Download The Proof of the Twin Primes Conjecture

The Proof of the Twin Primes Conjecture
Ye Zhijiu Guo Xiaoning
(Shaanxi Technical College of Finance & Economics, Shaanxi Xianyang, 712000, China)
Abstract: Use mathematical induction to prove a class of missing-item congruence equations
solution-less. If this class of missing-item congruence equations is solution-less, then the
conjecture of twin primes is established. There is infinite number of twin prime pairs in the
natural domains. The conjecture of twin primes is true.
Key words: Twin Primes Conjecture, Mathematical induction, Congruence equations,
Positive integer solution.
No. of Chinese Library Classification: O156.2 Algebraic number theory
Document identification code:
Article No:
孪生素数猜想的证明
叶雉鸠 郭晓宁
(陕西财经职业技术学院 陕西咸阳 712000)
摘
要：采用数学归纳法证明了一类缺项同余式方程组恒无解。若这一类缺项同余式方程组恒无解则孪生
素数猜想成立，即自然数域中存在无穷多对孪生素数。孪生素数猜想是成立的。
关
键
词：孪生素数猜想 数学归纳法
同余式方程组 正整数解
MR(2000)主题分类：11R04 中图分类号：O156.1 初等数论
O156.2 代数数论
文献标识码：
Twin Primes Conjecture is a famous unsolved problem in number theory. This conjecture
was officially proposed by Hilbert at International Congress of Mathematicians in 1900 in
Question No.8[1]. It can be described as this: there are infinite numbers of prime number p , and
( p  2) is also a prime number. This pair of prime numbers p and ( p  2) is called twin
primes.
Here define two sets:
odd
prime numbers  3,5,7,11,  P
odd
prime numbers 

2a  1  3,5, pb   Pb  P
1 Conversion of the problems
1.1 From the Sequence to the Congruence Equation
Let us first obverse the following sequence.
7×9、9×11、11×13、13×15……
(1-1)
Obviously, (1-1) is the sequence of adjacent odd numbers products, where the first three
items (1×3、3×5、5×7) are eliminated from the complete sequence. In order to prove the infinity
of twin primes, what we have to do is to prove that there is infinite number of twin prime pairs in
Sequence (1-1). Therefore, the elimination of the first three items does not affect the proof when
we use Sequence (1-1) to prove the infinity of twin primes.
The general term formulas of Sequence (1-1) are (2a  1)( 2a  1)  (2a)  1 ( a  4 ,
2
a  N ). The question is, how can we be sure that (2a) 2  1 is the product of a pair of twin
primes given a certain a ？
If (2a)  1 cannot be divided by odd prime number (no greater than 2a  1 ), then
2
(2a) 2  1 is the product of a pair of twin primes. If (2a) 2  1 can be divided by odd prime
number that no greater than 2a  1 , then (2a)  1 is not the product of a pair of twin primes.
2
In another word, if natural number factors of (2a)  1 include any odd prime number that no
2
greater than 2a  1 , then (2a)  1 is not the product of a pair of twin primes. If natural
2
number factors of (2a)  1 do not include any odd prime number that no greater than 2a  1 ,
2
then (2a)  1 is the product of a pair of twin primes.
2
The above can be represented by congruence equation (1-2).
(2a) 2  1  0(mod pu )
(1-2)
In (1-2), pu is odd prime number, pu 
2a  1 .
If (1-2) has solution, either 2a  1 or 2a  1 is composite number. If (1-2) has no solution,
then 2a  1 and 2a  1 are a pair of twin prime numbers.
Put Sequence (1-1) on the LHS of congruence equations, define modular region of
congruence equations within
2a  1 , we can obtain congruence equation (1-3).

(2  4) 2  1  0(mod p01 )

（2  5）2  1  0(mod p02 )


(2  6) 2  1  0(mod p03 )



2

[2(a - 1)]  1  0(mod p0 ( a4 ) )

(2a ) 2  1  0(mod p0 ( a41) )


max( p01, p02 , p03  p0 ( a41) )  pb  2a  1
（1-3）
Herein:
①The square factor of LHS of (1-3) congruence equations is continuous natural numbers (no less than 4).
②
pb
is the largest of those odd prime numbers which are no greater than
number of those odd prime numbers which are no greater than
③
p01, p02 , p03 , p0( a41)
totally different;
2a  1
,
b
is the total
2a  1 .
on the RHS of (1-3) equations can be totally same, or partially same, or
p01, p02 , p03 , p0( a41)
don’t pay attention to order, it does not mean that
p01  p02   p0 ( a  4 1) .
④The modular collection of RHS of (1-3) congruence equations is subset (or universal set) of all odd prime
numbers
Pb
which are no greater than
pb . It can be seen as modular domain.
1.2 The Implication of Congruence Equation (1-3)
In congruence equations (1-3), for the two factors 2asome  1 and 2asome  1 from
(2asome ) 2  1 on LHS of equation (1-3), if composite numbers or odd prime numbers that no
greater than
2a  1 exist, then there must be (2asome )2  1  0(mod psome b ) . Hence
psome b  Pb is true. However, if both 2asome  1 and 2asome  1 are odd prime numbers
which are greater than
2a  1 , i.e. twin prime numbers, then there must be
(2asome )2  1  0(mod psome b ) , hence psome b  Pb is false. If equations (1-3) have no
positive integer solutions, then 2asome  1 and 2asome  1 must be odd prime numbers which
are greater than
2a  1 , i.e. twin prime numbers.
People may doubt 1: If equations (1-3) have no positive integer solutions, can asome
approach   as a varies? The below is the deduction.
For certain a , if equations (1-3) have no positive integer solutions, then 2asome  1 and
2asome  1 must be odd prime numbers.
∵ 2asome  1 and 2asome  1 are odd prime numbers which are greater than
2a  1 .
∴ 2asome  1 ＞ 2a  1
asome ＞ ( 2a  1  1)  2 ，then asome is between ( 2a  1  1)  2 and a .
And ∵when a   ， ( 2a  1  1)  2  
And asome is between ( 2a  1  1)  2 and a
∴according to Squeeze Theorem, we can conclude that asome   .
As a   , if such congruence equation has no solution, which means that there must
always be twin prime numbers that greater than
2a  1 in equations (1-3). i.e. there are
infinite number of twin prime numbers in natural number domain (prime numbers with difference
of 2). Hence, the condition “such congruence equation has no solution” is the key to solve twin
prime number conjecture.
People may doubt 2: we only set one (not three) odd prime number which is no greater than
2a  1 in modulus space, can we be certain that Proposition 1 is a true statement? The answer
is yes.
If (2asome )  1  0 mod psome
2
b,
psome b  Pb cannot be true
So 2asome  1  0(mod psome b ) and 2asome  1  0(mod psome b ) cannot be true at the
same time
And since 2asome  1 and 2asome  1 are both no greater than 2a  1 (at most
2asome  1  2a  1 ), while psome b  Pb
So 2asome  1 and 2asome  1 are both prime numbers, in fact they are a pair of twin
primes.
Above all, as a   increases progressively, twin primes appear gradually. Hence, the
Twin Primes Conjecture is true.
1.3 Proposition 1 was presented
Proposition 1 For any natural number a ( a  11, a  N ) which is greater than 10 , the
congruence equations (1-3) have no positive integer solution.
We can understand “the congruence equations (1-3) have no positive integer solutions” in three ways.
(1)
a
(2) Let
modulus.
x
has no positive integer solutions.
x
be the difference between LHS of the congruence equations (1-3); let
divided by
y
y
be the respective
cannot all be positive integer numbers.
(3) The undetermined numbers in the congruence equations (1-3) cannot be positive integer numbers
simultaneously, i.e. the congruence equations (1-3) have no solutions.
The below is the proof of Proposition 1 using mathematical induction.
2 The proof of Proposition 1 using mathematical
induction
2.1 Substituting the initial value validation Proposition 1
When a  11, we can derive (1-4) from (1-3)

(2  4) 2  1  0(mod p01 )

(2  5) 2  1  0(mod p02 )


(2  6) 2  1  0(mod p03 )

(2  7) 2  1  0(mod p04 )


(2  8) 2  1  0(mod p05 )


(2  9) 2  1  0(mod p06 )


(2  10) 2  1  0(mod p07 )

(2  11) 2  1  0(mod p08 )

max( p01, p02 , p03  p08 )  p1  3  2  11  1
(1-4)
(1-4) can be further expanded into (1-5)
 (2  4) 2  1  0(mod 3) (1)

2
 (2  5)  1  0(mod 3) (2)
 (2  6) 2  1  0(mod 3) (3)

2
 (2  7)  1  0(mod 3) (4)

2
 (2  8)  1  0(mod 3) (5)
 (2  9) 2  1  0(mod 3) (6)

(2  10) 2  1  0(mod 3) (7)

2
 (2  11)  1  0(mod 3) (8)

max( p01, p02 , p03  p08 )  p1  3
(1-5)
After calculating, in (1-5), (3) and (6) have no positive integer solutions, so (1-5) has no
positive integer solutions, i.e. Proposition 1 is true.
When a  12 , we can derive (1-6) from (1-3)

(2  4) 2  1  0(mod p01 )

(2  5) 2  1  0(mod p02 )


(2  6) 2  1  0(mod p03 )



2

(2  12)  1  0(mod p09 )

max( p01, p02 , p03  p09 )  p2  5  2  12  1
(1-6)
(1-6) can be further expanded into (1-7)
 (2  4) 2  1  0(mod 3  5)(1)

2
 (2  5)  1  0(mod 3  5)(2)
 (2  6) 2  1  0(mod 3  5)(3)



2
(2  12)  1  0(mod 3  5)(9)


max( p01, p02 , p03  p09 )  p2  5
(1-7)
“  ” in (1-7) represents “otherwise”, it can be read as “extraction”. Equation (1-7) is in fact a
set of congruence equations. As the module of each equation has two possibilities, the
combination of nine equations produces 2  512 congruence equations. The modular domain
9
of (1-7) expands by one. In (1-7), (3) and (6) have no positive integer solutions, therefore (1-7) has
no positive integer solutions, i.e. Proposition 1 is true.
When a  13,14,15, 23 , the modular domain of congruence equations from (1-3) does
not expand, however the number of equations gradually increases. To add in new equations into
the unsolved equations makes the equations even more unsolvable. At this point, Proposition 1 is
true.
2.2 Suppose when a  n ( n  11 , n N ) Proposition 1 is
true
Condition: When a  n ( n  11 , n  N ), the congruence equations (1-8) have no positive
integer solutions.

(2  4) 2  1  0(mod p01 )

（2  5）2  1  0(mod p02 )


(2  6) 2  1  0(mod p03 )



2

[2(n - 1)]  1  0(mod p0 ( n4 ) )

(2n) 2  1  0(mod p0 ( n41) )


max( p01, p02 , p03  p0 ( n41) )  pb  2n  1
(1-8)
2.3 Prove that Proposition 1 is also true when a  n  1
Is “the congruence equations (1-3) have no positive integer solutions” still true when
a  n  1 ( n  11 , n  N )? The proof has to be done with two cases.
Case 1: a varying from n to n  1 causes no change to pb .
If a varying from n to n  1 causes no change of pb , i.e. a varying from n to
n  1 causes no change to the modular domain of equations. The modular domain does not
change, but the number of equations increases by one. To add in new equations into the unsolved
equations makes the equations even more unsolvable. At this point, Proposition 1 is true.
Case 2: a varying from n to n  1 causes a change to pb .
If a varying from n to n  1 causes a change to pb , i.e. a varying from n to
n  1 causes a change to the modular domain of equations. pb 1 is added in the modulus,
meanwhile the number of equations increases by one. At this time, the equations are shown as
(1-9).

(2  4) 2  1  0(mod p01 )

（2  5）2  1  0(mod p02 )


(2  6) 2  1  0(mod p03 )



2

(2n)  1  0(mod p0 ( n41) )

[2(n  1)]2  1  0(mod p0[( n1)41] )


max( p01, p02 , p03  p0[( n1)41] )  pb1  2(n  1)  1
At this time, pb 1 
2(n  1)  1
So pb 1  2(n  1)  1 , i.e. n  1 
2
pb21  1
2
(1-9)
In order to prove that (1-9) has no solutions when a  n  1 
pb21  1
, proof by
2
contradiction is applied.
Before we apply proof by contradiction, we need to adjust the equation numbers of (1-9) and
(1-8), so that the equation number of (1-9) is equal to that of (1-8).
Since the equation [2(n  1)]  1  0(mod p0[( n 1)  4 1] ) which we added in (1-9)
2
definitely has solutions, it can be omitted. As a result, the equation numbers of (1-9) and (1-8) are
now equal, which is n  4  1.
Prove [2(n  1)]  1  0(mod p0[( n 1)  4 1] ) definitely have solutions:
2
∵ [2(n  1)]  1  pb 1[2(n  1)  1]
2
2
∴ [2(n  1)]  1  0(mod pb 1 )
2
And ∵ max( p01 , p02 , p03  p0[( n 1)  41] )  pb 1
∴ [2(n  1)]  1  0(mod p0[( n 1)  4 1] ) definitely has solutions in (1-9).
2
Here define one set:
(2  4)
2
 

 1, (2  5)2  1,(2n)2  1  (2n)2  1n  11
Next, prove that (1-9) has no solutions using proof by contradiction.
Assumption: suppose that congruence equation (1-8) has no solutions while (1-9) has
solutions.
Recursion: when a  n  1 
pb21  1
2
We express the number, which cannot be expressed by modulus of congruence in Pb in the


left of (1-9), i.e. (2n)  1 n  11 and can only be expressed by modulus of congruence in
2
pb 1 , as (2nomit ) 2  1, and define “those involved natural numbers” in (2nomit ) 2  1 as
nomit , so (1-10) can be obtained from (1-9)
(2nomit ) 2  1  0 mod pb 1
(1-10)
Herein: nomit in (1-10) can also be interpreted as — — “those involved natural numbers in
(2n
omit

) 2  1 ”, having no solution in (1-8), but having solution in (1-9) in those equations.
∵ (2nomit )  1  (2nomit  1)( 2nomit  1)
2
∴From (1-10), we can derive (1-11) or (1-12)
2nomit  1  0 mod pb 1
(1-11)
2nomit  1  0 mod pb 1
(1-12)
Then 2nomit can be represented as 2nomit  (2r  1) pb 1  1 , where r are natural
numbers which make 2nomit no greater than pb 1  3 ( 2  4  2nomit  pb 1  3 ).
2
2
Substitute 2nomit  (2r  1) pb 1  1 into (1-10), we get (1-13)
(2nomit ) 2  1  (2r  1) pb 1[( 2r  1) pb 1  2]
(1-13)
Substitute 2nomit  (2r  1) pb 1  1 back into (1-11) or (1-12), we can get:
(2r  1) pb 1  0 mod pb 1
(1-14)
Since (1-13) and (1-14) have the common term 2r  1 , then let us discuss 2r  1 .
First, determine the numbers domain range of 2r  1
pb21  3
∵ 4  nomit  n 
2
∴ 8  2nomit  pb 1  3
2
(1-15)
And ∵ 2nomit  (2r  1) pb 1  1
∴ 8  (2r  1) pb 1  1  pb 1  3
2
(1-16)
Expand (1-16); we can get (1-17)
8  (2r  1) pb 1  1  pb21  3 or 8  (2r  1) pb 1  1  pb21  3
(1-17)
I.e. 7 / pb 1  2r  1  pb 1  4 / pb 1 or 9 / pb 1  2r  1  pb 1  2 / pb 1
(1-18)
So (1-19) can be derived from (1-18)
1  2r  1  pb 1  1
(1-19)
I.e. 2r  1 1,3,5,7,9,, pb 1  2
(1-20)
Next, classify 2r  1 into 2r  1  1 and 2r  1  1 , and discuss separately.
When 2r  1  1 , derive (1-21) from (1-13)
(2nomit ) 2  1  pb 1 ( pb 1  2)
(1-21)
If pb1  2 in (1-21) are composite numbers, then (1-22) is true
(2nomit )2  1  0 mod psome b , psome b  Pb
(1-22)
Herein: some b is some random b.
(1-22) show, when (1-8) has positive integer solutions, this contradicts with the assumption,
thus Proposition 1 is true.
(1-23)
If pb1  2 in (1-21) are prime numbers, this means that we can find a pair of twin primes
in this range, thus Twin Primes Conjecture is true.
(1-24)
When 2r  1  1 , 2r  1 3,5,7,9,, pb 1  2
We have 2r  1  0 mod psome
b
, psome
b
 Pb
(1-25)
(1-26) can be derived from (1-25)
I.e. (2r  1) pb 1[( 2r  1) pb 1  2]  0 mod psome
b
, psome
b
 Pb
(1-26)
(1-27) can be derived from (1-26) and (1-13)
(2nomit )2  1  0 mod psome b , psome b  Pb
(1-27)
When (1-8) has solutions, this contradicts with the assumption, i.e. Proposition 1 is true
under such circumstance.
Conclusion: According to mathematical induction, for any natural number n ( n  11 ,
n  N ), Proposition 1 is true.
3 Proposition of Loopholes and Single Solution-less
Theorem
3.1 One logic loophole
Although (1-24) is true, there is one logic loophole. Because proof by contradiction is always
based on upward solution-less recurrence, however (1-24) appears to have solution, it causes
upward solution-less recurrence to break, i.e. the upward recurrence cannot go towards infinitely
great natural number, thus ensuring Proposition 1 has solution in natural number range.
3.2 Proposition of single solution-less theorem
How to fix this loophole?
From the beginning of mathematic induction, we eliminate the two equations which have
(1-21) properties, in this way, the loophole is fixed. On top of Proposition 1, we have to propose
Proposition 2 which is “eliminating two equations which have (1-21) properties in (1-3)”.
Proposition 2: for any natural number a which is greater than 10( a  11, a  N ), after
eliminating two equations which have (1-21) properties in (1-3), the congruence equations have no
positive integer solutions.
Proposition 2 there is a stronger proposition Proposition 3.
Proposition 3: for any natural number a which is greater than 10( a  11, a  N ), after
eliminating two adjacent equations in (1-3), the congruence equations have no positive integer
solutions.
The numbers of equations in Proposition 2 and Proposition 1 are the partial and the whole,
if the partial has no solution, then the whole has no solution; if the whole has no solution, then the
partial either has or does not have solution. If Proposition 2 has no solution, then Proposition 1
has no solution; if Proposition 1 has no solution, then Proposition 2 either has or does not have
solution. Thus Proposition 2 is stronger than Proposition 1.
The eliminated equations in Proposition 3 and Proposition 2 are the random and the specific.
If the random has no solution, then the specific has no solution; if the specific has no solution,
then the random either has or does not have solution. If Proposition 3 has no solution, then
Proposition 2 has no solution; if Proposition 2 has no solution, then Proposition 3 either has or
does not have solution. Thus Proposition 3 is stronger than Proposition 2.
Let us name Proposition 3 as single solution-less theorem. However, single solution-less
theorem is not straightforward and can be hard to understand, thus we give the proof from (1-4) to
(1-27). After understanding the proof frames, it is easier now to understand the proof of single
solution-less theorem. Next, mathematic induction is used to prove Proposition 3.
4 Use Mathematical Induction to Prove Proposition 3
4.1 Substituting the initial value validation Proposition 3
When a  11, we can get (1-4) from (1-3), and (1-5) can be further obtained. In (1-5), (3) and
(6) have no positive integer solution. Since (3) and (6) are not adjacent, so after eliminating two
adjacent equations in (1-5), the congruence equations have no positive integer solutions. Hence,
Proposition 3 is true.
When a  12 , from (1-3) we can get (1-6), (1-6) can be further written as (1-7). The
modular domain of (1-7) expands by one. In (1-7), (3) and (6) have no positive integer solution.
Since (3) and (6) are not adjacent, so after eliminating two adjacent equations in (1-7), the
congruence equations have no positive integer solutions. Hence, Proposition 3 is true.
4.2 Assume when a  n（ n  11 ，n N ）, Proposition 3 is
true
Condition: When a  n （ n  11 ， n  N ）, after eliminating two adjacent equations in
(1-8), the congruence equations have no positive integer solutions.
4.3 Proposition 3 is true when a  n  1
When a  n  1（ n  11 , n  N ）, is “after eliminating two adjacent equations in (1-8), the
congruence equations have no positive integer solutions” still true? The discussion is under two
circumstances.
Circumstance 1: pb does not change when a varies from n to n  1
If pb does not change when a varies from n to n  1 , i.e. modular domain of
congruence equation does not change. However the number of equations increases by one when
modular domain does not change. The new equation set after adding one equation still has no
solution. Proposition 3 is true.
Circumstance 2: pb changes when a varies from n to n  1
If pb changes when a varies from n to n  1 , i.e. modular domain of congruence
equation changes. pb 1 is added in the modulus, meanwhile equation also increases by one. The
complete equation is shown as (1-9). The difference is that Proposition 3 has to face “congruence
equation set after eliminating two adjacent equations in (1-9)”.
Thus, pb 1 
2(n  1)  1 where n 
pb21  3
2
To prove that “congruence equation set after eliminating two adjacent equations in (1-9)” has
no solution when a  n  1 
pb21  1
, we have to use proof by contradiction.
2
The numbers of equations (1-9) and (1-8) have to be the same before we apply proof by
contradiction.
Since [2(n  1)]  1  0(mod p0[( n 1)  4 1] ) confirms to have solution, it can be eliminated.
2
In this way, the number of equation (1-9) is the same as that of (1-8), which is n  4  1
Next, we use proof by contradiction to prove that “congruence equation set after eliminating
two adjacent equations in (1-9)” has no solution
Assumption: Assume “congruence equation set after eliminating two adjacent equations in
(1-8)” has no solution, but “congruence equation set after eliminating two adjacent equations in
(1-9)” has solution
Recursion: When a  n  1 
pb21  1
, we can still get (1-10) to (1-27).However, (1-24) is
2
not the same. If pb 1  2 in (1-21) is prime number, (1-21) can be adjacent eliminated equation.
Even though it has solution, because of 2r  1  1 , “congruence equation set after eliminating
two adjacent equations in (1-9)” still has no solution
Conclusion: From Mathematic Induction, for any natural number n ( n  11 ，and n  N ),
Proposition 3 is always true.
Proposition 3 is true  Proposition 2 is true  Proposition 1 is true
Since Proposition 1 is true, as a   , twin prime numbers appear gradually, therefore
The conjecture of twin primes is true.
Reference:
[1] Bai Du Bai Ke [EB/OL]．http://baike.baidu.com/view/272607. htm, 2013-06-05．
About the author:
Ye Zhijiu(1965 - ), Male, Han(Race), Qianxian Shaanxi, Air Force Engineering University, Master of
Engineering, Associate Professor in Shaanxi Technical College of Finance & Economics, Major in Mathematics
and Economics.
About the translator:
Guo Xiaoning (1965 - ), Female, Han(Race), Xianyang Shaanxi, Xi'an International Studies University,
Associate Professor in Shaanxi Technical College of Finance & Economics, Major in English.
TELEPHONE: 13809102408
E-MAIL：[email protected]
MAILING ADDRESS：Shaanxi Technical College of Finance & Economics, NO.1 WENLIN ROAD
XIANYANG SHAANXI CHINA
POSTAL CODE:
712000
作者简介：叶雉鸠(1965-)，男，汉族，陕西乾县人，空军工程大学工程硕士，陕西财经职业技术学院
副教授，从事数学和经济学研究。
翻译人员简介：郭晓宁(1965-)，女，汉族，陕西咸阳人，西安外国语学院学士，陕西财经职业技术学
院副教授，从事英语教学与研究。
叶雉鸠电话号码：13809102408
E-mail：[email protected]
通讯地址：陕西省咸阳市文林路一号陕西财经职业技术学院
邮编：712000