Download The sum and number of primes between any two positive integers

The sum and number of primes between any two positive
integers
Zhen Liu
Abstract
Using the method for equation reconstruction of prime sequence, this paper gives
the proof that there is at least one prime between positive integers n 2 and n  1 .The sum
2
and number formulae of primes between any two positive integers are also given out.
1
Introduction
Because it can not be divisible except 1 and itself, primes are difficult to be described by
appropriate expressions. This property makes prime sequence be difficult to be described such
as arithmetic progression, geometric progression with the determined term formula. However,
this property can make prime number establish some diophantine equations. And prime
numbers can be decided by whether there is positive whole number solutions of these
diophantine equations. Therefore, the expressions for solutions of these diophantine equations
and its transform are used to describe the divisible property of prime number, and forming an
equivalent sequence for the property. Thus, this will be easy to find the key node and the law
implied to solve the problem. To this end, the theorem for equation reconstruction of prime
sequence is presented and proved. Using the method, this paper gives the proof that there is at
least one prime between positive integers n 2 and n  1 .The sum and number formulae of
2
primes between two positive integers are also given out. It could be hope to provide an idea
and methods to solve similar problems.
Zhen Liu
Doctor, Associate Professor of School of Marine Sciences, Sun Yat-sen University, China
Adds: Sun Yat-sen University, No.135, Xingang Xi Rd, Haizhu District, Guangzhou, Guangdong, 510275,
China
E-mail: [email protected]
1
In this paper, all parameters are positive whole number except where stated.
2
Proof of the theorem for equation reconstruction of prime sequence
Lemma: The prime sequence could be equivalent to the sequence with the determined general
term formula through equation reconstruction of prime number for the divisible property.
Proof.
Any prime number c could be expressed as 3a  1 ( a is an positive even), 4a  1 or
6a  1 .
Proof is carried out the following in the case of 3a  1 first.
If 3a  1 is a prime number, it certainly can not be written 3a  1  (3 x1  1)(3 x2  1) ,
otherwise, and vice versa.
Case 1: 3a  1
3a  1  (3 x1  1)(3 x2  1)  9 x1 x2  3( x1  x 2 )  1
or
3a  1  (3 x1  1)(3 x2  1)  9 x1 x 2  3( x1  x2 )  1
Where, let  x1  x1 ,  x2  x 2 .
Then there is a  3 x1 x2  ( x1  x2 ) .
It is easy to see that whether 3a  1 is a prime number depends entirely on the a .
Namely 3a  1 is a prime number that is equivalent x1 and x2 are both positive whole
number in a  3 x1 x 2  ( x1  x2 ) .
Let x1  x 2  q , x1 x 2  p
According to Vieta's formulas, equation (1) is established.
x 2  qx  p  0
Then there is x1, 2 
(1)
 q  q2  4 p
2
Therefore, if x1 and x2 of equation (1) roots are not both positive whole number, 3a  1
2
must be a prime number. Otherwise, it will be a composite number.
There is
a  3p  q
Obviously, if 3a  1 is a prime number, q and q 2  4 p are not both even numbers.
Therefore, in the divisible property of prime number, ci in prime sequence
cn 
is
equivalent to ai  3 pi  qi in sequence a n  , namely prime sequence cn  is equivalent to
sequence a n  .
Here qi and
qi2  4 pi are not both even numbers.
In order to facilitate the expression, let q  2s , p  2r .
Here s and r are real numbers.
∴ x1, 2  s  s 2  2r
Let
s 2  2r  t
There is
a  12st  12t 2  2s
Therefore, ai  3 pi  qi in sequence a n  ( qi and
qi2  4 pi are not both even numbers)
is equivalent to ai  12si t i  12t i2  2 si in sequence an  ( si and t i are not both positive
whole number solutions) .
Namely, in the divisible property of prime number, prime sequence cn  is equivalent to
sequence an  .
It is obvious that
s  s2 
t
Let s 2 
2s  a
3
2
2s  a
 e2
3
Then there is
3
a  3s 2  2s  3e 2
Therefore, ai  12si t i  12t i2  2 si in sequence an  ( si and t i are not both positive whole
number solutions) is equivalent to ai  3si2  2si  3ei2 in sequence an ( si and ei are
not both positive whole number solutions) .
Namely, in the divisible property of prime number, prime sequence cn  is equivalent to
sequence an .
It is obvious that
s
Let 9e 2  3a  1  3h  1
 1  9e 2 3a  1
3
2
Then there is
3a  1  3h  1  3e 
2
2
a  3h 2  2h  3e 2
Therefore, ai  3si2  2si  3ei2 in sequence an ( si and ei are not both positive whole
number solutions) is equivalent to 3a n  1  3hi  1  3ei  in sequence an ( ei and hi
2
2
are not both positive whole number solutions) .
Namely, in the divisible property of prime number, prime sequence cn  is equivalent to
sequence an .
Case 2: 3a  1
3a  1  (3 x1  1)(3 x2  1)  9 x1 x2  3( x2  x1 )  1
Where, let  x1  x1 , x2  x 2 .
Then there is a  3 x1 x2  ( x1  x2 ) .
Namely 3a  1 is a prime number that is equivalent x1 and x2 are both positive whole number
in a  3 x1 x 2  ( x1  x2 ) .
4
Let x1  x 2  q , x1 x 2  p . Here p is negative whole number.
According to Vieta's formulas, equation (2) is established.
x 2  qx  p  0
Then there is x1, 2 
(2)
 q  q2  4 p
.
2
Therefore, if x1 and x2 of equation (2) roots are not both positive whole number, 3a  1
must be a prime number. Otherwise, it will be a composite number.
There is
a  3 p  q
Obviously, if 3a  1 is a prime number, q and q 2  4 p are not both even numbers.
Therefore, in the divisible property of prime number, ci in prime sequence
cn 
is
equivalent to ai  3 pi  qi in sequence a n  , namely prime sequence cn  is equivalent to
sequence a n  .
Using the same argument as in the case 1, we can easily get
a  12t 2  12st  2s
Therefore, ai  3 pi  qi in sequence a n  ( qi and
qi2  4 pi are not both even numbers)
is equivalent to ai  12t i2  12 si t i  2 si in sequence an  ( si and t i are not both positive
whole number solutions) .
Namely, in the divisible property of prime number, prime sequence cn  is equivalent to
sequence an  .
It is obvious that
s  s2 
t
Let s 2 
2s  a
3
2
2s  a
 e2
3
Then there is
5
a  3e 2  3s 2  2s
Therefore, ai  12t i2  12 si t i  2 si in sequence an  ( si and t i are not both positive whole
number solutions) is equivalent to ai  3ei2  3si2  2s in sequence an ( si and ei are not
both positive whole number solutions) .
Namely, in the divisible property of prime number, prime sequence cn  is equivalent to
sequence an .
It is obvious that
s
Let 9e 2  3a  1  3h  1
 1  9e 2  3a  1
3
2
Then there is
3a  1  3e   3h  1
2
2
a  3e 2  3h 2  2h
Therefore, ai  3ei2  3si2  2s in sequence an ( si and ei are not both positive whole
number solutions) is equivalent to 3a n  1  3ei   3hi  1 in sequence an ( ei and hi
2
2
are not both positive whole number solutions) .
Namely, in the divisible property of prime number, prime sequence cn  is equivalent to
sequence an .
The prime sequence that prime number c could be expressed as 4a  1 or 6a  1 , have
equivalent methods that are similar to the case of 3a  1 . It can be proved in the same way as
shown in the case of 3a  1 before. Of course, some new equivalent sequences are
reconstructed through establishing other forms equations.
This completes the proof.
According to above proof, in the divisible property of prime number, the prime sequence
cn 
without term formula is analyzed by using the sequence a n  、 an  、 an 、 an with
6
term formula. This will be easy to find the key node and the law implied to solve the problem.
3
Proof of existing at least one prime between positive integers n 2 and n  1
2
Theorem1: There is at least one prime between positive integers n 2 and n  1 , where n
2
is any positive integer.
Proof.
It proves the Theorem with the reduction to absurdity follows.
If the Theorem is not true, it becomes: there is a positive integer n0 that makes all integers
between n02 and n0  1 be composite numbers.
2
According to the Lemma for equation reconstruction of prime sequence, there are
3a  1  3 x1  13 x2  1
(3)
3a  1  3 x1  13 x2  1 or 3a  1  3 x1  13 x2  1
(4)
Where, a  2l , l is positive whole number.
Therefore, when a is large enough, at least one of equation (9) and equation (10) has integer
solutions.
Using the same argument as in the proof of equation reconstruction of prime sequence, we
can easily get this statement fellows.
For equation (3), there are
a
 6t12  6s1t1  s1
2
t1 
Let s12 
s1  s12 
2s1  a
3
2
2s1  a
 e12
3
Then, there is
s1 
 1  9e12  6l  1
3
Let 9e12  6l  1  3h1  1 ，Namely it makes s1 be a positive whole number.
2
7
Then, there is
6l  1  9e12  3h1  1
2
a  2l  3e12  3h12  2h1
3a  1  9e12  3h1  1
2
(5)
For equation (4), there are
a
 6s 2 t 2  6t 22  s 2
2
t2 
Let s 22 
s 2  s 22 
2s2  a
3
2
2s2  a
 e22
3
Then, there is
s2 
 1  9e22  6l  1
3
Let 9e22  6l  1  3h2  1 ，Namely it makes s 2 be a positive whole number.
2
Then, there is
6l  1  3h2  1  9e22
2
a  2l  3h22  2h2  3e22
3a  1  3h2  1  9e22
2
(6)
According to the equation (5) and equation (6), there is
3a  1  3h  1  9e 2
2
Let 3a  1  n02  m , 0  m  2n0  1 .
Then, if the Theorem is not true, it becomes: there is a positive integer n0 that makes
n02  m be composite numbers. Namely there is a positive integer n0 that makes
n02  m  3h  1  3e 2 .
2
For n0 , there are 3 cases: n0  3k 0  1 , n0  3k 0 and n0  3k 0  1 .
8
Let when k 0 is a even, k 0  2d 0 . Let when k is odd, k 0  2d 0  1 .
Then,
Case n0  3k 0  1 :
when k 0  2d 0 , n02  36d 02  12d 0  1 ,
when k 0  2d 0  1 , n02  36d 02  48d 0  16 .
Csae n0  3k 0 :
when k 0  2d 0 , n02  36d 02 ,
when k 0  2d 0  1 , n02  36d 02  36d 0  9 .
Case n0  3k 0  1 :
when k 0  2d 0 , n02  36d 02  12d 0  1 ,
when k 0  2d 0  1 , n02  36d 02  24d 0  4 .
∵Any prime could be expressed as 6b  1 .
∴Just consider only a situation that n02  m could be expressed as 6b  1 , while the rest are
composite numbers.
Because of n02  m corresponding to be expressed as 6b  1 and 6b  1 , there are 6 cases
as follows:
when n02  36d 02  12d 0  1 , m  6c or m  6c  4 ,
when n02  36d 02  48d 0  16 , m  6c  1 or m  6c  1 ,
when n02  36d 02 , m  6c  1 or m  6c  1 ,
when n02  36d 02  36d 0  9 , m  6c  4 or m  6c  2 ,
when n02  36d 02  12d 0  1 , m  6c  4 or m  6c ,
when n02  36d 02  24d 0  4 , m  6c  1 or m  6c  3 .
When n02  m  3h  1  3e  , let
2
2
9
n02  m0  3a 0  1  3h0  1  3e0   A
2
2
Where, m0 is the minimum of m .
Then the equation n02  m0  m  3h0   h   1  3e0   e 
2
2
has integer solutions for
any m .
There is
h 
1
 3h0  1 
3 
3h0  12  92e  18e0  e  m 

∴ 3h0  1  92e  18e0  e  m  is a square number.
2
Then let
3h0  12  92e  18e0  e  m  
f
2
3h0  12  B
There is
e 

1
 3e0  9e02  m  B  f
3
2

∴ 9e02  m  B  f 2 is also a square number.
Then, there is 9e02  m  B  f 2   c20  c for arbitrary continuous m .
And there is
 c2  c   c2  m  6c
0
0
It is easy to see that 6c are not all the difference between square numbers. This is also in
contradiction with the difference between square numbers  m20  m   m20 .
When n02  m  3e   3h  1 , let
2
2
n02  m0  3a 0  1  3e0   3h0  1  A
2
2
Where, m0 is the minimum of m .
Then the equation n02  m0  m  3e0   e   3h0   h   1
2
2
has integer solutions for
any m .
There is
10
h 
3h0  12  92e  18e0  e  m 
1
 3h0  1 
3 

∴ 3h0  1  92e  18e0  e  m  is a square number.
2
Then let
3h0  12  92e  18e0  e  m  
f
2
3h0  12  B
There is
e 
∴ 9e02  m  B  f
2

1
 3e0  9e02  m  B  f
3
2

is also a square number.
Then, there is 9e02  m  B  f 2   c20  c for arbitrary continuous m .
And there is
 c2  c   c2  m  6c
0
0
It is easy to see that 6c are not all the difference between square numbers. This is also in
contradiction with the difference between square numbers  m20  m   m20 .
It is now obvious that the theorem holds.
There is at least one prime between positive integers n 2 and n  1 .
2
This completes the proof.
4. Proof of the sum and number formulae of primes between two positive integers
Theorem2: The sum formula   ,   and number formula   ,   of primes between
positive integers  and  , where    .
Proof.
According to the Lemma for equation reconstruction of prime sequence, any prime could be
expressed as 3a  1 , where   p   . And
3a  1  3h  1  3e 
2
2
Then
11
 1
3
∴The equation 3a  1  3h  1  3e 
2
3a  1 between 
2
a
 1
3
does not have integer solutions for all primes
and  , while the equation 3a  1  3h  1  3e 
2
2
has integer
solutions for composite numbers.
∴ The sum formula   ,   of primes between positive integers  and  is
an
hn
en
a1
h1
e1
  ,     3ai  1   3hi  12  3ei 2
   1
   1
2
2
Where, a1  
 1 , an  
, 3a1  1  3h1  1  3e1  ,


 3 
 3 
3a n  1  3hn  1  3en  .
2
2
3hi  12  3ei 2
And
 3h j  1  3e j  . if
2
2
3hi  12  3ei 2
 3h j  1  3e j  , it
2
2
could only take one of ( hi , ei ) and ( h j , e j ) in the calculations.
And there are 3a  1  3h2  1  9e22 , 3a  1  9e12  3h1  1 .
2
2
∴ The number formula   ,   of primes between positive integers  and  is
  ,     3ai  1   9h  6hi  9e  
hn
an
en
2
i
h1
a1
2
i
e1
an
 3a
a1
i
hn
en
h1
e1

 1   9ei2  9hi2  6hi

It could also be written as
an
hn
en
a1
h1
e1
  ,     3ai  1   9hi2  6hi  9ei2  2n
5. Proof of the upper and lower limit formula of number of primes between two positive
integers
Theorem3: The upper and lower limit formula of number   ,   for primes between
positive integers  and  , where    .
Proof.
Using the same method as in the proof of Theorem1, for  , there are 3 cases:   3  1 ,
12
  3 and   3  1 .
Let when  is a even,   2 . Let when  is odd,   2  1 .
Then,
Case   3  1 :
when   2 ,   6  1 ,
when   2  1 ,   6  4 .
Case   3 :
when   2 ,   6 ,
when   2  1 ,   6  3 .
Case   3  1 :
when   2 ,   6  1 ,
when   2  1 ,   6  2 .
∵Any prime could be expressed as 6b  1 .
∴Just consider only a situation that    could be expressed as 6b  1 , while the rest are
composite numbers. Where      .
Because of    corresponding to be expressed as 6b  1 and 6b  1 , there are 6 cases
as follows:
when   6  1 ,   6  2 or   6 ,
when   6  4 ,   6  5 or   6  3 ,
when   6 ,   6  1 or   6  5 ,
when   6  3 ,   6  4 or   6  2 ,
when   6  1 ,   6 or   6  4 ,
when   6  2 ,   6  3 or   6  1 .
13
Let   6   , then it could be marked
when   6  1 ,  1,  2   2, 0 ,
when   6  4 , 1,  2   5, 3 ,
when   6 ,  1,  2   1, 5 ,
when   6  3 ,  1,  2   4, 2 ,
when   6  1 , 1,  2   0, 4 ,
when   6  2 , 1,  2   3, 1 .
Using the same method as in the proof of Theorem1, there are
When     3h  1  3e  , let
2
2
   0  3a 0  1  3h0  12  3e0 2  A
Where,  0 is the minimum of  .
If    is a composite number, the equation    0    3h0   h   1  3e0   e 
2
2
has integer solutions.
There is
h 
1
 3h0  1 
3 
3h0  12  92e  18e0  e   

∴ 3h0  1  92e  18e0  e    is a square number.
2
Then let
3h0  12  92e  18e0  e    
f
2
3h0  12  B
There is
e 
∴ 9e02    B  f
2

1
 3e0  9e02    B  f
3
2

is also a square number.
And  max    
14
      1  1
∴ The number of  that make    be expressed as 6b  1 is 
.
6


The number of  that make 9e02    B  f
2
be square number and be divided evenly by
     1  1 
6 is 
.
6


 1     1  1 
The number of  that make  e be positive integers is 
.
6
 3

But this  could not make  h be positive integers.
∴ The number of primes be expressed as 6b  1 between  and  is
      1  1  1      1  1 
      1  1

  1    ,    



6
6
6

  3



When     3e   3h  1 , let
2
2
   0    3a 0  1  3e0 2  3h0  12  A
Where,  0 is the minimum of  .
If    is a composite number, the equation    0    3e0   e   3h0   h   1
2
2
has integer solutions.
There is
h 
1
 3h0  1 
3 
3h0  12  92e  18e0  e   

∴ 3h0  1  92e  18e0  e    is a square number.
2
Then let
3h0  12  92e  18e0  e    
f
2
3h0  12  B
There is
e 
∴ 9e02    B  f
2

1
 3e0  9e02    B  f
3
2

is also a square number.
15
And  max    
      2  1
∴ The number of  that make    be expressed as 6b  1 is 
.
6


The number of  that make 9e02    B  f
2
be square number and be divided evenly by
    2 1
6 is 
.
6


1      2  1 
The number of  that make  e be positive integers is 
.
6
 3

But this  could not make  h be positive integers.
∴ The number of primes be expressed as 6b  1 between  and  is
      2  1  1      2  1 
      2  1









1
,



 3


6
6
6

 



∴The upper and lower limit of number   ,   for primes between positive integers 
and  is
      1  1       2  1


6
6

 

  ,    
      1  1       2  1  1      1  1   1      2  1 

2

  3
6
6
6
6

 
 
  3

  ,    
This completes the proof.
References
1. Vinogradov Ivan Matveevich: Elements of Number Theory. State izdatelstvo technical and theoretical
literature (1952). (Qiu Guangming Translate, Harbin Institute of Technology Press (2011))
2. Pan Chengdong, Pan Chengbiao: Elementary proof of the prime number theorem. Shanghai Science and
Technology Press (1988).
3. Wang Yuan. Dictionary of Mathematics. Science Press of China (2010).
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