LXXXI.3 (1997)
Diagonal cubic equations
by
Hongze Li (Jinan)
1. Introduction. Let λ
jbe positive integers. We shall be concerned with solutions of the diophantine equation
(1.1) λ
1x
31+ λ
2x
32+ . . . + λ
7x
37= 0
which are small in the following sense. Let Λ = λ
1λ
2. . . λ
7. Then we seek for solutions to (1.1) which satisfy
(1.2) 0 <
X
7 i=1λ
i|x
i|
3Λ
gwith 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]
(1.6) G(α, X) = X
M1<p1≤2M1
X
M2<p2≤2M2
g(α(p
1p
2)
3, Y ), where, here and throughout, p, p
idenote prime numbers.
Lemma 1. In the notation introduced above, for any ε > 0 we have
1
\
0
|G(α, X)|
6dα X
3+2θ1+ε. This is the case l = 3 of the Theorem in Br¨ udern [2].
2. Preliminaries. Now define (2.1) χ
3(X)
= {p
1p
2y : 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
1x
31+ µ
2p
3(y
13+ y
23) = µ
1x
32+ µ
2p
3(y
33+ y
34) in integers x
1, x
2, y
1, y
2, y
3, y
4and primes p satisfying
B < x
1, x
2≤ 2B, p - µ
1x
1x
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
2be positive integers, and (2.6) µ
2E
3µ
1B
3µ
2E
3, µ
1µ
2, E
θ≤ B
1/7. Then
(2.7) S(µ
1, µ
2, B, E) B
1+εE
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
εS
0, where S
0is 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
2contribute to S
0at most BE
θE
2(1−θ)+ε= BE
2−θ+ε. For the solutions with x
16= x
2we again follow through the argument in Vaughan [7] and find that there are at most 2S
1remaining solutions where S
1equals the number of solutions to
µ
1h(3x
2+ h
2p
6) = 4µ
2(y
13+ y
23− y
33− y
43),
where x ≤ 4B, E
θ< p ≤ 2E
θ, 0 < h ≤ H, y
i∈ χ
3(E
1−θ), and H = CBE
−3θfor some constant C. On writing
Ω(α) = X
Eθ<p≤2Eθ
X
h≤H
X
x≤4B
e(αh(3x
2+ h
2p
6))
we have, by H¨older’s inequality, S
1=
1
\
0
Ω(µ
1α)|G(4µ
2α, E
1−θ)|
4dα
≤
1\
0
|Ω(µ
1α)|
3dα
1/31\
0
|G(α, E
1−θ)|
6dα
2/3.
The argument of Vaughan [8], p. 39 shows that for B
1/8< E
θ≤ B
1/7we have
1
\
0
|Ω(α)|
3dα H
2(E
θB)
3/2+ε. Since Ω(α) has period 1, we have
1
\
0
|Ω(µ
1α)|
3dα = 1 µ
1µ
\
10
|Ω(α)|
3dα =
1
\
0
|Ω(α)|
3dα and by Lemma 1 we have
1
\
0
|G(α, E
1−θ)|
6dα (E
1−θ)
3+2θ1+ε. 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
7are 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
0Λ
g, P = N
1/3, P
j= P λ
−1/3j,
where C
0is 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
1x
3− λ
2p
3y
3+ λ
3p
3p
331p
332p
333z
33+ λ
4p
3p
341p
342p
343z
43+ λ
5u
3+ λ
6p
361p
362p
363v
63+ λ
7p
371p
372p
373v
37= 0
in integers x, y, z
3, z
4, u, v
6, v
7and primes p, p
ijsatisfying (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
1P
11/7, W = P
2Y
−1, R
i= 1
200 (P
iY
−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
iY
−1)
233/1815, Z
i2= 2
−i(P
iY
−1)
210/1815,
Z
i3= 2
−i(P
iY
−1)
196/1815(i = 3, 4), (3.11) Z
i1= 2
−iP
i233/1815, Z
i2= 2
−iP
i210/1815,
Z
i3= 2
−iP
i196/1815(i = 6, 7), where C
1is sufficiently large.
The inequalities
(3.12) N λ
1P
13, λ
2Y
3W
3, λ
3Y
3Z
313Z
323Z
333R
33, λ
4Y
3Z
413Z
423Z
433R
34, λ
5U
3, λ
6Z
613Z
623Z
633V
63, λ
7Z
713Z
723Z
733V
73N 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
73and let
(3.14) A =
Y
4 i=3Y
3 j=1p
ij, B = Y
7 i=6Y
3 j=1p
ij,
(3.15) F (p, α) = f
p(P
1, λ
1α)f
A(W, λ
2p
3α)f (R
3, λ
3p
3p
331p
332p
333α)
× f (R
4, λ
4p
3p
341p
342p
343α)
× f
B(U, λ
5α)f (V
6, λ
6p
361p
362p
363α)f (V
7, λ
7p
371p
372p
373α),
(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
−1LN
−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
2/211Λ
−1/3−εN
26/21(logN )
−13, (3.20)
\
m
F (α) dα λ
2/211Λ
−1/3N
26/21−ε. (3.21)
It follows from (3.19)–(3.21) and (3.17) that
(3.22) R(λ
1, . . . , λ
7) λ
2/211Λ
−1/3−εN
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=1e(br
3/q), J(β, X) =
2X
\
X
e(βx
3) dx.
The estimate
(3.24) S(q, b) q
1+εψ(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.
Lemma 3. Let d be an integer with 1 divisors. Let (3.27) s
d(q, c) = q
−1X
b|d
µ(d) S(q, b
3c) 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
1a)J(λ
1(α − a/q), P
1), (3.29)
g(p, α) = s
A(q, λ
2p
3a)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
1α) − s(p, α) q
1/2+ε∆
1/2, (3.32)
f
A(W, λ
2p
3α) − 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
2P
1Y
−3, Q = P
1Y
−1,
where C
2is 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
2α)
× f (R
3, λ
3p
331p
332p
333α)f (R
4, λ
4p
341p
342p
343α)
2
and
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
1α
4 (3hy
2+ h
3p
6)
.
Let n denote the set of real numbers in (0, 1] with the property that whenever |α − a/q| ≤ q
−1LN
−1and (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
5α)
× f (V
6, λ
6p
361p
362p
363α)f (V
7, λ
7p
371p
372p
373α)
2dα.
Then J
1P
5P
62θP
72+ε.
P r o o f. By considering the underlying diophantine equation we have J
11
\
0
X
p61,p62,p63
(3.7)
X
p71,p72,p73
(3.7)
f
p61p71(U, λ
5α)f (V
6, λ
6p
361p
362p
363α)
× f (V
7, λ
7p
371p
372p
373α)
2dα.
From the inequality |zw| ≤
12|z|
2+
12|w|
2, we have J
1X
7 k=6 1\
0
X
p61
(3.7)
X
p71
(3.7)
|f
p61p71(U, λ
5α)|
X
pk2,pk3
(3.7)
f (V
k, λ
kp
3k1p
3k2p
3k3α)
2 2dα.
Hence, by Cauchy’s inequality, J
1Z
61Z
71X
7 k=6 1\
0
X
p61
(3.7)
X
p71
(3.7)
|f
p61p71(U, λ
5α)|
2×
X
pk2,pk3 (3.7)
f (V
k, λ
kp
3k1p
3k2p
3k3α)
4dα
Z
712Z
61S(λ
5, λ
6, P
5, P
6) + Z
612Z
71S(λ
5, λ
7, P
5, P
7).
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
1P
5P
62θP
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
1α)f
A(W, λ
2p
3α)
× f (R
3, λ
3p
3p
331p
332p
333α)f (R
4, λ
4p
3p
341p
342p
343α)
2dα.
Then J
2Y 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
34βy
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
34λ
1αhy
2, (4.8)
Ψ
p(α) = 2Re X
h≤H
F
p(α; h)e
14λ
1αh
3p
6, (4.9)
D
p(α; h) = X
2P1/p+hp2<y≤4P1/p−hp2 y≡h (mod 2)
e
34λ
1αhp
2y
2, (4.10)
Ξ
p(α) = 2Re X
h≤H
D
p(α; h)e
14λ
1αh
3p
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)
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
1Y T
3)
from (4.5).
We estimate T
3via Lemma 2. By (3.1), (3.2), (3.8) and (3.10) we have (P
3Y
−1)
θ≤ (P
4Y
−1)
θ≤ (P
2Y
−1)
1/7.
Just as in Lemma 5, we have
(4.17) T
3(P
2Y
−1)(P
3Y
−1)
2θ(P
4Y
−1)
2+εP
2P
32θP
42+εY
−3−2θ. Let
(4.18) T
5(γ, θ) = \
n
X
h≤H
|F (αλ
1h, γ; h)G
h(αλ
1h
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)/21Λ
−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)|
2P
1δq
−1HP
121 + 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
−1Q
−3H
3/4, (a, q) = 1, q ≤ Q
3H
−3/4. Then
X
h≤H
|G
h(αh
3, γh)|
2P
1εHY
2q
−1/3(1 + Q
3|α − a/q|)
1/3+ H
3/4Y
2.
This is essentially Lemma 8 of [7].
Lemma 9. Let γ ∈ R and θ ∈ {−1, 1}. Then
T
5(γ, θ) P
11+δP
2P
32θP
42+εY
−2−2θ+ λ
1+ε1P
12+εP
31+εP
41+εY
−1+ λ
1/21P
11+εP
21/2+εP
35/4+εP
45/4+εY
−1/2.
P r o o f. Let n
1denote the set of α in n with the property that whenever (4.20) |λ
1α − r/b| ≤ r
−1H
7/4Q
−3, (b, r) = 1,
one has r > H
7/4. Let α ∈ n
1. We apply Lemma 7 with λ
1α in place of α. By Dirichlet’s theorem there exist integers b, r satisfying (4.20) with r ≤ Q
3H
−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)|
2P
1δ(P
12H
−3/4+ HP
1+ Q
3H
−7/4) (4.21)
P
12+δH
−3/4.
We may choose c, s so that (c, s) = 1, s ≤ Q
3H
−3/4, and
|λ
1α − c/s| ≤ s
−1H
3/4Q
−3.
Then |λ
1α − c/s| ≤ s
−1H
7/4Q
−3, so that s > H
7/4> H
3/4. By Lemma 8, applied to λ
1α in place of α, we have
X
h≤H
|G
h(λ
1αh
3, θγh)|
2P
1εH
3/4Y
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(λ
1αh
3, θγh)|S(α) dα
P
11+δY T
3P
11+δP
2P
32θP
42+εY
−2−2θ. It remains to consider n \ n
1. Let α ∈ n \ n
1. There are integers b, r satisfying
(4.23) |λ
1α − b/r| ≤ r
−1H
7/4Q
−3, (b, r) = 1, r ≤ H
7/4.
We write b/(λ
1r) = 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
−1d
−1LN
−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
−1d
−1LN
−1, N
2(d, r, a) =
α :
α − a
rd
≤ r
−1H
7/4Q
−3λ
−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).
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
−1and (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
1a/(rd) may be written in the form b/r with (b, r) = 1. Since
(λ
1a, rd) = d(λ
1a/d, r) = d, by Lemma 7, with λ
1α in place of α, and
λ
1α − b r = λ
1α − a
rd
≤ r
−1H
7/4Q
−3≤ 1 24rHP
1, we have
X
h≤H
|F (αhλ
1, γ; h)|
2P
1δr
−1HP
121 + Q
3|λ
1α − b/r| + HP
1+ r
(4.26)
r
−1HP
12+δ1 + N Y
−3|α − a/(rd)| . Here we require the observations that
Q
3λ
1= N Y
−3, HP
1+ r HP
1r
−1HP
12, (HP
1+ r)rQ
3|λ
1α − b/r| HP
1H
7/4HP
12.
Choose c, s so that |λ
1α − c/s| ≤ s
−1H
3/4Q
−3, (c, s) = 1 and s ≤ Q
3H
−3/4. If b/r = c/s, then by the definition of L(d, r, a), we know that either
|λ
1α − c/s| > s
−1d
−1LN
−1λ
1≥ s
−1LN
−1,
or s > M d
−1, or both. By (3.1), (3.2) we have either s > H
3/4or Q
3s|λ
1α − c/s| ≥ H
3/4, or both. If b/r 6= c/s, then
1 rs ≤
b
r − c s
≤ (H
3/4s
−1+ H
7/4r
−1)Q
−3≤ 1
2sr + H
7/4r
−1Q
−3,
whence s >
12Q
3H
−7/4> H
3/4once more. Now we may apply Lemma 8 with λ
1α in place of α, obtaining
(4.27) X
h≤H
|G
h(λ
1αh
3, θγh)|
2P
1εH
3/4Y
2.
Therefore, by (4.26) and Cauchy’s inequality (4.28) X
h≤H
|F (αhλ
1, γ; h)G
h(λ
1αh
3, θγh)| Y H
7/8P
11+δr
−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(λ
1αh
3, θγh)|S(α) dα
P
11+δH
7/8Y \
N2(d)\N1(d)
r
−1/21 + N Y
−3α − a
rd
−1/2S(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
2α) = s
A(q, λ
2a)J(λ
2(α − a/q), W ) + O(Γ ) = Γ
1+ O(Γ ).
Hence by (3.26) we have
|f
A(W, λ
2α)|
2Γ
12+ Γ
2|s
A(q, λ
2a)J(λ
2(α − a/q), W )|
2+ Γ
2s
2A(q, λ
2a)W
2(1 + N Y
−3|α − a/q|)
2+ Γ
2. Hence
(4.30) \
N2(d)\N1(d)
r
−1/21 + N Y
−3α − a
rd
−1/2S(α) dα
\
N2(d)\N1(d)
r
−1/21 + N Y
−3α − a
rd
−1/2(Γ
12+ Γ
2)
×
X
p31,p32,p33 (3.7)
X
p41,p42,p43 (3.7)
f (R
3, λ
3p
331p
332p
333α)f (R
4, λ
4p
341p
342p
343α)
2dα.
By Hua’s inequality we have
1
\
0
X
p31,p32,p33
(3.7)
f (R
3, λ
3p
331p
332p
333α)
8dα (P
3Y
−1)
5+ε,
1
\
0
X
p41,p42,p43
(3.7)
f (R
4, λ
4p
341p
342p
343α)
8dα (P
4Y
−1)
5+εand
1
\
0
X
p31,p32,p33
(3.7)
X
p41,p42,p43
(3.7)
f (R
3, λ
3p
331p
332p
333α)f (R
4, λ
4p
341p
342p
343α)
2dα
1
\
0
X
p31,p32,p33
(3.7)
f (R
3, λ
3p
331p
332p
333α)
4dα
1/2×
1\
0
X
p41,p42,p43
(3.7)
f (R
4, λ
4p
341p
342p
343α)
4dα
1/2((P
3Y
−1)
2+ε(P
4Y
−1)
2+ε)
1/2. By the definition of Γ ,
Γ
2= q
1+ε1 + N Y
−3α − a
q
= (rd)
1+ε1 + N Y
−3α − a
rd
. For α ∈ N
2(d) \ N
1(d),
r
1/2+εd
1+ε1 + N Y
−3α − a
rd
1/2d
1+εH
7/8+ελ
1+ε1H
7/8+ε. Hence
(4.31) \
N2(d)\N1(d)
r
−1/21 + N Y
−3α − a
rd
−1/2Γ
2× | X
p31,p32,p33 (3.7)
X
p41,p42,p43 (3.7)
f (R
3, λ
3p
331p
332p
333α)f (R
4, λ
4p
341p
342p
343α)|
2dα
\
N2(d)\N1(d)
r
1/2+εd
1+ε1 + N Y
−3α − a
rd
1/2×
X
p31,p32,p33 (3.7)
X
p41,p42,p43 (3.7)
f (R
3, λ
3p
331p
332p
333α)f (R
4, λ
4p
341p
342p
343α)
2dα
λ
1+ε1H
7/8+ε(P
3P
4Y
−2)
1+ε.
Now we consider
(4.32) \
N2(d)\N1(d)
Γ
14dα = \
N2(d)\N1(d)
s
4A(q, λ
2a)W
4(1 + N Y
−3|α − a/(rd)|)
4dα
W
4X
r≤H7/4
X
a≤rd
\
L(d,r,a)
s
4A(q, λ
2a)
(1 + N Y
−3|α − a/(rd)|)
4dα.
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
−1S(q, a) κ(q) when (a, q) = 1.
Now we consider X
q≤Ξ
qκ
∗q
(q, λ
2)
4Y
p≤Ξ
1 +
X
∞ l=1p
lκ
∗p
l(p
l, λ
2)
4Y
p-λ2 p≤Ξ
1 + 16p
−1+ p
−1+ X
∞ l=3p
−l/3× Y
pkλ2
p≤Ξ
1 + p + 16 + 1 + 1 + X
∞l=1
p
−l/3× Y
p2kλ2 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
p2kλ2
p≤Ξ
(p
2+ 22p + 1) Y
p2kλ2
p
2+ε.
The constant implied by depends only on ε. So we have
(4.33) X
q≤Ξ
qκ
∗q
(q, λ
2)
4λ
1+ε2(log Ξ)
20. By (3.9) of [1],
s
A(q, λ
2a) q
εκ
q
(q, λ
2a)
q
εκ
∗q
(q, λ
2)
, so we have
(4.34) \
N2(d)\N1(d)
Γ
14dα W
4Y
3N
−1+ελ
2.
For α ∈ N
2(d) \ N
1(d), we have r
−1/21 + N Y
−3α − a
rd
−1/2M
−1/2d
1/2. Hence by Schwarz’s inequality,
(4.35) \
N2(d)\N1(d)
r
−1/21 + N Y
−3α − a
rd
−1/2Γ
12×
X
p31,p32,p33
(3.7)
X
p41,p42,p43
(3.7)
f (R
3, λ
3p
331p
332p
333α)f (R
4, λ
4p
341p
342p
343α)
2dα
(W
4Y
3N
−1+ελ
2)
1/2M
−1/2d
1/2(P
3Y
−1)
(5+ε)/4(P
4Y
−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(λ
1αh
3, θγh)|S(α) dα
P
11+δH
7/8Y [λ
1+ε1H
7/8+ε(P
3P
4Y
−2)
1+ε+ (W
4Y
3N
−1+ελ
2)
1/2M
−1/2d
1/2(P
3P
4Y
−2)
(5+ε)/4]
λ
1+ε1P
12+εP
31+εP
41+εY
−1+ λ
1/21P
11+εP
21/2+εP
35/4+εP
45/4+εY
−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
1α) and f
A(W, λ
2p
3α) replaced by suitable approximations. Let
D(α) = q
1/2(1 + N |α − a/q|)
1/2.
By Lemma 4,
f
p(P
1, λ
1α) − s(p, α) q
1/2+ε∆
1/2D(α)q
ε, f
A(W, λ
2α) − g(p, α) q
1/2+ε∆
1/2D(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
1α) − s(p, α))f
A(W, λ
2p
3α)
× f (R
3, λ
3p
3p
331p
332p
333α)f (R
4, λ
4p
3p
341p
342p
343α)
2dα, 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
2α) − g(p, α))
× f (R
3, λ
3p
3p
331p
332p
333α)f (R
4, λ
4p
3p
341p
342p
343α)
2dα.
We use Cauchy’s inequality and note that D(α) L
1/2for α ∈ M . Hence by (4.17),
v
1Y L
1+εX
Y <p≤2Y 1
\
0
X
p31,p32,p33
(3.7)
X
p41,p42,p43
(3.7)
f
A(W, λ
2p
3α) (5.1)
× f (R
3, λ
3p
3p
331p
332p
333α)f (R
4, λ
4p
3p
341p
342p
343α)
2dα
Y
2L
1+ε1
\
0
X
p31,p32,p33 (3.7)
X
p41,p42,p43 (3.7)
f
A(W, λ
2α)
× f (R
3, λ
3p
331p
332p
333α)f (R
4, λ
4p
341p
342p
343α)
2dα
Y
2L
1+εT
3L
1+εP
2P
32θP
42+εY
−1−2θ.
Now we consider v
2. By Cauchy’s inequality and Schwarz’s inequality we have
v
2Y P
εX
Y <p≤2Y
\
M
s(p, α)D(α)
× X
p31,p32,p33 (3.7)
X
p41,p42,p43 (3.7)