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arXiv:1511.03409v2 [math.NT] 17 Dec 2015
Explicit Chen’s theorem
Tomohiro Yamada
Abstract
We show that every even number > exp exp 36 can be represented
as the sum of a prime and a product of at most two primes.
1 Introduction
In a letter of 1742 to Euler, Goldbach conjectured that every integer greater
then 2 is the sum of three primes including 1, which is equivalent that every
even integer N ≥ 4 is the sum of two primes (not including 1) or of the
form p + 3 with p prime.
Euler replied that this is equivalent to the statement that every even
integer N ≥ 4 is the sum of two primes.
An weaker conjecture is that every odd integer N ≥ 7 can be represented
as the sum of three primes. Vinogradov[19][20][14, Chapter 8] showed that
every sufficiently large odd integer can be represented as the sum of three
primes. His student K. Borozdin[1] proved that 3
315 is large enough. Chen
and Wang[4] reduced the constant to exp exp 11.503, Chen and Wang[5]
to exp exp 9.715 and Liu and Wang[12] to exp 3100. Deshouillers, Effinger,
te Riele and Zinoviev[6] showed that the Generalized Riemann Hypothesis
gives the weaker conjecture. Recently, Harald Helfgott claimed to have
proved three prime conjecture unconditionally.
Contrastly, the ordinary Goldbach’s conjecture is still unsolved. A well-
known partial result is the theorem of Chen[2][3], who proved that every
sufficiently large even number can be represented as the sum of a prime
and the product of at most two primes. Ross[16] gave a simpler proof.
∗ 2010 Mathematics Subject Classification: 11N35. † Key words and phrases: Linear sieve, Rosser-Iwaniec sieve.
1 INTRODUCTION 2
Nathanson[14, Chapter 10] gave another proof based on Iwaniec’s unpub-
lished lecture note. However, they did not give an explicit constant above
which every even number can be represented as p + P 2. The purpose of this
paper is to give an explicit constant for Chen’s theorem; every even number
exp exp 36 can be represented as the sum of a prime and a product of at
most two primes. Indeed, we shall prove the following result:
Theorem 1.1. Let π 2 (N) denote the number of representations of a given
integer N as the sum of a prime and a product of at most two primes. If
N is an even integer > exp exp 36, then we have
π 2 (N) >
0. 007 UN N
log
2 N
where
UN = 2e
−γ
p> 2
(p − 1)
2
p> 2 ,p|N
p − 1
p − 2
Our argument is based on Nathanson’s one, which used Rosser-Iwaniec
linear sieve to give upper and lower bounds for numbers of sifted primes,
combining explicit error terms for the disribution of primes in arithmetic
progressions and explicit Rosser-Iwaniec linear sieve, which are given in
other papers by the author [21][22].
However, possible existence of a Siegel zero prevents from making the
size of error term in Rosser-Iwaniec linear sieve explicit. There are two
cases — the exceptional modulus is large or small. If the exceptional mod-
ulus is small, then we can see that the contribution of the Siegel zero can be
absorbed into error estimates concerning the distribution of primes in arith-
metic progression (see Lemma 2.5). In the other case, when the exceptional
modulus is large, it is easy to avoid a possible Siegel zero in the argument
to estimate upper bounds since we can exclude a prime dividing the excep-
tional modulus from sifting primes. However, we cannot directly avoid a
possible Siegel zero in the argument to estimate lower bounds. In order to
overcome this obstacle, we use a variant of inclusion-exclusion principle and
both upper bound and lower bound sieves, as performed in Section 5. Thus
we can obtain explicit bounds.
So that, our argument can be divided into four parts: error estimates
involving the number of primes in arithmetic progressions based on esti-
mates in [21], explicit error terms in Rosser-Iwaniec linear sieve shown in
[22], upper bounds and lower bounds for various sets of sifted primes, and
the final conclusion.
For calculations of constants, we used PARI-GP. Our script is available
from http://tyamada1093.web.fc2.com/math/files/prim0009pari.txt
2 PRELIMINARY RESULTS 4
Henthforth we shall give explicit estimates for various quantities involv-
ing the error terms concerning to the number of primes in arithmetic pro-
gressions. Let Ef (x; k, l) denote the error function f (x; k, l) −
f (x) ϕ(k)
for f = π
(i.e. f (x) = π(x) and f (x; k, l) = π(x; k, l)), θ or ψ.
Lemma 2.3. Let x > X 1 = exp exp 11. 7 and k < log
10 x be an integer.
Let E 0 = 1 and β 0 denote the Siegel zero modulo k if it exists and E 0 = 0
otherwise. Then we have
ϕ(q)
x
|Eψ (x; k, l)| <
log
8 x
+ E 0
x
β 0 − 1
β 0
Proof. This is Theorem 1.1 of [21] with (α 1 , α, Y 0 ) = (10, 8 , 11 .7).
Define Π(s, q) =
χ (mod q) L(s, χ) and let^ R^0 = 6.397 and^ R^1 = 2.^0452 · · ·^.
Theorem 1.1 of Kadiri[11] states that the function Π(s, q) has at most one
zero ρ = β + it in the region 0 ≤ β < 1 − 1 /R 0 log max{q, q |t|}, which
must be real and simple and induced by some nonprincipal real primitive
character χ˜ (mod ˜q) with 987 ≤ q˜ ≤ x. Moreover, Theorem 1.3 of [11] im-
plies that, for any given Q 1 , such zero satisfies β < 1 − 1 / 2 R 1 log Q 1 except
possibly one modulus below Q 1. Henceforth let k 0 be a such a modulus if it
exists and call this modulus and the corresponding character to be excep-
tional. Furthermore, we set δ = 7/5 and define k 1 = k 0 if k 0 ≥ log
δ x and
k 1 = 0 otherwise, so that k 1 | k is equivalent to both k 0 | k and k 0 ≥ log
δ x
hold.
Corollary 2.4. Assume that x is a real number > X 2 = exp exp 32, x 2 =
e−^100 x log^4 x
, K 0 = log
δ x 2 and let Q 1 = log
10 x 2. Moreover, let k 1 = k 0 if k 0 ≥ K 0
and k 1 = 0 otherwise, If k is a modulus ≤ Q 1 not divisible by k 1 , then we
have
ϕ(k)
x
|Eπ(x; k, l)| <
e
− 14
log
4 x
Proof. We begin by observing that β 0 ≤ 1 −π/ 0. 4923 K
1 / 2 0 log
2 K 0. It is clear
that either k 0 ∤ k or k 0 < log
δ x 2 holds if there exists the Siegel zero. In the
case k 0 ∤ k, it is clear that β 0 < 1 − 1 / 2 R 1 log Q 1 = 1 − 1 / 20 R 1 log log x 2 <
1 − π/ 0. 4923 K
1 / 2 0 log
2 K 0. In the other case k 0 < K 0 , Theorem 3 of [12]
gives β 0 ≤ 1 − π/ 0. 4923 k
1 / 2 0 log
2 k 0 ≤ 1 − π/ 0. 4923 K
1 / 2 0 log
2 K 0.
Let y be an arbitrary real number with x 2 < y ≤ x. Since 1/ 2 < β 0 ≤
1 − π/ 0. 4923 K
1 / 2 0 log
2 K 0 , we can see that
yβ^0 −^1 β 0
e−^15 log^3 y
from x > X 2 if the
Siegel zero exists. Thus, by Lemma 2.3,
ϕ(q)
y
ψ(y; k, l) −
y
ϕ(k)
- 1 e
− 15
log
3 y
2 PRELIMINARY RESULTS 5
The rough estimate |ψ(y; k, l) − θ(y; k, l)| < y
1 (^2) log
2 y/ log 2 is enough to
give
ϕ(k)
y
θ(y; k, l) −
y
ϕ(k)
- 2 e
− 15
log
3 y
Since
ϕ(k)
θ(y)
y
- 1 e
− 15
log
3 y
for y > X 2 by Theorem 2 of [18], we have
ϕ(k)
y
|Eθ(y; k, l)| <
- 3 e
− 15
log
3 y
Now, partial summation yields
Eπ(x; k, l) = Eπ(x 2 ; k, l) +
Eθ(x; k, l)
log x
Eθ(x 2 ; k, l)
log x 2
∫ (^) x
x 2
Eθ (t; k, l)
t log
2 t
dt, (14)
and
ϕ(k)
∫ (^) x
x 2
Eθ (t; k, l)
t log
2 t
dt <
- 3 e
− 15
5 log x 2
∫ (^) x
x 2
log
5 t
log t
dt
- 3 e
− 15
5 log x 2
x
log
5 x
x 2
log
5 x 2
Hence we obtain
|Eπ(x; k, l)| <
- 3 e
− 15 x
log
4 x
- 6 e
− 15 x 2
log
4 x 2
− 15
x
log
5 x
x 2
log
5 x 2
The right-hand side does not exceed
e−^14 log^4 x
if x > X 2 , which proves the
corollary.
Lemma 2.5. Assume that x is a any real number > X 2. Let x 2 , Q 1 , k 1 be
as in Corollary 2.4 and Q =
√ x 2
log^10 x 2
. Then we have
k≤Q,k 1 ∤k
μ
2 (k) max l (mod k)
|Eπ(x; k, l)| <
e
− 8 x
log
3 x
Proof. Let y be an arbitrary real number with x 2 < y ≤ x. We begin by
showing that
k≤Q,k 1 ∤k
μ
2 (k) max l (mod k)
|Eψ (y; k, l)| <
- 9 e
− 9 y
log
2 y
2 PRELIMINARY RESULTS 7
where β 0 denotes the Siegel zero modulo k 0. Theorem 3 of [12] gives β 0 ≤
1 − π/ 0. 4923 k
1 / 2 0 log
2 k 0 ≤ 1 − π/ 0. 4923 K
1 / 2 0 log
2 K 0 and we see that
y
1 − π
- 4923 K
1 2 0 log
(^2) K 0
π
- 4923 K
1 2 0 log
(^2) K 0
e
− 11. 5 y
log
3 y log log y
A similar argument to the proof of Theorem 1.4 of [21] using this inequality
instead of (52) in [21] gives (18). Similarly to (53), for each k ≤ Q 1 , we
have
χ (mod k)
|ψ(y, χ)| <
(C 0 + e
− 18 )y
log
7 y
e
− 11. 5 y
log
3 y log log y
2 e
− 12 y
log
3 y log log y
and, similarly to (54) in [21], we obtain
q≤Q 1 ,q 0 ∤q
ϕ(q)
χ (mod q)
|ψ(x, χ)| ≤
- 01 c 1 e
− 12 y(1 + 10 log log y)
log
3 y log log y
- 95 e
− 9 y
log
3 y
This gives
k≤Q,k 1 ∤k
μ
2 (k) max l (mod k)
|Eψ (y; k, l)| <
- 95 c 1 e
− 9 y
2 log
2 y
- 9 e
− 9 y
log
2 y
Now we estimate Eπ(x; k, l). We begin by observing that partial sum-
mation gives
Eπ(x; k, l) = Eπ(x 2 ; k, l) +
Eθ(x; k, l)
log x
Eθ(x 2 ; k, l)
log x 2
∫ (^) x
x 2
Eθ (t; k, l)
t log
2 t
dt. (29)
We would like to majorize the four terms in the right-hand side.
From the argument in the proof of Theorem A. 17 of [14], we see that ∑
k≤Q
1 ϕ(k)
< c 1 (1 + log Q) <
c 1 2
log x and therefore (18) yields
k≤Q,k 1 ∤k
μ
2 (k) max l (mod k)
|Eθ(y; k, l)|
k≤Q,k 1 ∤k
μ
2 (k) max l (mod k)
|Eψ (y; k, l)|
2 y
1 (^2) log
2 y
log 2
k≤Q,k 1 ∤k
μ
2 (k)
ϕ(k)
- 9 e
− 9 y
log
2 y
c 1 y
1 (^2) log
2 y log x
log 2
2 PRELIMINARY RESULTS 8
for x 2 ≤ y ≤ x. Hence we obtain
k≤Q
μ
2 (k) max l (mod k)
|Eθ(x; k, l)| <
2 e
− 9 x
log
2 x
k≤Q
μ
2 (k) max l (mod k)
|Eθ(x 2 ; k, l)| <
- 9 e
− 9 x 2
log
2 x 2
c 1 x
1 (^2) log
3 x
log 2
and
∫ (^) x
x 2
maxl (mod k) |Eθ (t; k, l)|
t log
2 t
dt <
∫ (^) x
x 2
- 9 e
− 9
log
4 t
c 1 log x
t
1 (^2) log 2
dt
1 5 log x 2
∫ (^) x
x 2
- 9 e
− 9
log
4 t
log t
dt + 2c 1 x
1 (^2) log x
- 9 e
− 9
1 5 log x 2
x
log
4 x
x 2
log
4 x 2
1 (^2) log x
2 e
− 9 x
log
4 x
Moreover, we use a trivial estimate Eπ (x 2 ; k, l) ≤ x 2 to obtain
k≤Q
μ
2 (k) max l (mod k)
Eπ(x 2 ; k, l) < x 2
k≤Q
μ
2 (k)
ϕ(k)
< c 1 x 2 log x. (34)
Combining (31)-(34), we have
k≤Q,k 1 ∤k
μ
2 (k) max l (mod k)
|Eπ(x; k, l)| <
e
− 8 x
log
3 x
This proves the lemma.
Now we introduce a extention of the previous lemma, which plays an
important role in our argument to avoid the exceptional modulus.
Lemma 2.6. Let x > X 2 = exp exp 36 and Q, k 1 as in the previous lemma.
If k divides k 1 , then we have
d≤Q/k
μ
2 (k) max l (mod k)
π(x; kd, l) −
π(x; k, l)
ϕ(d)
e
− 8 x
log
3 x
where d runs over integers such that k 1 ∤ kd if k 6 = k 1.
4 FRAMEWORK OF OUR SIEVE ARGUMENT 10
Using Selberg’s sieve, Jurkat and Richert[10] gave upper and lower bounds
for the linear sieve. Using Rosser’s combinatorial argument in his un-
published manuscript, Iwaniec[9] improved their upper and lower bounds.
Moreover, an explicit version of Rosser-Iwaniec linear sieve is given in Chap-
ter 9 in [14] although it requires an additional condition. In [22], the author
gave another explicit version of Rosser-Iwaniec linear sieve which can be
applied in more general cases. Here we shall introduce Theorem 1.2 in [22].
The assumption in this theorem is satisfied, for instance, when A is the set
of odd integers of the form aq + b with a, b fixed coprime integers and q odd
prime and Ωp consists at most one congruent class modulo p for each prime
p. In particular, A can be taken to be sets of integers the form N − q with
N even and q odd prime, which we shall consider in the following sections.
Theorem 3.1. Assume that ρ(p) ≤ p/(p − 1) and ρ(2) = 0. Then, for
every D, s > 0 , we have
S(A, P (D
1 s (^) )) > X
V (P (D
1 s (^) )) − 2
f 1 (s)
log D
log
3 (^2) D
− |R(D, P (z))|
and
S(A, P (D
1 s (^) )) < X
V (P (D
1 s (^) )) + 2
F 1 (s)
log D
log
3 (^2) D
+|R(D, P (z))|.
where f 1 (s), F 1 (s) are functions such that F 1 (s) = 2e
γ − s for 0 ≤ s ≤ 3 and
F
′ 1 (s) =^ −^
f (s − 1)
s − 1
for s ≥ 3 ,
f
′ 1 (s) =^ −^
F 1 (s − 1)
s − 1
for s ≥ 2 ,
and R(D, P ) =
d|P,d≤D
|μ
2 (d)r(d)|.
4 Framework of our sieve argument
We use the following notation. As we assumed in Theorem 1.1, let N be
an even integer ≥ X 2 = exp exp 36. Let z = N
1 (^8) and y = N
1 (^3). Then
z > exp exp 33 and y > exp exp 34.
We shall consider the set A = {N − p : p ≤ N, p ∤ N}. If A contains
at least one prime, then N could be represented by the sum of two primes.
We set Ωp to be the congruent class 0 (mod p), so that Aq = {m : q | m}.
4 FRAMEWORK OF OUR SIEVE ARGUMENT 11
Moreover, let r(d) = |Ad| −
|A| ϕ(d)
and rk(d) = |Akd| −
|Ak | ϕ(d)
denote error terms.
Clearly we have |A| = π(N) − ω(N) and |Ak| = π(N; k, N) − ω(N; k, N),
where ω(n; q, a) denotes the number of prime factors of n which is equivalent
to a (mod q).
As Chen and other authors did, we introduce the other set B = {N −
p 1 p 2 p 3 : z ≤ p 1 < y ≤ p 2 ≤ p 3 , p 1 p 2 p 3 < N, (p 1 p 2 p 3 , N) = 1} and obtain the
following lower bound, which is Theorem 10.2 in [14].
Lemma 4.1.
π 2 (N) > S(A, P (z)) −
z≤q<y
S(Aq, P (z)) −
S(B, P (y)) − 2 N
7 (^8) − 2 N
1 (^3).
So that, it suffices to give an lower bound for S(A, P (z)) and upper
bounds for S(B, P (y)) and S(Aq, P (z)) for each primes q with z ≤ q < y.
We set x 2 =
e−^100 N log^4 N
, which coincides x 2 in Section 2 with x = N, K 0 =
log
δ x 2 and let Q 1 = log
10 x 2. Moreover, we set k 0 to be the exceptional
modulus defined as in Section 2 and k 1 = k 0 if k 0 ≥ K 0 and k 1 = 0
otherwise,
Let q 1 > q 2 > · · · > ql be all prime factors of k 1 and mj = q 1 q 2 · · · qj , A
(j)
Amj and P
(j) (x) =
p<x,p∤N,p 6 =q 1 ,q 2 ,...,qj
p for j = 0, 1 , 2 ,... , l. We note
that m 0 = 1, A
(0) = A, P
(0) (x) = P (x). Moreover, we write for brevity
V (x) = V (P (x)) amd V
(j) (x) = V (P
(j) (x)).
As in the proof of Theorem 10.3 in [14], we deduce from (6) that
V
(j) (x) =
U
(j) N
log x
θ
5 log
2 x
8 θ log x
x
for j = 0, 1 ,... , l and x ≥ z, where
U
(j) N = 2e
−γ
p> 2
(p − 1)
2
p> 2 ,p|N mj
p − 1
p − 2
so that U
(0) N =^ UN^. Moreover, we have
l <
1 .3841 log(10 log N)
log log(10 log N)
e log log N
log log log N
by (6). Moreover, since k 1 ≥ log
δ x 2 > 3 × 5 × 7 × 11 × · · · × 53, we have
q 1 ≥ max{ 59 , 2 + log log N} (46)
5 LOWER BOUNDS FOR SOME SUMS OVER PRIMES 13
Recalling that |Ak| = π(N; k, N)−ω(N; k, N) and |Akd| = π(N; kd, N)−
ω(N; kd, N), we see that
rk(d) = π(N; kd, N) −
π(N; k, N)
ϕ(d)
- 3841 θ log N
log log N
and therefore
∑#
d<Q/k
μ
2 (kd) |rk(d)| <
- 3841 Q log N
log log N
d<Q/k
π(N; kd, N) −
π(N; k, N)
ϕ(d)
Now we apply Lemma 2.6 with x = N and obtain
d<Q/k
μ
2 (kd) |rk(d)| <
- 3841 D log N
log log N
e
− 8 x
log
3 x
- 1 e
− 8 x
log
3 x
5 Lower bounds for some sums over primes
The purpose of this section is to obtain an lower bound for S(A, P (z)):
Theorem 5.1.
S(A, P (z)) >
UN |A|
log N
×
4 e
γ log 3 − 0. 5198 ǫ 0 (N) −
log
1 (^2) N
As mentioned in the introduction, we cannot directly estimate S(A, P (z))
due to possible existence of the exceptional zero k 1. However, the following
inclusion-exclusion identity allows us to overcome this obstacle.
Lemma 5.2.
S(A, P (z)) =
l− 1 ∑
i=
i S(A
(i) , P
(i+1) (z)) + (−1)
l S(A
(l) , P
(l) (z)). (57)
Proof. S(A, P
1 (z))−S(A, P (z)) counts the number of integers in A divisible
by q 1 but not divisible by any other primes below z. S(A
(1) , P
(2) (z)) −
S(A, P
1 (z)) + S(A, P (z)) counts the number of integers in A divisible by
q 1 , q 2 but not divisible by any other primes below z. Iterating this argument,
we see that
∑l− 1
i=0(−1)
l− 1 −i S(A
(i) , P
(i+1) (z)) + (−1)
l S(A, P (z)) counts the
number of integers in A divisible by q 1 , q 2 ,... , ql but not divisible by any
other primes below z, which is equal to S(A
(l) , P
(l) (z)).
5 LOWER BOUNDS FOR SOME SUMS OVER PRIMES 14
As we will see below, each quantity can be estimated by sieve argument
without encountering the exceptional character.
Theorem 3.1 immediately gives the following estimates:
Lemma 5.3. Let
Ej =
d|P (j+1)(z),d<D rmj (d)^ if^ j^ = 0,^1 ,... , l^ −^1 ∑
d|P (l)(z),d<D
rml (d) if j = l.
Then we have
S(A
(j) , P
(j+1) (z))
∣A(j)
[
V
(j+1) (z) − U
(j+1) N
f (sj )
log D
log
3 (^2) D
)]
− |Ej |
and
S(A
(j) , P
(j+1) (z))
∣A(j)
[
V
(j+1) (z) + U
(j+1) N
F (sj )
log D
log
3 (^2) D
)]
for j = 0, 1 ,... , l − 1. Moreover, we have
S(A
(l) , P
(l) (z))
∣A(l)
[
V
(l) (z) − U
(l) N
f (sl)
log D
log
3 (^2) D
)]
− |El|
and
S(A
(l) , P
(l) (z))
∣A(l)
[
V
(l) (z) + U
(l) N
F (sl)
log D
log
3 (^2) D
)]
Let D =
√ x 2
k 1 log^10 x 2
and sj =
log D/mj
log z
. Since mj < k 1 < log
10 N, we have
log D >
log N
− 22 log log N − 50 (63)
and
8(32 log log N + 50)
log N
< s 1 < 4. (64)
5 LOWER BOUNDS FOR SOME SUMS OVER PRIMES 16
we obtain
S(A, P (z)) >
∑^ l−^1
j=
j
∣A(j)
∣ (^) V (j+1)(z) + (−1)l^
∣A(l)
∣ (^) V (l)(z)
− |A| U
(1) N
f (s 1 )
log D
log
3 (^2) D
∑l−^1
j=
∣A(j)
∣ (^) U(j+1) N +^
∣A(l)
∣ (^) U(l) N
×
log D
log
3 (^2) D
e
− 23 N
log
- 5 N
Now we shall evaluate each line in (69). We shall begin by showing that
l− 1 ∑
j=
j
∣A(j)
∣ (^) V (j+1)(z) + (−1)l^
∣A(l)
∣ (^) V (l)(z) = |A| V (z)
e
− 27 θ
log
1 (^2) N
We divide each term in the left-hand side of (70) and obtain
l− 1 ∑
j=
j
∣A(j)
∣ (^) V (j+1)(z) + (−1)l^
∣A(l)
∣ (^) V (l)(z)
l− 1 ∑
j=
j |A|
ϕ(mj )
V
(j+1) (z) + (−1)
l |A|
ϕ(ml)
V
(l) (z)
l− 1 ∑
j=
j r(mj )V
(j+1) (z) + (−1)
l r(ml)V
(l) (z).
We write ϕ
∗ (N) = N
p|N
p− 2 p
. Since
|A|
ϕ(mj )
V
(j+1) (z) =
|A| V (z)
ϕ∗(mj )
qj+1 − 2
we have
∑^ l−^1
j=
j |A|
ϕ(mj )
V
(j+1) (z) = |A| V (z)
j
ϕ
∗ (mj+1)
5 LOWER BOUNDS FOR SOME SUMS OVER PRIMES 17
and therefore
∑^ l−^1
j=
j
|A|
ϕ(mj )
V
(j+1) (z) + (−1)
l
|A|
ϕ(ml)
V
(l) (z) = |A| V (z). (74)
Substituting (74), (49), (50) into (71), we obtain
∑^ l−^1
j=
j
∣A(j)
∣ (^) V (j+1)(z) + (−1)l^
∣A(l)
∣ (^) V (l)(z)
= |A| V (z) +
- 193 θN
log
- 3 N
V
(l) (z)
Using Theorem 15, (3.41-42) of [17] again, we have
V
(l) (z)
V (z)
l ∏
j=
qj − 1
qj − 2
l ∏
j=
qj
qj − 1
p
(qj − 1)
2
qj (qj − 2)
e
γ log log k 1 +
2 log log k 1
<4 log log log N.
Hence, (48) and (75) gives
∑^ l−^1
j=
j
∣A(j)
∣ (^) V (j+1)(z) + (−1)l^
∣A(l)
∣ (^) V (l)(z)
= |A| V (z) +
- 772 θN log log log N
log
- 3 N
V (z)
= |A| V (z) +
e
− 27 θN
log
- 5 N
V (z)
= |A| V (z)
e
− 27 θ
log
1 (^2) N
which is (70).
Next we shall estimate the second line of (69). Since Lemma 2.2 gives
V (z) −
UN f (s)
log D
> UN
2 e
γ log(s − 1)
log D
4 log
3 z
We see that log D > (0. 5 − e
− 16 ) log N and log(s 1 − 1) > log 3 −
1
e^9 log
1 (^2) N
by
6 UPPER BOUNDS FOR SOME SUMS OVER PRIMES 19
we conclude that
( ∑l−^1
j=
∣A(j)
∣ (^) U(j+1) N +^
∣A(l)
∣ (^) U(l) N
log D
log
3 (^2) D
< UN |A| ǫ 0 (N)
log N
log
3 (^2) N
Substituting (70), (79) and (83) into (69), we obtain Theorem 5.1.
6 Upper bounds for some sums over primes
In this section, we shall obtain an upper bound for
z≤q<y
S(Aq, P, z):
Theorem 6.1.
z≤q<y
S(Aq, P (z)) <
UN |A|
log N
×
4 e
γ log 6(1 + ǫ 0 (N)) +
log
1 (^2) N
In this mission, it is much easier to break the obstacle due to possi-
ble existence of exceptional modulus; it suffices to give an upper bound for ∑
z≤q<y
S(Aq, P
(1) (z)) since it is clear that
z≤q<y
S(Aq, P (z)) ≤
z≤q<y
S(Aq, P
(1) (z)).
In this section, we set D =
x
1 2 2 log^10 x 2
and sq =
log(D/q) log z
. Theorem 3.
immediately gives
S(Aq, P
(1) (z)) < |Aq|
[
V
(1) (z) + U
(1) N
F (sq)
log
D q
log
3 2 D q
)]
d<D/q,d|P (1)(z)
|rq(d)| ,
where rq(d) = |Aqd| − |Aq| /ϕ(d), and therefore the sum
z≤q<y S(Aq, P, z)
can be bounded from above by
z≤q<y
|Aq|
[
V
(1) (z) + U
(1) N
F (sq)
log
D q
log
3 2 D q
)]
z≤q<y
d<D/q,d|P (1)(z)
|rq(d)|
6 UPPER BOUNDS FOR SOME SUMS OVER PRIMES 20
Using Lemma 2.5, we obtain that the sum over the error terms is
z≤q<y
d<D/q,d|P (1)(z)
|rq(d)|
z≤q<y
d<D/q,d|P (1)(z)
π(N; qd, N) −
π(N; q, N)
ϕ(d)
z≤q<y
d<D/q,d|P (1)(z)
|Eπ(N; qd, N)| +
Eπ(N; q, N)
ϕ(d)
- 2 e
− 8 N
log
3 N
Since N
1 (^8) < D y
D q
D z
< N
3 (^8) , we have 1 < sq < 3 and therefore
F (sq) = 2e
γ − sq. Thus we have
V
(1) (z) + U
(1) N
F (sq)
log
D q
=U
(1) N
2 e
γ
log
D q
+ V
(1) (z) −
U
(1) N
log z
<U
(1) N
2 e
γ
log
D q
5 log
2 z
This gives
z≤q<y
|Aq|
[
V
(1) (z) + U
(1) N
F (sq)
log
D q
log
3 2 D q
)]
≤ U
(1) N
z≤q<y
|Aq|
2 e
γ
log
D q
log
3 2 D q
5 log
2 z
Since |Aq | ≤
|A|+ω(N ) q− 1
z≤q<y
|Aq|
2 e
γ
log
D q
log
3 2 D q
5 log
2 z
z(|A| + ω(N))
z − 1
z≤q<y
2 e
γ
q log
D q
q log
3 2 D q
5 q log
2 z
2 e
γ
log
D y
log
3 2 D y
5 log
2 z
z≤q<y
Eπ(N; q, N).
Using Lemma 2.5, we have
z≤q<y
Eπ (N; q, N) <
e−^8 N log^3 N
and therefore
