()(1)  P  P JPP ()[1()]  k

P. O. Box 3924, Beijing 100854, P. R. China [email protected]

Abstract: Using Jiang function we prove that the new prime theorems (141)- （ 190) contain infinitely many prime solutions and no prime solutions.

[Chun-Xuan Jiang. The New Prime theorems（141）-（190）. Academ Arena 2015;7(1s): 186-236]. (ISSN 1553-992X). http://www.sciencepub.net/academia. 49

Keywords: prime; theorem; function; number; new

The New Prime theorem（141）

, 202 ( 1, , 1)

P jP   k j j   k 

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 202   k j contain infinitely many prime solutions and no prime solutions.

Keywords: prime; theorem; function; number; new

Theorem. Let k be a given odd prime.

, 202 ( 1, , 1)

P jP   k j j   k 

. （ 1 ）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 202

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

We prove that for k  3 (1) contain infinitely many prime solutions The New Prime theorem（142）

, 204 ( 1, , 1)

P jP   k j j   k 

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 204   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 204 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 204

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²⁰⁴  ² 1 ¹

( , 2) : ~ ( )

(204) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

We prove that for k  3, 5, 7,13,103 , (1) contain infinitely many prime solutions The New Prime theorem（143）

, 206 ( 1, , 1)

P jP   k j j   k 

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 206   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 206 ( 1, , 1)

P jP   k j j   k 

. （ 1 ）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（2）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 206

1 0 (mod ), 1, , 1

k

j jq k j P q P



  

        

（3）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（4）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（ 5 ）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 

1

206 2

1

( , 2) : ~ ( )

(206) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3 . From (2) and(3) we have

2 ( ) 0

J  

（ 7 ）

Abstract: Using Jiang function we prove that jP   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 208 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions or no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 208

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²⁰⁸  ² 1 ¹

( , 2) : ~ ( )

(208) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5,17, 53 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3, 5,17, 53 (1) contain no prime solutions.

Example 2. Let k  3, 5,17, 53 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 5,17, 53 (1) contain infinitely many prime solutions

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 210   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 210 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 210

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²¹⁰  ² 1 ¹

( , 2) : ~ ( )

(210) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 7,11, 31, 71, 211 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3, 7,11, 31, 71, 211 , (1) contain no prime solutions.

Example 2. Let k  3, 7,11, 31, 71, 211 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 7,11, 31, 71, 211 , (1) contain infinitely many prime solutions

Abstract

Using Jiang function we prove that

jP 212   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 212 ( 1, , 1)

P jP   k j j   k  . （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 212

1 0 (mod ), 1, , 1

k

j jq k j P q P



  

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（ 5 ）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²¹²  ² 1 ¹

( , 2) : ~ ( )

(212) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（6）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5,107 . From (2) and(3) we have

2 ( ) 0

J  

（ 7 ）

We prove that for k  3, 5,107 , (1) contain no prime solutions.

Example 2. Let k  3, 5,107 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 5,107 , (1) contain infinitely many prime solutions

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 214   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 214 ( 1, , 1)

P jP   k j j   k 

, （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 214

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²¹⁴  ² 1 ¹

( , 2) : ~ ( )

(214) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3 , (1) contain no prime solutions.

Example 2. Let k  3 . From (2) and (3) we have

2 ( ) 0

J  

（ 8 ）

We prove that for k  3 , (1) contain infinitely many prime solutions

Abstract

Using Jiang function we prove that

jP 216   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 216 ( 1, , 1)

P jP   k j j   k  , （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 216

1 0 (mod ), 1, , 1

k

j jq k j P q P



  

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（ 5 ）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²¹⁶  ² 1 ¹

( , 2) : ~ ( )

(216) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（6）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5, 7,13,19, 37, 73,109 . From (2) and(3) we have

2 ( ) 0

J  

（ 7 ）

We prove that for k  3, 5, 7,13,19, 37, 73,109 , (1) contain no prime solutions.

Example 2. Let k  3, 5, 7,13,19, 37, 73,109 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 5, 7,13,19, 37, 73,109 , (1) contain infinitely many prime solutions

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 218   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 218 ( 1, , 1)

P jP   k j j   k 

, （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 218

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²¹⁸  ² 1 ¹

( , 2) : ~ ( )

(218) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3 , (1) contain no prime solutions.

Example 2. Let k  3 . From (2) and (3) we have

2 ( ) 0

J  

（ 8 ）

We prove that for k  3 , (1) contain infinitely many prime solutions

Abstract

Using Jiang function we prove that

jP 220   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 220 ( 1, , 1)

P jP   k j j   k  , （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 220

1 0 (mod ), 1, , 1

k

j jq k j P q P



  

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（ 5 ）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²²⁰  ² 1 ¹

( , 2) : ~ ( )

(220) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（6）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5,11, 23 . From (2) and(3) we have

2 ( ) 0

J  

（ 7 ）

We prove that for k  3, 5,11, 23 , (1) contain no prime solutions.

Example 2. Let k  3, 5,11, 23 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 5,11, 23 , (1) contain infinitely many prime solutions

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 222   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 222 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 222

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²²²  ² 1 ¹

( , 2) : ~ ( )

(222) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 7, 223 . From (2) and(3) we have

2 ( ) 0

J  

（7）

we prove that for k  3, 7, 223 , (1) contain no prime solutions Example 2. Let k  3, 7, 223 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 7, 223 (1) contain infinitely many prime solutions

Abstract

Using Jiang function we prove that

jP 224   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 224 ( 1, , 1)

P jP   k j j   k  . （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 224

1 0 (mod ), 1, , 1

k

j jq k j P q P



  

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（ 5 ）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²²⁴  ² 1 ¹

( , 2) : ~ ( )

(224) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（6）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5,17, 29,113 . From (2) and(3) we have

2 ( ) 0

J  

（ 7 ）

We prove that for k  3, 5,17, 29,113 (1) contain no prime solutions.

Example 2. Let k  3, 5,17, 29,113 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 5,17, 29,113 , (1) contain infinitely many prime solutions

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 226   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 226 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 226

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²²⁶  ² 1 ¹

( , 2) : ~ ( )

(226) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 227 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3, 227 , (1) contain no prime solutions.

Example 2. Let k  3, 227 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 227 (1) contain infinitely many prime solutions

Abstract: Using Jiang function we prove that jP   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 228 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions or no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 228

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²²⁸  ² 1 ¹

( , 2) : ~ ( )

(228) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5, 7,13, 229 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3, 5, 7,13, 229 (1) contain no prime solutions.

Example 2. Let k  3, 5, 7,13, 229 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 5, 7,13, 229 (1) contain infinitely many prime solutions

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 230   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 230 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 230

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²³⁰  ² 1 ¹

( , 2) : ~ ( )

(230) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3,11, 47 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3,11, 47 , (1) contain no prime solutions.

Example 2. Let k  3,11, 47 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3,11, 47 , (1) contain infinitely many prime solutions

Abstract: Using Jiang function we prove that jP   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 232 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 232

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²³²  ² 1 ¹

( , 2) : ~ ( )

(232) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5, 59, 233 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3, 5, 59, 233 , (1) contain no prime solutions.

Example 2. Let k  3, 5, 59, 233 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 5, 59, 233 , (1) contain infinitely many prime solutions

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 234   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 234 ( 1, , 1)

P jP   k j j   k 

, （ 1 ）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（2）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 234

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（3）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（4）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（ 5 ）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²³⁴  ² 1 ¹

( , 2) : ~ ( )

(234) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 7,19, 79 . From (2) and(3) we have

2 ( ) 0

J  

（ 7 ）

We prove that for k  3, 7,19, 79 , (1) contain no prime solutions.

Example 2. Let k  3, 7,19, 79 . From (2) and (3) we have

2 ( ) 0

J  

（ 8 ）

We prove that for k  3, 7,19, 79 , (1) contain infinitely many prime solutions

Abstract: Using Jiang function we prove that jP   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 236 ( 1, , 1)

P jP   k j j   k 

, （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 236

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²³⁶  ² 1 ¹

( , 2) : ~ ( )

(236) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3, 5 , (1) contain no prime solutions.

Example 2. Let k  5 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  5 , (1) contain infinitely many prime solutions

Chun-Xuan Jiang [email protected] Abstract

Using Jiang function we prove that

jP 238   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 238 ( 1, , 1)

P jP   k j j   k  , （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 238

1 0 (mod ), 1, , 1

k

j jq k j P q P



  

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（ 5 ）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²³⁸  ² 1 ¹

( , 2) : ~ ( )

(238) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（6）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 239 . From (2) and(3) we have

2 ( ) 0

J  

（ 7 ）

We prove that for k  3, 239 , (1) contain no prime solutions.

Example 2. Let k  3, 239 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 239 , (1) contain infinitely many prime solutions

Abstract

Using Jiang function we prove that

jP 240   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 240 ( 1, , 1)

P jP   k j j   k  , （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 240

1 0 (mod ), 1, , 1

k

j jq k j P q P



  

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（ 5 ）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²⁴⁰  ² 1 ¹

( , 2) : ~ ( )

(240) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（6）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5, 7,11,13,17, 31, 41, 61, 241 . From (2) and(3) we have

2 ( ) 0

J  

（ 7 ）

We prove that for k  3, 5, 7,11,13,17, 31, 41, 61, 241 , (1) contain no prime solutions.

Example 2. Let k  3, 5, 7,11,13,17, 31, 41, 61, 241 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 5, 7,11,13,17, 31, 41, 61, 241 , (1) contain infinitely many prime solutions

Chun-Xuan Jiang [email protected] Abstract

Using Jiang function we prove that

jP 242   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 242 ( 1, , 1)

P jP   k j j   k  . （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 242

1 0 (mod ), 1, , 1

k

j jq k j P q P



  

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（ 5 ）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²⁴²  ² 1 ¹

( , 2) : ~ ( )

(242) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（6）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 23 . From (2) and(3) we have

2 ( ) 0

J  

（ 7 ）

we prove that for k  3, 23 , (1) contain no prime solutions Example 2. Let k  3, 23 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 23 (1) contain infinitely many prime solutions

Abstract: Using Jiang function we prove that jP   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 244 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 244

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²⁴⁴  ² 1 ¹

( , 2) : ~ ( )

(244) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3, 5 (1) contain no prime solutions.

Example 2. Let k  5 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  5 , (1) contain infinitely many prime solutions

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 246   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 246 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 246

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²⁴⁶  ² 1 ¹

( , 2) : ~ ( )

(246) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 7,83 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3, 7,83 , (1) contain no prime solutions.

Example 2. Let k  3, 7,83 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 7,83 (1) contain infinitely many prime solutions

Abstract: Using Jiang function we prove that jP   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 248 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions or no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 248

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²⁴⁸  ² 1 ¹

( , 2) : ~ ( )

(248) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3, 5 (1) contain no prime solutions.

Example 2. Let k  5 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  5 (1) contain infinitely many prime solutions

Chun-Xuan Jiang [email protected] Abstract: Using Jiang function we prove that

jP 250   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 250 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 250

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²⁵⁰  ² 1 ¹

( , 2) : ~ ( )

(250) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3,11, 251 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3,11, 251 , (1) contain no prime solutions.

Example 2. Let k  3,11, 251 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3,11, 251 , (1) contain infinitely many prime solutions

Abstract: Using Jiang function we prove that jP   k j contain infinitely many prime solutions and no prime solutions.

Theorem. Let k be a given odd prime.

, 252 ( 1, , 1)

P jP   k j j   k 

. （1）

contain infinitely many prime solutions and no prime solutions.

Proof. We have Jiang function [1,2]

2 ( ) [ 1 ( )]

J    P P    P

（ 2 ）

where    ^P P

，  ( ) P is the number of solutions of congruence

1 252

1 0 (mod ), 1, , 1

k

j jq k j P q P





 

        

（ 3 ）

If  ( ) P  P  2 then from (2) and (3) we have

2 ( ) 0

J  

（ 4 ）

We prove that (1) contain infinitely many prime solutions.

If  ( ) P  P  1 then from (2) and (3) we have

2 ( ) 0

J  

（5）

We prove that (1) contain no prime solutions [1,2]

If J ² ( )   0

then we have asymptotic formula [1,2]

 ²⁵²  ² 1 ¹

( , 2) : ~ ( )

(252) ( ) log

k

k k k k

J N

N P N jP k j prime

N

  

 



     

（ 6 ）

where ( ) ( 1)

P P

    

.

Example 1. Let k  3, 5, 7,13,19, 29, 37, 43,127 . From (2) and(3) we have

2 ( ) 0

J  

（7）

We prove that for k  3, 5, 7,13,19, 29, 37, 43,127 , (1) contain no prime solutions.

Example 2. Let k  3, 5, 7,13,19, 29, 37, 43,127 . From (2) and (3) we have

2 ( ) 0

J  

（8）

We prove that for k  3, 5, 7,13,19, 29, 37, 43,127 , (1) contain infinitely many prime solutions