let < be the unit interval [LN

−1

, 1 + LN

−1

]. Then clearly (3.17) R(λ

1

, . . . , λ

7

) = \

<

F (α) dα.

Here

(3.18) L = P

1

/10.

When 1 ≤ a ≤ q ≤ L and (a, q) = 1, we take M (q, a) to be the interval {α : |α − a/q| ≤ q

−1

LN

−1

}, and let M denote the union of all such M (q, a).

Then M (q, a) are disjoint subsets of M , as we easily verify using (3.2) and (3.18). Let

(3.19) m = < \ M.

We shall prove that

\

M

F (α) dα  λ

2/211

Λ

−1/3−ε

N

26/21

(logN )

−13

, (3.20)

\

m

F (α) dα  λ

2/211

Λ

−1/3

N

26/21−ε

. (3.21)

It follows from (3.19)–(3.21) and (3.17) that

(3.22) R(λ

1

, . . . , λ

7

)  λ

2/211

Λ

−1/3−ε

N

26/21

(log N )

−13

. Thus the Theorem will follow once (3.20)–(3.21) have been proved.

We recall the standard notations (3.23) S(q, b) =

X

q r=1

e(br

3

/q), J(β, X) =

2X

\

X

e(βx

3

) dx.

The estimate

(3.24) S(q, b)  q

1+ε

ψ(q), (q, b) = 1, where ψ(q) is the multiplicative function with

(3.25) ψ(p

3h+r

) = p

−h−r/2

(h = 0, 1, . . . ; 0 ≤ r ≤ 2), follows from Lemmas 4.3 and 4.4 of [5]. By partial integration,

(3.26) J(β, X)  X

1 + X

3

|β|

for positive X and real β.

(6)

Lemma 3. Let d be an integer with  1 divisors. Let (3.27) s

d

(q, c) = q

−1

X

b|d

µ(d) S(q, b

3

c) b and let λ be a positive integer. Then

f

d

(X, λα) = s

d

(q, λa)J(λ(α − a/q), X) (3.28)

+ O(q

1/2+ε

(1 + λX

3

|α − a/q|)

1/2

) for any X > 0, real α and rational number a/q.

This is Lemma 2 of [1].

For α ∈ M (q, a), we introduce the notations s(p, α) = s

p

(q, λ

1

a)J(λ

1

(α − a/q), P

1

), (3.29)

g(p, α) = s

A

(q, λ

2

p

3

a)J(λ

2

(α − a/q)p

3

, W ).

(3.30)

It is convenient to write

(3.31) ∆ = ∆(α, a, q) = 1 + N |α − a/q|.

Lemma 4. Let p satisfy (3.6) and (3.7). Let α ∈ M (q, a) where 1 ≤ a ≤ q ≤ L, (a, q) = 1. Then

f

p

(P

1

, λ

1

α) − s(p, α)  q

1/2+ε

1/2

, (3.32)

f

A

(W, λ

2

p

3

α) − g(p, α)  q

1/2+ε

1/2

. (3.33)

P r o o f. The proof is similar to that of Lemma 3 of [1]; see [1] and [6].

4. The minor arcs. In this section we use the notations (4.1) H = C

2

P

1

Y

−3

, Q = P

1

Y

−1

,

where C

2

is a sufficiently large absolute constant; also

(4.2) M = L(2Y )

−3

.

Further, let (4.3) S(α) =

X

p31,p32,p33

(3.7)

X

p41,p42,p43

(3.7)

f

A

(W, λ

2

α)

× f (R

3

, λ

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

341

p

342

p

343

α)

2

and

(7)

Φ

p

(α) = X

P1<y≤2P1 p|y

1 (4.4)

+ 2Re X

h≤H

X

2P1+hp3<y≤4P1−hp3 p-y, y≡h (mod 2)

e

 λ

1

α

4 (3hy

2

+ h

3

p

6

)

 .

Let n denote the set of real numbers in (0, 1] with the property that whenever |α − a/q| ≤ q

−1

LN

−1

and (a, q) = 1, we have q > M . Let

(4.5) T = \

n

X

p (3.6)

Φ

p

(α)S(α) dα.

Lemma 5. Let J

1

=

1

\

0

X

p61,p62,p63

(3.7)

X

p71,p72,p73

(3.7)

f

B

(U, λ

5

α)

× f (V

6

, λ

6

p

361

p

362

p

363

α)f (V

7

, λ

7

p

371

p

372

p

373

α)

2

dα.

Then J

1

 P

5

P

6

P

72+ε

.

P r o o f. By considering the underlying diophantine equation we have J

1



1

\

0

X

p61,p62,p63

(3.7)

X

p71,p72,p73

(3.7)

f

p61p71

(U, λ

5

α)f (V

6

, λ

6

p

361

p

362

p

363

α)

× f (V

7

, λ

7

p

371

p

372

p

373

α)

2

dα.

From the inequality |zw| ≤

12

|z|

2

+

12

|w|

2

, we have J

1



X

7 k=6 1

\

0

 X

p61

(3.7)

X

p71

(3.7)

|f

p61p71

(U, λ

5

α)|

X

pk2,pk3

(3.7)

f (V

k

, λ

k

p

3k1

p

3k2

p

3k3

α)

2



2

dα.

Hence, by Cauchy’s inequality, J

1

 Z

61

Z

71

X

7 k=6 1

\

0

X

p61

(3.7)

X

p71

(3.7)

|f

p61p71

(U, λ

5

α)|

2

×

X

pk2,pk3 (3.7)

f (V

k

, λ

k

p

3k1

p

3k2

p

3k3

α)

4

 Z

712

Z

61

S(λ

5

, λ

6

, P

5

, P

6

) + Z

612

Z

71

S(λ

5

, λ

7

, P

5

, P

7

).

(8)

By (3.1) and (3.2) we have

P

6

≤ P

7

, P

6θ

≤ P

7θ

≤ P

51/7

. By (3.11), (3.12) and Lemma 2,

J

1

 P

5

P

6

P

72+ε

. This completes the proof of Lemma 5.

Lemma 6. Let J

2

= \

m

X

p (3.6)

X

p31,p32,p33

(3.7)

X

p41,p42,p43

(3.7)

f

p

(P

1

, λ

1

α)f

A

(W, λ

2

p

3

α)

× f (R

3

, λ

3

p

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

3

p

341

p

342

p

343

α)

2

dα.

Then J

2

 Y T.

P r o o f. The proof is similar to that of Lemma 15 of [1].

Let

F (β, γ; h) = X

2P1<y≤4P1

y≡h (mod 2)

e

34

βy

2

− γy  , (4.6)

G

h

(%, σ) = X

Y <p≤2Y, p≡2 (mod 3) p≤(2P1/h)1/3

e

14

%p

6

+ σp

3

 , (4.7)

F

p

(α; h) = X

2P1+hp3<y≤4P1−hp3 y≡h (mod 2)

e

34

λ

1

αhy

2

 , (4.8)

Ψ

p

(α) = 2Re X

h≤H

F

p

(α; h)e

14

λ

1

αh

3

p

6

 , (4.9)

D

p

(α; h) = X

2P1/p+hp2<y≤4P1/p−hp2 y≡h (mod 2)

e

34

λ

1

αhp

2

y

2

 , (4.10)

Ξ

p

(α) = 2Re X

h≤H

D

p

(α; h)e

14

λ

1

αh

3

p

6

 , (4.11)

T

1

(p) = \

n

Ψ

p

(α)S(α) dα, (4.12)

T

2

(p) = \

n

Ξ

p

(α)S(α) dα, (4.13)

T

3

=

1

\

0

S(α) dα.

(4.14)

(9)

By (4.4),

(4.15) Φ

p

(α) = Ψ

p

(α) − Ξ

p

(α) + O(P

1

).

Hence

(4.16) T = X

p (3.6)

(T

1

(p) − T

2

(p)) + O(P

1

Y T

3

)

from (4.5).

We estimate T

3

via Lemma 2. By (3.1), (3.2), (3.8) and (3.10) we have (P

3

Y

−1

)

θ

≤ (P

4

Y

−1

)

θ

≤ (P

2

Y

−1

)

1/7

.

Just as in Lemma 5, we have

(4.17) T

3

 (P

2

Y

−1

)(P

3

Y

−1

)

(P

4

Y

−1

)

2+ε

 P

2

P

3

P

42+ε

Y

−3−2θ

. Let

(4.18) T

5

(γ, θ) = \

n

X

h≤H

|F (αλ

1

h, γ; h)G

h

(αλ

1

h

3

, θγh)|S(α) dα.

By a very minor adaptation of the proof of (5.26) of [7], one finds that

(4.19) X

p (3.6)

T

1

(p)  (log P

1

) sup

0≤γ≤1 θ=±1

T

5

(γ, θ) + N

(26+12θ)/21

Λ

−1

.

We omit the details.

Lemma 7. Suppose that α ∈ R and

α − a q

1

24qHP

1

, (a, q) = 1.

Then

X

h≤H

|F (αh, γ; h)|

2

 P

1δ

 q

−1

HP

12

1 + Q

3

|α − a/q| + HP

1

+ q

 . This is Lemma 16 of [1].

Lemma 8. Let α, γ ∈ R. Suppose that

α − a q

≤ q

−1

Q

−3

H

3/4

, (a, q) = 1, q ≤ Q

3

H

−3/4

. Then

X

h≤H

|G

h

(αh

3

, γh)|

2

 P

1ε

 HY

2

q

−1/3

(1 + Q

3

|α − a/q|)

1/3

+ H

3/4

Y

2



.

This is essentially Lemma 8 of [7].

(10)

Lemma 9. Let γ ∈ R and θ ∈ {−1, 1}. Then

T

5

(γ, θ)  P

11+δ

P

2

P

3

P

42+ε

Y

−2−2θ

+ λ

1+ε1

P

12+ε

P

31+ε

P

41+ε

Y

−1

+ λ

1/21

P

11+ε

P

21/2+ε

P

35/4+ε

P

45/4+ε

Y

−1/2

.

P r o o f. Let n

1

denote the set of α in n with the property that whenever (4.20)

1

α − r/b| ≤ r

−1

H

7/4

Q

−3

, (b, r) = 1,

one has r > H

7/4

. Let α ∈ n

1

. We apply Lemma 7 with λ

1

α in place of α. By Dirichlet’s theorem there exist integers b, r satisfying (4.20) with r ≤ Q

3

H

−7/4

; here we must have r > H

7/4

. Now the condition of Lemma 7 is easily verified, thus

X

h≤H

|F (αhλ

1

, γ; h)|

2

 P

1δ

(P

12

H

−3/4

+ HP

1

+ Q

3

H

−7/4

) (4.21)

 P

12+δ

H

−3/4

.

We may choose c, s so that (c, s) = 1, s ≤ Q

3

H

−3/4

, and

1

α − c/s| ≤ s

−1

H

3/4

Q

−3

.

Then |λ

1

α − c/s| ≤ s

−1

H

7/4

Q

−3

, so that s > H

7/4

> H

3/4

. By Lemma 8, applied to λ

1

α in place of α, we have

X

h≤H

|G

h

1

αh

3

, θγh)|

2

 P

1ε

H

3/4

Y

2

. Hence by Cauchy’s inequality, (4.14), (4.17) and (4.21), (4.22) \

n1

X

h≤H

|F (αhλ

1

, γ; h)G

h

1

αh

3

, θγh)|S(α) dα

 P

11+δ

Y T

3

 P

11+δ

P

2

P

3

P

42+ε

Y

−2−2θ

. It remains to consider n \ n

1

. Let α ∈ n \ n

1

. There are integers b, r satisfying

(4.23)

1

α − b/r| ≤ r

−1

H

7/4

Q

−3

, (b, r) = 1, r ≤ H

7/4

.

We write b/(λ

1

r) = a/q in lowest terms. Clearly q = rd where d = λ

1

/(λ

1

, b).

Also, (r, λ

1

/d) = (r, (λ

1

, b)) = 1. Because α ∈ n, we know that either

|α − a/(rd)| > r

−1

d

−1

LN

−1

, or rd > M , or both.

Let d be a given divisor of λ

1

. Let N

1

(d, r, a) =

 α :

α − a

rd

≤ r

−1

d

−1

LN

−1

 , N

2

(d, r, a) =

 α :

α − a

rd

≤ r

−1

H

7/4

Q

−3

λ

−11

 .

It is clear from (3.2), (3.8), (3.18), (4.1) that N

2

(d, r, a) contains N

1

(d, r, a).

(11)

Let N

1

(d) denote the union of N

1

(d, r, a) with 1 ≤ a ≤ rd, (a, rd) = (r, λ

1

/d) = 1, (4.24)

r ≤ M d

−1

. (4.25)

Let N

2

(d) denote the union of N

2

(d, r, a) with (4.24) and r ≤ H

7/4

. By the discussion following (4.22) we have, modulo one,

n \ n

1

[

d|λ1

{N

2

(d) \ N

1

(d)}.

Let L(d, r, a) denote N

2

(d, r, a) when M d

−1

< r ≤ H

7/4

, and (4.24) holds;

and denote N

2

(d, r, a) \ N

1

(d, r, a) when r ≤ M d

−1

and (4.24) holds. For fixed d, we have

N

2

(d) \ N

1

(d) = [

r≤H7/4;a

L(d, r, a).

Let α ∈ L(d, r, a). The fraction λ

1

a/(rd) may be written in the form b/r with (b, r) = 1. Since

1

a, rd) = d(λ

1

a/d, r) = d, by Lemma 7, with λ

1

α in place of α, and

λ

1

α − b r = λ

1

α − a

rd

≤ r

−1

H

7/4

Q

−3

1 24rHP

1

, we have

X

h≤H

|F (αhλ

1

, γ; h)|

2

 P

1δ

 r

−1

HP

12

1 + Q

3

1

α − b/r| + HP

1

+ r

 (4.26)

 r

−1

HP

12+δ

1 + N Y

−3

|α − a/(rd)| . Here we require the observations that

Q

3

λ

1

= N Y

−3

, HP

1

+ r  HP

1

 r

−1

HP

12

, (HP

1

+ r)rQ

3

1

α − b/r|  HP

1

H

7/4

 HP

12

.

Choose c, s so that |λ

1

α − c/s| ≤ s

−1

H

3/4

Q

−3

, (c, s) = 1 and s ≤ Q

3

H

−3/4

. If b/r = c/s, then by the definition of L(d, r, a), we know that either

1

α − c/s| > s

−1

d

−1

LN

−1

λ

1

≥ s

−1

LN

−1

,

or s > M d

−1

, or both. By (3.1), (3.2) we have either s > H

3/4

or Q

3

s|λ

1

α − c/s| ≥ H

3/4

, or both. If b/r 6= c/s, then

1 rs

b

r c s

≤ (H

3/4

s

−1

+ H

7/4

r

−1

)Q

−3

1

2sr + H

7/4

r

−1

Q

−3

,

(12)

whence s >

12

Q

3

H

−7/4

> H

3/4

once more. Now we may apply Lemma 8 with λ

1

α in place of α, obtaining

(4.27) X

h≤H

|G

h

1

αh

3

, θγh)|

2

 P

1ε

H

3/4

Y

2

.

Therefore, by (4.26) and Cauchy’s inequality (4.28) X

h≤H

|F (αhλ

1

, γ; h)G

h

1

αh

3

, θγh)|  Y H

7/8

P

11+δ

r

−1/2

(1 + N Y

−3

|α − a/(rd)|)

1/2

. It follows that

(4.29) \

N2(d)\N1(d)

X

h≤H

|F (αhλ

1

, γ; h)G

h

1

αh

3

, θγh)|S(α) dα

 P

11+δ

H

7/8

Y \

N2(d)\N1(d)

r

−1/2



1 + N Y

−3

α − a

rd



−1/2

S(α) dα.

We write

Γ = Γ (α, a, q) = q

1/2+ε

(1 + N Y

−3

|α − a/q|)

1/2

. By Lemma 3, we have

f

A

(W, λ

2

α) = s

A

(q, λ

2

a)J(λ

2

(α − a/q), W ) + O(Γ ) = Γ

1

+ O(Γ ).

Hence by (3.26) we have

|f

A

(W, λ

2

α)|

2

 Γ

12

+ Γ

2

 |s

A

(q, λ

2

a)J(λ

2

(α − a/q), W )|

2

+ Γ

2

 s

2A

(q, λ

2

a)W

2

(1 + N Y

−3

|α − a/q|)

2

+ Γ

2

. Hence

(4.30) \

N2(d)\N1(d)

r

−1/2



1 + N Y

−3

α − a

rd



−1/2

S(α) dα

 \

N2(d)\N1(d)

r

−1/2



1 + N Y

−3

α − a

rd



−1/2

12

+ Γ

2

)

×

X

p31,p32,p33 (3.7)

X

p41,p42,p43 (3.7)

f (R

3

, λ

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

341

p

342

p

343

α)

2

dα.

By Hua’s inequality we have

1

\

0

X

p31,p32,p33

(3.7)

f (R

3

, λ

3

p

331

p

332

p

333

α)

8

dα  (P

3

Y

−1

)

5+ε

,

(13)

1

\

0

X

p41,p42,p43

(3.7)

f (R

4

, λ

4

p

341

p

342

p

343

α)

8

dα  (P

4

Y

−1

)

5+ε

and

1

\

0

X

p31,p32,p33

(3.7)

X

p41,p42,p43

(3.7)

f (R

3

, λ

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

341

p

342

p

343

α)

2





1

\

0

X

p31,p32,p33

(3.7)

f (R

3

, λ

3

p

331

p

332

p

333

α)

4



1/2

×



1

\

0

X

p41,p42,p43

(3.7)

f (R

4

, λ

4

p

341

p

342

p

343

α)

4



1/2

 ((P

3

Y

−1

)

2+ε

(P

4

Y

−1

)

2+ε

)

1/2

. By the definition of Γ ,

Γ

2

= q

1+ε



1 + N Y

−3

α − a

q



= (rd)

1+ε



1 + N Y

−3

α − a

rd

 . For α ∈ N

2

(d) \ N

1

(d),

r

1/2+ε

d

1+ε



1 + N Y

−3

α − a

rd



1/2

 d

1+ε

H

7/8+ε

 λ

1+ε1

H

7/8+ε

. Hence

(4.31) \

N2(d)\N1(d)

r

−1/2



1 + N Y

−3

α − a

rd



−1/2

Γ

2

× | X

p31,p32,p33 (3.7)

X

p41,p42,p43 (3.7)

f (R

3

, λ

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

341

p

342

p

343

α)|

2

 \

N2(d)\N1(d)

r

1/2+ε

d

1+ε



1 + N Y

−3

α − a

rd



1/2

×

X

p31,p32,p33 (3.7)

X

p41,p42,p43 (3.7)

f (R

3

, λ

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

341

p

342

p

343

α)

2

 λ

1+ε1

H

7/8+ε

(P

3

P

4

Y

−2

)

1+ε

.

(14)

Now we consider

(4.32) \

N2(d)\N1(d)

Γ

14

dα = \

N2(d)\N1(d)

s

4A

(q, λ

2

a)W

4

(1 + N Y

−3

|α − a/(rd)|)

4

 W

4

X

r≤H7/4

X

a≤rd

\

L(d,r,a)

s

4A

(q, λ

2

a)

(1 + N Y

−3

|α − a/(rd)|)

4

dα.

Let κ(q) and κ

(q) be the multiplicative functions defined by κ(p

3k

) = p

−k

, κ(p

3k+1

) = 2p

−k−1/2

, κ(p

3k+2

) = p

−k−1

, κ

(p) = 2p

−1/2

, κ

(p

2

) = p

−3/4

, κ

(p

l

) = p

−l/3

(l ≥ 3).

Then κ(q) ≤ κ

(q), and by Lemmas 4.3, 4.4 and Theorem 4.2 of Vaughan [5], q

−1

S(q, a)  κ(q) when (a, q) = 1.

Now we consider X

q≤Ξ

 q

(q, λ

2

)



4

 Y

p≤Ξ

 1 +

X

l=1

p

l

κ

 p

l

(p

l

, λ

2

)



4



 Y

p-λ2 p≤Ξ



1 + 16p

−1

+ p

−1

+ X

l=3

p

−l/3



× Y

pkλ2

p≤Ξ



1 + p + 16 + 1 + 1 + X

l=1

p

−l/3



× Y

p22 p≤Ξ



1 + p + p

2

+ 16p + p + p + p

2/3

+ p

1/3

+ X

l=0

p

−l/3

 .

We have Y

pkλ2 p≤Ξ

(20 + p)  Y

pkλ2 p≤C(ε)

(20 + p) Y

pkλ2 p≥C(ε)

(20 + p)  Y

pkλ2

p

1+ε

.

Here we use the fact that for p ≥ C(ε) we have 20 + p ≤ p

1+ε

. Similarly Y

p22

p≤Ξ

(p

2

+ 22p + 1)  Y

p22

p

2+ε

.

(15)

The constant implied by  depends only on ε. So we have

(4.33) X

q≤Ξ

 q

(q, λ

2

)



4

 λ

1+ε2

(log Ξ)

20

. By (3.9) of [1],

s

A

(q, λ

2

a)  q

ε

κ

 q

(q, λ

2

a)



 q

ε

κ

 q

(q, λ

2

)

 , so we have

(4.34) \

N2(d)\N1(d)

Γ

14

dα  W

4

Y

3

N

−1+ε

λ

2

.

For α ∈ N

2

(d) \ N

1

(d), we have r

−1/2



1 + N Y

−3

α − a

rd



−1/2

 M

−1/2

d

1/2

. Hence by Schwarz’s inequality,

(4.35) \

N2(d)\N1(d)

r

−1/2



1 + N Y

−3

α − a

rd



−1/2

Γ

12

×

X

p31,p32,p33

(3.7)

X

p41,p42,p43

(3.7)

f (R

3

, λ

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

341

p

342

p

343

α)

2

 (W

4

Y

3

N

−1+ε

λ

2

)

1/2

M

−1/2

d

1/2

(P

3

Y

−1

)

(5+ε)/4

(P

4

Y

−1

)

(5+ε)/4

. By (4.29)–(4.35) we have

(4.36) \

n\n1

X

h≤H

|F (αhλ

1

, γ; h)G

h

1

αh

3

, θγh)|S(α) dα

 P

11+δ

H

7/8

Y [λ

1+ε1

H

7/8+ε

(P

3

P

4

Y

−2

)

1+ε

+ (W

4

Y

3

N

−1+ε

λ

2

)

1/2

M

−1/2

d

1/2

(P

3

P

4

Y

−2

)

(5+ε)/4

]

 λ

1+ε1

P

12+ε

P

31+ε

P

41+ε

Y

−1

+ λ

1/21

P

11+ε

P

21/2+ε

P

35/4+ε

P

45/4+ε

Y

−1/2

. By (4.22), (4.36) the lemma follows.

As in Section 8 of [1], for the estimation of P

p,(3.6)

T

2

(p) we have the same upper bound for T

5

(γ, θ).

5. First steps on the major arcs. In this section we show that R(λ

1

, . . . , λ

7

) is well approximated by integrals similar to (3.17), but with f

p

(P

1

, λ

1

α) and f

A

(W, λ

2

p

3

α) replaced by suitable approximations. Let

D(α) = q

1/2

(1 + N |α − a/q|)

1/2

.

(16)

By Lemma 4,

f

p

(P

1

, λ

1

α) − s(p, α)  q

1/2+ε

1/2

 D(α)q

ε

, f

A

(W, λ

2

α) − g(p, α)  q

1/2+ε

1/2

 D(α)q

ε

. Here

∆ = 1 + N |α − a/q|.

Now we introduce the numbers v

1

= \

M

X

p (3.6)

X

p31,p32,p33 (3.7)

X

p41,p42,p43 (3.7)

(f

p

(P

1

, λ

1

α) − s(p, α))f

A

(W, λ

2

p

3

α)

× f (R

3

, λ

3

p

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

3

p

341

p

342

p

343

α)

2

dα, v

2

= \

M

X

p (3.6)

X

p31,p32,p33

(3.7)

X

p41,p42,p43

(3.7)

s(p, α)(f

A

(W, λ

2

α) − g(p, α))

× f (R

3

, λ

3

p

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

3

p

341

p

342

p

343

α)

2

dα.

We use Cauchy’s inequality and note that D(α)  L

1/2

for α ∈ M . Hence by (4.17),

v

1

 Y L

1+ε

X

Y <p≤2Y 1

\

0

X

p31,p32,p33

(3.7)

X

p41,p42,p43

(3.7)

f

A

(W, λ

2

p

3

α) (5.1)

× f (R

3

, λ

3

p

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

3

p

341

p

342

p

343

α)

2

 Y

2

L

1+ε

1

\

0

X

p31,p32,p33 (3.7)

X

p41,p42,p43 (3.7)

f

A

(W, λ

2

α)

× f (R

3

, λ

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

341

p

342

p

343

α)

2

 Y

2

L

1+ε

T

3

 L

1+ε

P

2

P

3

P

42+ε

Y

−1−2θ

.

Now we consider v

2

. By Cauchy’s inequality and Schwarz’s inequality we have

v

2

 Y P

ε

X

Y <p≤2Y

\

M

s(p, α)D(α)

× X

p31,p32,p33 (3.7)

X

p41,p42,p43 (3.7)

f (R

3

, λ

3

p

3

p

331

p

332

p

333

α)f (R

4

, λ

4

p

3

p

341

p

342

p

343

α)

2

 Y P

ε

(I

1

I

2

)

1/2

,

Cytaty

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