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(1)

itf (n)

γ

γ

p>2

2

2<p|γ

2

h

γ

γ

[193]

(2)

h

h

h

γ

h

h

h

λ∈R

2

p≤x

2

h

h

−1/2

h

γ

γ

γ

γ

0<γ≤x

0<m,n≤x m−n=γ

p>2

2

p|γ p>2

2

−c

0<γ≤x

0<m,n≤x+γ m−n=γ

2

0<γ≤x

q=p+γp≤x

2

3/2

0<γ≤x

γ

3/2

(3)

0<γ≤x

γ

2

c

0<γ≤x

γ

γ

γ

h

h

−1/2

1/10

h

h

p|a(γ−a)

h

γ

γ

γ≤x

γ

−1

γ≤x

γ

h

h

|t|≤T

p≤x

−1

it

n≤x

p|D, p|n p-γ

p|D, p|n p|γ

−1/4

(4)

p|n

p|n

γ

γ

γ

γ

γ

γ

ω(k)

m,n

## = Y

m<p≤n p-a, p-(a−γ)

d

1

γ

p|d

γ

−1

k

d<x (d,k)=1

2

dj|P j=1,2

d1

d2

γ

1

2

1

2

dj

j

d

p|d

γ

−1

d

1

−11

−11

p|P

γ

λ

(5)

w,z

1/10

1

1

2

1

p|Pw,z

p|γ

w<p≤z p|a(a−γ)

1

B

Q<p≤M

n≤x n≡r (mod D)

n≤x (n,D)=1

1/8−ε

β

0

y,w

1/4

(6)

n≤x n≡r (mod D)

p|∆

p|n, p-γ

p|∆

p|n, p|γ

n≤x (n,D)=1

p|∆

p|n, p-γ

p|∆

p|n, p|γ

1/8−ε

β

0

0

l

## (m, γ, n) = Y

p|m, p-γ p|n, p-l

##  Y

p|m, p|γ p|n, p-l

1

l

d|n

k

d=1

α

p

α

(7)

n≤x n≡r (D)

l

n≤x (n,D)=1

l

d≤x (d,D)=1

m≤x/d m≡rd−1(D)

m≤x/d (m,D)=1

d≤√ x (d,D)=1

1/8−ε

1/8−ε

√x<d≤x (d,D)=1

3/4

1/8−ε

1

2

1

1

k

k

2

2

k

k

k−1

2

k

n≤x

2

α

d≤x

α

p≤x/d

2

1

2

n≤x n≡r (D)

d≤x (d,D)=1

2

m≤x/d m≡rd−1(D)

1

d

(8)

n≤x (n,D)=1

d≤x (d,D)=1

2

m≤x/d (m,D)=1

1

d

d≤x (d,D)=1

2

m≤x/d m≡rd−1(D)

1

d

m≤x/d (m,D)=1

1

d

1

3/4

3/4

3/4

1/8−ε

d≤x3/4 (d,D)=1

2

1/8−ε

3/4

p|∆

2

3

y,w

w

3

−3/8

x3/4<d≤x

2

5/8

7

2

3/4

1/8−ε

1/4

(9)

p|m p-γ

p|m p|γ

14/15

y,w

0

γ

0

j

γ

0

## Then either X

n≤x (Pγ(n),Py,w)=1

y,w

n≤x

y,w

−1/20

## or X

n≤x (Pγ(n),Py,w)=1

y,w

n≤x

y,w

y,w

γ

0

%γ

(D0) j=1

j

n≤x

y,w

0

−21/20

γ

0

0

0

p|D0 p-γ

p|D0 p|γ

0

0

y,w

0

0

251

3

(10)

d|n

d|n ω(d)≤r−1

d|n ω(d)=r

γ

y,w

d|Py,w

d≤Q ω(d)≤r−1

n≤x Pγ(n)≡0 (d)

d|Py,w

d≤Q ω(d)=r

n≤x Pγ(n)≡0 (d)

d|Py,w Q<d≤Pγ(x)

ω(d)≤r−1

n≤x Pγ(n)≡0 (d)

d|Py,w Q<d≤Pγ(x)

ω(d)=r

n≤x Pγ(n)≡0 (d)

γ

3

y,w

y,w

γ

d|Py,w Q<d≤√

x

2

γ

d|Py,w Q<d≤√

x

2

γ

d|Py,w

ω(d)

1/2

d|Py,w Q<d≤√

x

1/2

2

k

k+1

d|Py,w

2kU <d≤2k+1U

k

d|Py,w

2kU <d≤2k+1U

−1/15

(11)

d|Py,w

Q<d≤√ x

1/2

1/2

3/2

γ

d|Py,w

d≤Q

γ

n≤x (n,d)=1

d|Py,w

d≤Q

2

γ

1/8−ε

3

1/8−ε

y<p≤w

1/10

3/2

n≤x

d|Py,w

(d,n)=1

γ

3/2

d|Py,w (d,n)=1

γ

y,w

y,w

0

0

y,w

y,w

0

0

0

0

y,w

0

0

251

(12)

0

y,w

γ

0

0

ω(D0)

−1/19

5

1/19

−1/20

0

0

y,w

0

251

y,w

1

0

2

1

1

2

2

0

1

2

1

1

2

2

0

0

## Proceeding as in the case where there were no exceptional moduli, we get X

n≤x (Pγ(n),Py,w)=1

d|Py,w

d≤Q

n≤x Pγ(n)≡0 (d)

3/2

d|Py,w d≤Q

## µ(d) X

n≤x Pγ(n)≡0 (d1) Pγ(n)≡0 (d2)

ml

γ

m

γ

m

d|Py,w d≤Q

1

%γ

(d1) j=1

χ1

1

1j

n≤x Pγ(n)≡0 (d2)

1

0

2

1

1

1

1

1

2

0

(13)

1

1

0

2

1

1

1

2

1

1

d|Py,w d≤Q

1

%γ

(d1) j=1

χ1

1

1j

γ

2

2

n≤x (n,d2)=1

1

d|Py,w

d≤Q

1

%γ

(d1) j=1

χ1

1

1j

γ

2

2

1/8−ε

1/20

%γ

(d1) j=1

## X

n≤x (n,d2)=1 n≡c1j(d1)

## g(n) = X

n≤x (n,d2)=1 Pγ(n)≡0 (d1)

2

3/2

0

y,w

0

## X

n≤x (Pγ(n),Py,w)=1

n≤x

d1|D0

d1|Pγ(n)

1

d2|P0

(d2,n)=1

2

γ

2

2

1/20

(14)

2

0

m|D0

n≤x Pγ(n)≡0 (m)

0

mj

γ

γ

n≤x Pγ(n)≡0 (m)

0

%

γ(m) j=1

n≤x n≡bmj(m)

0

0

0

m|D0

m<D0

%γ

(m) j=1

n≤x (n,m)=1

0

0

0

%γ

(D0) j=1

n≤x n≡bD0,j(D0)

0

0

1/10

0

γ

0

0

n≤x (n,D0)=1

0

0

n≤x

0

m|D0

(m,n)=1

γ

0

0

0

y,w

0

0

0

y,w

0

0

y,w

0

0

0

%γ

(D0) j=1

χ0(mod D0)

0

D0,j

n≤x

0

0

(15)

0

0

0

0

0

%γ

(D0) j=1

χ0

0

j

n≤x (n,D0)=1

0

0

0

0

0

0

y,w

γ

0

0

0

n≤x

0

0

−1/9

0

0

1/9

0

0

γ

0

0

## X

χ06=χ nonprincipal

0

251

−1/20

1

−1

itu

1

0

2

2

itf

1

−1

−ith

(16)

y,z

h

γ

## X

n≤x (n(n+γ),Py,z)=1

1

−1

−ith

γ

d

d

1

h

1

−1

−ith

γ

dj≤z dj|Pw,z

j=1,2

d1

d2

## X

n≤x (n(n+γ),Py,w)=1 n(n+γ)≡0 ([d1,d2])

γ

1

1

k

k

1

1

1

1

1

ump=n u≤logBx

p≤b

1

2

urp=n u≤logBx

r≤b

1

1

2

6A+15

j

(B+1)/2

ε

1

ump=n u≤logBx

p≤b

ump=n u≤logBx

p≤b

0

2

urp=n u≤logBx

r≤b

0

0

0

0

2

(17)

0

k

1

2

1

2

1

k

u+v=k

u

v

1/10

h

γ

1

−1

−ith

dj≤z dj|Pw,z

d1

d2

## X

n≤x (n,Py,w)=1 n(n+γ)≡0 ([d1,d2])

2 m=1

γ

1

−1

−ith

dj≤z dj|Pw,z

d1

d2

## × X

n≤x (n,Py,w)=1 n(n+γ)≡0 ([d1,d2])

m

1/10

γ

0

1

2

0

dj≤z dj|Pw,z

d1

d2

## X

n≤x (n(n+γ),Py,w)=1 n(n+γ)≡0 ([d1,d2])

dj≤z dj|Pw,z

d1

d2

γ

1

2

1

2

## X

n≤x (n(n+γ),Py,w)=1

(n+a,[d1,d2])=1

0

1

−1

−ith

γ

dj≤z dj|Pw,z

d1

d2

γ

1

2

1

2

## × X

n≤x (n(n+γ),Py,w)=1

(n+a,[d1,d2])=1

1

−1

−ith

γ

1

2

j

j

w,z

(18)

1

2

1

2

ω(D)

dj

B+1/2

## E(t)  X

D≤z2 p|D⇒p>w (D,a|a−γ|)=1

ω(D)

## X

n≤x (Pγ(n),Py,w)=1

Pγ(n)≡0 (D)

γ

## X

n≤x (Pγ(n),Py,w)=1

(n,D)=1

1/10

81/2

##  X

D≤z2 p|D⇒p>w (D,a|a−γ|)=1

γ

1/3

4/3

3/4

## F (t, D, x) = X

n≤x (Pγ(n),Py,w)=1

Pγ(n)≡0 (D)

γ

## X

n≤x (Pγ(n),Py,w)=1

(n,D)=1

γ

1/3

## X

n≤x (Pγ(n),Py,w)=1

Pγ(n)≡0 (D)

γ

## X

n≤x (Pγ(n),Py,w)=1

(n,D)=1

1/3

1/3

(B+1)/6

1/4

(B+1)/8

##  X

D≤z2 p|D⇒p>w (D,a|a−γ|)=1

## X

n≤x (Pγ(n),Py,w)=1

Pγ(n)≡0 (D)

γ

## X

n≤x (Pγ(n),Py,w)=1

(n,D)=1

3/4

6A+15

(19)

D1D2≤xδ

(r,D

1D2)=1

n≤x n≡r (D1D2)

2

## X

n≤x (n,D2)=1 n≡r (D1)

−A

2

−1

2

−1/2

5/2

5/2−B/2

1

2

D1D2≤xδ

## X

n≤x Pγ(n)≡0 (D1D2)

γ

2

2

## X

n≤x (n,D2)=1 Pγ(n)≡0 (D1)

(4−A)/2

A+6

4

13/4

1/2

(13−B)/4

1

1,w

2

2

j

γ

1

2

γ

1

2

ik

γ

i

γ

i

γ

1

2

γ

1

2

γ

1

γ

2

j

1k

2l

1

2

1

2

(20)

γ

2

2

## X

n≤x (n,D2)=1 Pγ(n)≡0 (D1)

2

%γ

(D1) k=1

%γ

(D2) i=1

## X

n≤x (n,D2)=1 n≡c1k(D1)

2

%γ(D

1D2) j=1

## X

n≤x (n,D2)=1 n≡bj(D1)

1

2

j

1

2

1

2

1

2

D1D2≤xδ

γ

1

2

(b,D1D2)=1

n≤x n≡b (D1D2)

2

## X

n≤x (n,D2)=1 n≡b (D1)

4

D1D2≤xδ

1

2

(b,D1D2)=1

n≤x n≡b (D1D2)

2

## X

n≤x (n,D2)=1 n≡b (D1)

2

1

2

n≤x n≡b (D1D2)

2

## X

n≤x (n,D2)=1 n≡b (D1)

4

D1D2≤xδ

(b,D

1D2)=1

n≤x n≡b (D1D2)

2

## X

n≤x (n,D2)=1 n≡b (D1)

2A+14

1/4

(B+1)/8

3−A/2

3/4

3−A/8

## .

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