江苏选调生-中秋节日祝福

此文发表在：Advances in Theoretical and Applied Mathematics (ATAM), ISSN 0793-4554, Vol. 7, №
4, 2012, pp.417-424
Proving Goldbach’s Conjecture by Two Number Axes’ Positive
Half Lines which Reverse from Each Other’s Directions
Zhang Tianshu
Nanhai west oil corporation,China offshore Petroleum,
Zhanjiang city, Guangdong province,

Email:
tianshu_zhang507@
;

Abstract
We know that every positive even number 2n(n
≥
3) can express in a sum which 3
plus an odd number 2k+1(k
≥
1) makes. And then, for any odd point 2k+1 (k
≥
1)at
the number axis, if 2k+1 is an odd prime point, of course even number 3+(2k+1) is
equal to the sum which odd prime number 2k+1 plus odd prime number 3 makes;
If 2k+1 is an odd composite point, then let 3point, and enable line segment B(2k+1) to equal line segment 3C. If C is an odd
prime point, then even number 3+(2k+1) is equal to the sum which odd prime
number B plus odd prime number C makes. So the proof for Goldbach’s
Conjecture is converted to prove there be certainly such an odd prime point B at the
number axis’s a line segment which take odd point 3 and odd point 2k+1 as ends, so
as to prove the conjecture by such a method indirectly.
Keywords
Number theory, Goldbach’s Conjecture, Even number, Odd prime number,
Mathematical induction, Two number axes’ positive half lines which reverse
from each other’s direction, OD, PL, CL, and RPL.
Basic Concepts
1

Goldbach’s conjecture states that every even number 2N is a sum of two
prime numbers, and every odd number 2N+3 is a sum of three prime
numbers, where N≥2.
We shall prove the Goldbach’s conjecture thereinafter by odd points at two
number axes’ positive half lines which reverse from each other’s directions
and which begin with odd point 3.

First we must understand listed below basic concepts before the proof of the
conjecture, in order to apply them in the proof.

Axiom．Each and every even number 2n (n≥3) can express in a sum which
3 plus each odd number 2k+1 (k≥1) makes.

Definition 1．A line segment which takes two odd points as two ends at the
number axis’s positive half line which begins with odd point 3 is called an
odd distance. “OD” is abbreviated from “odd distance”.
The OD between odd point N and odd point N+2t is written as OD N(N+2t),
where N≥3, and t≥1.
A integer which the length of OD between two consecutive odd points
expresses is 2.
A length of OD between odd point 3 and each odd point is unique.

Definition 2．An OD between odd point 3 and each odd prime point at the
number axis’s positive half line which begins with odd point 3, otherwise
called a prime length. “PL” is abbreviated from “prime length”, and “PLS”
2

denotes the plural of PL.
An integer which each length of from small to large PL expresses be
successively 2, 4, 8, 10, 14, 16, 20, 26. . .

Definition 3．An OD between odd point 3 and each odd composite point at
the number axis’s positive half line which begins with odd point 3,
otherwise called a composite length. “CL” is abbreviated from “composite
length”.
An integer which each length of from small to large CL expresses be
successively 6, 12, 18, 22, 24, 30, 32. . .

We know that positive integers and positive integers’ points at the number
axis’s positive half line are one-to-one correspondence, namely each integer’s
point at the number axis’s positive half line represents only a positive integer.
The value of a positive integer expresses the length of the line segment
between point 0 and the positive integer’s point here. When the line segment
is longer, it can express in a sum of some shorter line segments;
correspondingly the positive integer can also express in a sum of some
smaller integers.
Since each and every line segment between two consecutive integer’s points
and the line segment between point 0 and point 1 have an identical length,
hence when use the length as a unit to measure a line segment between two
integer’s points or between point 0 and any integer’s point, the line segment
has some such unit length, then the integer which the line segment expresses
is exactly some.
3

Since the proof for the conjecture relate merely to positive integers which are
not less than 3, hence we take only the number axis’s positive half line which
begins with odd point 3. However we stipulate that an integer which each
integer’s point represents expresses yet the length of the line segment between
the integer’s point and point 0. For example, an odd prime value which the
right end’s point of any PL represents expresses yet the length of the line
segment between the odd prime point and point 0 really.
We can prove next three theorems easier according to above- mentioned some
relations among line segments, integers’ points and integers.

Theorem 1．If the OD which takes odd point F and odd prime point P
S
as
two ends is equal to a PL, then even number 3+F can express in a sum of
two odd prime numbers, where F>P
S
.


Proof．Odd prime point P
S
represents odd prime number P
S
, it expresses the
length of the line segment from odd prime point P
S
to point 0.
Though lack the line segment from odd point 3 to point 0 at the number
axis’s positive half line which begins with odd point 3, but odd prime point
P
S
represents yet odd prime P
S
according to above-mentioned stipulation;
Let OD P
S
F=PL 3P
b
, odd prime point P
b
represents odd prime number P
b
, it
expresses the length of the line segment from odd prime point P
b
to point 0.
Since PL 3P
b
lack the segment from odd point 3 to point 0, therefore the
integer which the length of PL 3P
b
expresses is even number P
b
-3, namely
the integer which the length of OD P
S
F expresses is even number P
b
-3.
Consequently there is F=P
S
+(P
b
-3), i.e. 3+F=odd prime P
S
+ odd prime P
b
.

4

Theorem 2．If even number 3+F can express in a sun of two odd prime
numbers, then the OD which takes odd point 3 and odd point F as ends can
express in a sun of two PLS, where F is an odd number which is more than
3.
Proof．Suppose the two odd prime numbers are P
b
and P
d
, then there be
3+F= P
b
+P
d
.
It is obvious that there be OD 3F=PL 3P
b
+ OD P
b
F at the number axis’s
positive half line which begins with odd point 3.
Odd prime point P
b
represents odd prime number P
b
according to
above-mentioned stipulation, then the length of line segment P
b
(3+F) is
precisely P
d
, nevertheless P
d
expresses also the length of the line segment
from odd prime point P
d
to point 0. Thereupon cut down 3 unit lengths of
line segment P
b
(3+F), we obtain OD P
b
F; again cut down 3 unit lengths of
the line segment from odd prime point P
d
to point 0, we obtain PL 3P
d
, then
there be OD P
b
F=PL 3P
d
.
Consequently there be OD 3F=PL 3P
b
+ PL 3P
d
.

Theorem 3．If the OD between odd point F and odd point 3 can express in a
sum of two PLS, then even number 3+F can express in a sum of two odd
prime numbers, where F is an odd number which is more than 3.
Proof．Suppose one of the two PLS is PL 3P
S
, then there be F＞P
S
, and the
OD between odd point F and odd prime point P
S
is another PL.
Consequently even number 3+F can express in a sum of two odd prime
numbers according to theorem 1.
The Proof
5

First let us give ordinal number K to from small to large each and every odd
number 2k+1, where k≥1, then from small to large each and every even
number which is not less than 6 is equal to 3+(2k+1).

We shall prove this conjecture by the mathematical induction thereinafter.

1．When k=1, 2, 3 and 4, we getting even number be orderly
3+(2*1+1)=6=3+3, 3+(2*2+1)=8=3+5, 3+(2*3+1)=10=3+7 and
3+(2*4+1)=12=5+7. This shows that each of them can express in a sum of
two odd prime numbers.

2．Suppose k=m, the even number which 3 plus №m odd number makes, i.e.
3+(2m+1) can express in a sum of two odd prime numbers, where m≥4.

3．Prove that when k=m+1, the even number which 3 plus №(m+1) odd
number makes, i.e. 3+(2m+3) can also express in a sum of two odd prime
numbers.
Proof
．In case 2m+3 is an odd prime number, naturally even number 3+
（2m+3）is the sum of odd prime number 3 plus odd prime number 2m+3
makes.
When 2m+3 is an odd composite number, suppose that the greatest odd
prime number which is less than 2m+3 is P
m
, then the OD between odd
prime point P
m
and odd composite point 2m+3 is either a PL or a CL.
When the OD between odd prime point P
m
and odd composite point 2m+3 is
a PL, the even number 3+（2m+3）can express in a sum of two odd prime
numbers according to theorem 1.
6

If the OD between odd prime point P
m
and odd composite point 2m+3 is a
CL, then we need to prove that OD 3（2m+3）can express in a sum of two
PLS, on purpose to use the theorem 3.
When OD P
m
(2m+3) is a CL, from small to large odd composite number
2m+3 be successively 95, 119, 125, 145. . .

First let us adopt two number axes’ positive half lines which reverse from
each other’s directions and which begin with odd point 3.
At first, enable end point 3 of either half line to coincide with odd point 2m+1
of another half line. Please, see first illustration:

3 5 7 2m-3 2m+1

2m+1 2m-3 7 5 3

First Illustration
Such a coincident line segment can shorten or elongate, namely end point 3
of either half line can coincide with any odd point of another half line.
This proof will perform at some such coincident line segments. And for
certain of odd points at such a coincident line segment, we use usually
names which mark at the rightward direction’s half line.
We call PLS which belong both in the leftward direction’s half line and in a
coincident line segment “reverse PLS”. “RPLS” is abbreviated from
“reverse PLS”, and “RPL” denotes the singular of RPLS.
The RPLS whereby odd point 2k+1 at the rightward direction’s half line
acts as the common right endmost point are written as RPLS
2k+1
, and
RPL
2k+1
denotes the singular, where k>1.
This is known that each and every OD at a line segment which takes odd
7

point 2m+1 and odd point 3 as two ends can express in a sum of a PL and a
RPL according to preceding theorem 2 and the supposition of №2 step of
the mathematical induction.

We consider a PL and the RPL
2k+1
wherewith to express together the length
of OD 3(2k+1) as a pair of PLS, where k >1. One of the pair’s PLS is a PL
which takes odd point 3 as the left endmost point, and another is a RPL
2k+1

which takes odd point 2k+1 as the right endmost point. We consider the
RPL
2k+1
and another RPL
2k+1
which equals the PL as twin RPLS
2k+1
.
For a pair of PLS, the PL is either unequal or equal to the RPL
2k+1
. If the PL
is unequal to the RPL
2k+1
, then longer one is more than a half of OD 3(2k+1),
yet another is less than the half. If the PL is equal to the RPL
2k+1
, then either
is equal to the half. A pair of PLS has a common end’s point.

Since each of RPLS
2k-1
is equal to a RPL
2k+1
, and their both left endmost
points are consecutive odd points, and their both right endmost points are
consecutive odd points too. So seriatim leftwards move RPLS
2k+1
to become
RPLS
2k-y
, then part left endmost points of RPLS
2k+1
plus RPLS
2k-y

coincide
monogamously with part odd prime points at OD 3(2k+1), where y=1, 3,
5, ...

Thus let us begin with odd point 2m+1, leftward take seriatim each odd
point 2m-y+2 as a common right endmost point of RPLS
2m-y+2
, where y= 1,
3, 5, 7 ... ỹ ...
Suppose that y increases orderly to odd number ỹ,

and ∑ part left endmost
8

points of RPLS
2m-y+2

(1≤y≤ỹ+2)
coincide just right with all odd prime points
at OD 3(2m+1) monogamously, then there are altogether (ỹ+3)2 odd points
at OD (2m-ỹ)(2m+1), and let μ=(ỹ+3)2.

Let us separate seriatim OD 3(2m-y+2) (y=1, 3, 5,…) from each coincident
line segment of two such half lines, and arrange them from top to bottom
orderly. After that, put an odd prime number which each odd prime point at
the rightward direction’s half line expresses to on the odd prime point, and
put another odd prime number which each left endmost point of RPLS
2m-y+2

at the leftward direction’s half line expresses to beneath the odd prime point.

For example, when 2m+3=95, 2m+1=93, 2m-1=91 and 2m-ỹ =89，μ=3.
For the distributer of odd prime points which coincide monogamously with
left endmost points of RPLS
95
, RPLS
93
, RPLS
91
, RPLS
89
and RPLS
87
, please
see second illustration:

OD 3(95)

19 31 37 61 67 79

79 67 61 37 31 19
OD 3(93)
7 13 17 23 29 37 43

53 59 67 73 79 83 89

89 83 79 73 67 59 53 43 37 29 23 17 13 7
OD 3(91)
5 11 23 41 47 53 71 83 89
89 83 71 53 47 41 23 11 5
OD 3(89)
13 19 31 61 73 79

79 73 61 31 19 13
OD 3(87)
7 11 17 19 23 29 31 37 43 47 53 59 61 67 71 73 79 83
83 79 73 71 67 61 59 53 47 43 37 31 29 23 19 17 11 7

Second Illustration
Two left endmost points of twin RPLS
2m-y+2
at OD 3(2m-y+2) coincide
monogamously with two odd prime points, they assume always bilateral
symmetry whereby the centric point of OD 3(2m-y+2) acts as symmetric
9

centric. If the centric point is an odd prime point, then it is both the left
endmost point of RPL
2m-y+2
and the odd prime point which coincides with
the left endmost point, e.g. centric point 47 of OD 3(91) in above-cited that
example.

We consider each odd prime point which coincides with a left endmost point
of RPLS
2m-ỹ
alone as a characteristic odd prime point, at OD 3(2m+1).
Thus it can seen, there is at least one characteristic odd prime point at OD
3(2m-ỹ) according to aforesaid the way of making things, e.g. odd prime
points 19, 31 and 61 at OD 3(89) in above-cited that example.
Whereas there is not any such characteristic odd prime point in odd prime
points which coincide monogamously with left endmost points of RPLS
2m+1
plus RPLS
2m-1
... plus RPLS
2m-ỹ+2.
In other words, every characteristic odd prime point is not any left endmost
point of RPLS
2m+1
plus RPLS
2m-1
... plus RPLS
2k-ỹ+2
.



Moreover left endmost points of RPLS
2m-y
are №1 odd points on the lefts of
left endmost points of RPLS
2m-y+2
monogamously, where y is an odd
number≥1.

Consequently, №1 odd point on the left of each and every characteristic odd
prime point isn’t any left endmost point of RPLS
2m-1
plus RPLS
2m-3
…plus
RPLS
2m-ỹ
. — {1}

Since each RPL
2m-y+2
is equal to a PL at OD 3(2m-y+2).
10

In addition, odd prime points which coincide monogamously with left
endmost points of RPLS
2m+1
plus RPLS
2m-1
... plus RPLS
2m-ỹ
are all odd
prime points at OD 3(2m+1).
Hence considering length, at OD 3(2m+1) RPLS whose left endmost points
coincide monogamously with all odd prime points are all RPLS at OD
3(2m+1), irrespective of the frequency of RPLS on identical length.
Evidently the longest RPL at OD 3(2m+1) is equal to PL 3P
m
.

When OD P
m
(2m+3) is a CL, let us review aforesaid the way of making
thing once again, namely begin with odd point 2m+1, leftward take seriatim
each odd point 2m-y+2 as a common right endmost point of RPLS
2m-y+2
,
and ∑ part left endmost points of RPLS
2m-y+2

(1≤y≤ỹ+2)
coincide just right
with all odd prime points at OD 3(2m+1) monogamously.
Which one of left endmost points of RPLS
2m-y+2

(1≤y≤ỹ+2)
coincides first
with odd prime point 3? Naturally it can only be the left endmost point of
the longest RPL
P
m

whereby odd prime point P
m
acts as the right endmost
point.
Besides all coincidences for odd prime points at OD 3(2m+1) begin with
left endmost points of RPLS
2m+1
, whereas left endmost points of RPLS
2m-ỹ
are final one series in the event that all odd prime points at OD 3(2m+1) are
coincided just right by left endmost points of RPLS.


Therefore odd point 2m-ỹ as the common right endmost point of RPLS
2m-ỹ

cannot lie on the right of odd prime point P
m
, then odd point 2m-ỹ-2 can
only lie on the left of odd prime point P
m
. This shows that every RPL
2m-ỹ-2
at
OD 3(2m-ỹ-2) is shorter than PL 3 P
m
.
11

In addition, №1 odd point on the left of a left endmost point of each and
every RPL
2k-ỹ
is a left endmost point of RPLS
2k-ỹ-2.

Therefore

each and every RPL
2m-ỹ-2
can extend contrary into at least one
RPL
2m-y+2,
where y is a positive odd number ≤ỹ+2.
That is to say, every left endmost point of RPLS
2m-ỹ-2
is surely at least one
left endmost point of RPLS
2k-y+2
.
Since left endmost points of RPLS
2m-ỹ-2
lie monogamously at №1 odd point
on the left of left endmost points of RPLS
2m-ỹ
including characteristic odd
prime points.
Consequently, №1 odd point

on the left of each and every characteristic odd
prime point is surely a left endmost point of RPLS
2m-y+2
,
where
1≤y≤ỹ+2.—— {2}

So we draw inevitably such a conclusion that №1 odd point on the left of
each and every characteristic odd prime point can only be a left endmost
point of RPLS
2m+1
under these qualifications which satisfy both
above-reached conclusion {1}

and above-reached conclusion {2}.

Such being the case, let us rightwards move a RPL
2m+1
whose left endmost
point lies at №1 odd point on the left of any characteristic odd prime point
to adjacent odd points, then the RPL
2m+1
is moved into a RPL
2m+3.

Evidently the left endmost point of the RPL
2m+3
is the characteristic odd
prime point, and its right endmost point is odd point 2m+3.

So OD 3(2m+3) can express in a sum of two PLS, and the common
12

endmost point of the two PLS is exactly the characteristic odd prime point.
Thus far we have proven that even if OD P
m
(2m+3) is a CL, likewise OD
3(2m+3) can also express in a sum of two PLS.

Consequently even number which 3 plus №(m+1) odd number makes, i.e.
3+(2m+3) can also express in a sum of two odd prime numbers according to
aforementioned theorem 3.

Proceed from a proven conclusion to prove a larger even number for each
once, then via infinite many an once, namely let k to equal each and every
natural number, we reach exactly a conclusion that every even number
3+(2k+1) can express in a sum of two odd prime numbers, where k≥1.
To wit every even number 2N can express in a sum of two odd prime
numbers, where N>2.

In addition let N =2, get 2N=4=even prime number 2+even prime number 2.
Consequently every even number 2N can express in a sum of two prime
numbers, where N≥2.

Since every odd number 2N+3 can express in a sum which a prime number
plus the even number makes, consequently every odd number 2N+3 can
express in a sum of three prime numbers, where N≥2.

To sum up, we have proven that two propositions of the Goldbach’s
conjecture are tenable, thus Goldbach’s conjecture holds water.
13

附，翻译成汉语：
利用互为反向数轴的正射线证明哥德巴赫猜想
张天树
Tianshu_zhang507@
摘 要
我们知道,依次增大 的每一个正偶数2n(n≧3)可以表示成3分别与依次增大的一
个奇数2k+1（k≧1）之和
.
于是,对于数轴上的任意一个奇数点2k+1（k≧1）,如果2k+1
是一个奇素数点, 当然,偶数3+(2k+1）可等于奇素数 2k+1 与奇素数 3 之和;如果
2k+1是一个奇合数点,那么,取 3
2k+1,这里,
B
是一个奇素数点,且使线段
B(2k+1)等于线段3C
.
如果C是一个奇素数点,那么, 偶数3+（2k+1）等于奇素数B与
奇素数 C 之和.
于是对哥德巴赫猜想的证明就变换成了去证明：在数轴的以奇数点3
和2k+1为端点的 线段上，总是存在着这样的奇素数点B，以此方法来间接地证明哥德
巴赫猜想
.

关键词
数论、哥德巴赫猜想、数学归纳法、偶数、奇素数、互为反方向的数轴
的正射 线、奇距、素长、合长和反向素长.
基 本 慨 念
哥德巴赫猜想表为:每一个偶数2N 都是两个素数之和,每一个奇数
2N+3都是三个素数之和,这里 N
≧2
.


本文将用互为反方向的两条数轴的从奇数点3开始的正方向射线上
的奇数点来证明这个猜想.

在证明这个猜想之前，我们先要熟知下述的基本慨念，以便在证明
的过程中应用它们.
14

公理.

依次增 大的每一个正偶数2n(n≧3)可以表示成3分别与依次增大
的一个奇数2k+1（k≧1）之和.

定义1. 在数轴的从奇数点3开始的正方向射线上，以任意两个奇数点为端点的线段,称为这两个奇数点之间的距离，简称奇距．我们用符号
“OD”表示奇距.
奇数点N与奇数点N+2t之间的奇距,写成 OD N(N+2t)，这里 N
≧3,
t
≧1
.
相邻两个奇数点之间的奇距是2，奇数点3与各个奇数点之间的奇距长度,
都是唯一的.

定义2. 在数轴的从奇数点3开始的正方向射线上，以任意一个奇素
数点 和奇数点3为端点的奇距，也被称为这个奇素数点的素长，并用符号
“PL”表示一条素长，和符号“P LS” 表示至少两条素长，即素长的复数.
从小到大的素长依次是: 2、4、8、10、14、16、20、26. . . . . .

定义3. 在数轴的 从奇数点3开始的正方向射线上，以任意一个奇合
数点和奇数点3为端点的奇距，又被称为这个奇合数点 的合长，并用符号
“CL” 表示一条合长.
从小到大的合长依次是: 6、12、18、22、24、30、32. . . . . .

我们知道，在数轴的正 方向射线上的整数点与正整数是一一对应的.
也就是说，在数轴的正方向射线上，每一个整数点只代表一 个整数，这
个整数的值在这里表示这个整数点距0点的线段长度. 当这条线段较长
时，它可表为若干条线段之和. 因此，这个整数也可表为若干个整数之
15

和. 且 因为0点与1点、以及相邻两个整数点之间的线段长度都是相等的 ,
当我们把这些相等线段每条的长度作为1个长度单位去度量任意一个整
数点距0点的线段或任 意两个整数点之间的线段长度时,该线段有多少个
这样的长度单位，它就代表多大的整数.
因为本文的证明仅仅用到不小于3的整数，所以，我们只取数轴的从
奇数点3开始的正方向射线，但我们 规定：在这射线上的每个整数点代表
的整数值仍然是指这个整数点距0点的线段长度. 例如，在这射线 上任意
一条素长的右端点所代表的奇素数值，实际上是指这个奇素数点距0点的
线段长度.
根据以上所述，我们很容易地证明以下三条定理:

定理1. 如果以奇数点F和奇素数点P
S
为端点的奇距等于一条素长，
那么，偶数3+F可表为两个奇素数之和，这里 F＞P
S
.
证明：奇素数点P
S
代表奇素数P< br>S
，P
S
的值表示奇素数点P
S
到0点的长
度.在数 轴的从奇数点3开始的正方向射线上虽然缺少从奇数点3到0点的
一段，但是按照前面的规定，奇素数点 P
S
仍然代表奇素数P
S
；
令 OD P
S
F=PL 3P
b
，奇素数点P
b
代表奇素数P
b
，P
b
的值表示奇素数点P
b
到0点的长度. 但是，PL 3P
b
缺少从奇数点3到0点的一段，因此，PL 3P
b
的长度表示的整数是偶数P
b
-3，即OD P
S
F的长度表示的整数是偶数P
b
-3.
所以, F=P
S
+(P
b
-3)，即 3+F=奇素数P
S
+奇素数P
b
.

定理2. 如果偶数3+F可表为两个奇素数之和，那么，以奇数点F和奇
数点3为端点的奇距可表为两 条素长之和，这里 F是一个大于3的奇数.


证明：假设这两个奇素数为P
b
和P
d
，则有3+F= P
b
+P
d
. 显然，在数
轴的从奇数点3开始的正方向射线上，OD 3F=PL 3P
b
+ OD P
b
F. 根据前面
16

的规定，奇素数点P
b
代表奇素数P
b
，那么，线段 P
b
(3+F)的长度就是P
d
，
该长度是指从奇素数点P
d
到 0点的线段长度.在线段 P
b
(3+F)上去掉3个单
位长度，得到OD P
b
F；在从奇素数点P
d
到0点的线段上去掉3个单位长度，
得到PL 3P
d
，于是，OD P
b
F=PL 3P
d
.

所以, OD 3F=PL 3P
b
+ PL 3P
d
.



定理3. 如果奇数点F与奇数点3

之间的奇距可表为两条素长之和，那
么，偶数3+F可表为两个奇素数之和，这里 F是一个大于3的奇数.


证明：假设这两条素长中的一条是PL 3P
S
，那么，F＞P
S
，且奇数点
F与奇素数点P
S
之间的奇距是另一条素长. 根据定理1，偶数3+F可表为两
个奇素数之和.
证 明

首先,让我们以自然数k给每一个依次增大的奇数2k+1编上序号，
这里 k
≧1
,那么，每一个不小于6的依次增大的偶数等于3+(2k+1). 在下文
中, 我们将运用数学归纳法来证明这个猜想.

1.当 k= 1、2、3和4时，我们依次得到的偶数: 3+（2×1+1）=6=3+3，
3+（ 2×2+1）=8=3+5，3+（2×3+1）=10=3+7或5+5，3+（2×4+1）=12=5+7
都可表为两个奇素数之和.

2. 假设当 k=m时，3加上第m个奇数所得的偶数、即 3+（2m+1）可
表为两个奇素数之和，这里 m
≧4
.

3.证明: 当k=m+1时，3加上第m+1个奇数所得的偶数、即 3+（2m+3）
也可表为两个奇素数之和.

17

证明 如果2m+3是一个奇素数，当然，偶数 3+（2m+3）可表为奇
素数3与奇素数2m+3之和.
当2m+3是一个奇合数时，假设小于2m+3的最大奇素数为P
m
，

那么，
奇合数点2m+3与奇素数点P
m

之间的奇距或是一条素长，或是一条合长.

当奇合数点2m+3与奇素数点P
m

之间的奇距是一条素长时，根据定理
1，偶数 3+（2m+3）可表为两个奇素数之和.
如果奇合数点2m+3与奇素数点P
m

之间的奇距是一条合长，我们则需
要证明 OD 3（2m+3）可表为两条素长之和，以便应用定理3.

当OD P
m
( 2m+3)是一条合长时，2m+3从小到大的值依次是95、119、
125、145. . . . . .

首先，让我们采用两条互为相反方向的数轴的从奇数点3开始的正方
向射 线，最初,让一条射线上的奇数点3与另一条射线上的奇数点2m+1重
合. 请参看第一图：

3 5 7 2m-3 2m+1

2m+1 2m-3 7 5 3



第一图
这两条射线互相重合的线段能够缩短或伸长，也就是说，一条射线
的端点3能够重合另 一条射线的任意一个奇数点. 本文的整个证明将在这
样互相重合、且可伸缩的线段上施行，并且，对这 线段上互相重合后的
一个奇数点，我们主要使用它在向右方向射线上的名称.

我们把既属于向左方向射线，又在两条射线互相重合的线段上的素长
18

称为“反向素长” . 一条反向素长用符号“RPL”表示，它的复数表为
“RPLS”. 以向右方向射线上的奇数点2k+ 1作为共同右端点的至少两条反
向素长被写作RPLS
2k+1
,以及RPL
2k+1
表示其中的一条，这里 k＞
1
.

根据前列的 定理2和数学归纳法中第二步的假设，我们知道，在以奇
数点2m+1和奇数点3为端点的线段上的各条 奇距，都能够表为一条素长
与一条反向素长之和. 我们把表示 OD 3(2k+1)为两条素长之和的两条素
长看作是一对素长，这里 k＞1. 在这对素长中，一条是以奇数点3为左
端点的素长，另一条则是以奇数点2k+1为右端点的反向素长. 我们把这
对素长中的反向素长和在长度上等于这对素长中素长的另一条以奇数点
2k+1为右端 点的反向素长看作是孪生的反向素长.

对于共表 OD 3(2k+1)长度的一对素长，如果它们不等，那么，较长
的一条比OD 3（2k+1）的一半长，而另一条则比这一半短. 如果这两条
素长相等，那么，每一条都等于OD 3（2k+1）的一半. 共表OD 3（2k+1）
长度的一对素长，它们有一个共同的端点.
因为在长度上每一条RPL
2k-1
等于一条RPL
2k+1
. 并且，它们的左端点
是相邻的奇数点，右端点也是相邻的奇数点. 于是,逐点向左移动
RPLS
2k+1
使之成为RPLS
2k-y,

那么，RPLS
2k+1
和RPLS
2k-y
的部分左端点就一
对一地重合OD 3（2k+1）上的奇素数点，这儿 y=1, 3, 5, …...

因此,让我们在数轴的从奇数点3开始的正方向射线上 ，从奇数点
2m+1开始，向左依次取每一个奇数点2m-y+2作为反向素长的共同右端
点， 这里 y=1、3、5、7、...ỹ、...
假设y依次增大到奇数ỹ，且 RPLS
2m+1
、RPLS
2m-1
... RPLS
2m-ỹ+2
和
19

RPLS
2m-ỹ
的部份左端点一对一地恰好重合完OD 3（2m+1）上的全部奇素
数点，那么，在 OD（2m- ỹ）（2m+1）上共有（ỹ+3）2 个奇数点，并 且
令 μ =（ỹ+3）2.

然后，我们按照从长到短的次序，从两条射线的重合段中依次分离出
OD 3（2m-y+2） ，并把它们从上到下依次排列起来，还把向右方向射线
上表示奇素数点的奇素数放在这个奇素数点的上方 ，把重合这奇素数点
的表示反向素长左端点的奇素数放在它的下方.

例如:当2m+3=95时，2m+1=93, 2m-1=91和2m-ỹ =89，μ=3. 对于OD
3(95)、OD 3(93)、OD 3(91) 、OD 3(89) 和OD 3(87)上反向素长左端点重
合奇素数点的分布情况，如下图：

OD
5(95)

19 31 37 61 67 79

79 67 61 37 31 19
OD
5(93)
7 13 17 23 29 37 43

53 59 67 73 79 83 89

89 83 79 73 67 59 53 43 37 29 23 17 13 7
OD
5(91)

5 11 23 41 47 53 71 83 89
89 83 71 53 47 41 23 11 5
OD
5(89)
13 19 31 61 73 79

79 73 61 31 19 13
OD
5(87)

7 11 17 19 23 29 31 37 43 47 53 59 61 67 71 73 79 83
83 79 73 71 67 61 59 53 47 43 37 31 29 23 19 17 11 7
第 二 图

在OD 3（2m-y+2）上，孪生反向素长的 两个左端点一对一重合的两
个奇素数点，以该奇距的中点为对称中心左右对称. 如果该奇距的中点是一个奇素数点，那么,它既是反向素长的左端点，又是这个左端点重合
的奇素数点，例如，在上面 引用例子中，OD 3（91）的中点 47.
在OD 3（2m+1）上，我们仅仅把唯一由RPLS
2m-ỹ
的左端点重合的奇
素数点， 即在RPLS
2m+1
、RPLS
2m-1
、...和 RPLS
2m-ỹ+2
的左端点一一重合的
20

奇素数点中没有的奇素数点看作是特有的奇素数点. 那么，根据上述的
作法，在RPLS
2m-ỹ
的左端点一一 重合的奇素数点中，有至少一个特有的
奇素数点. 例如，上面所举那个例子中，在OD 3(89)上的奇素数点19、31
和61.

因为在RPLS
2m+1
、RPLS
2m-1
、...和RPLS
2m-ỹ+2
的左 端点一一 重合的奇
素数点中，没有一个特有的奇素数点. 换言之，每一个特有的奇素数点
都 不是RPLS
2m+1
、RPLS
2m-1
、...和RPLS
2m -ỹ+2
的左端点.
又 RPLS
2m-y+2
的左端点左边的第一 个奇数点是RPLS
2m-y
的左端点，这
里 y是≧1的正奇数.
所 以,在每一个特有奇素数点左边的第一个奇数点不是RPLS
2m-1
、
RPLS2m-3
、...和RPLS
2m-ỹ
的左端点. --- {1}

因为在OD 3(2m-y+2)上，每一条反向素长等于一条素长. 加之，
RPLS
2m+1
、RPLS
2m-1
、...和 RPLS
2m-ỹ
的左端点一一重合的奇素数点是OD 3
（2m+1）上全部的奇素数点.
因此，仅以长度而言，在OD 3(2m+1)上，其左端点一对一地重合完
全部奇素数点的反向素长是OD 3(2m+1)上全部的反向素长，而不与具有
同一长度的反向素长的条数有关.
显然，在OD 3(2m+1)上的全部反向素长中，最长的反向素长等于PL 3P
m
.

当奇合数点2m+3与奇素数点P
m

之间的奇距是一条合长时，让我们回
顾前述的作法，即 从2m+1开始，向左依次取每一个奇 数点作为反向素
长的共同右端点，并且，RPLS
2m+1
、RPLS
2m- 1
、...和 RPLS
2m-ỹ
的左端点一
对一重合的奇素数点是OD 3（2m+1）上全部的奇素数点.
21

究竟哪一条反向素长 的左端点首先重合奇素数点3呢？显然，它仅仅
能够是以奇素数点P
m
为右端点的最长 的反向素长的左端点.

又 从RPLS
2m+1
的左端点开始，RPLS
2m-ỹ
的左端点是重合完OD 3
（2m+1）上全部奇素数点的最后一组.
所以，作为RPLS
2m-ỹ
的共同右端点的奇数点2m-ỹ不能位于奇素数点
P
m
的右边，于是，奇数点2m-ỹ -2位于奇素数点P
m
的左边. 由此可知，在
OD 3（2m- ỹ-2）上的每一条反向素长都比PL 3

P
m
短.

又 因为在RPLS
2m-ỹ
的左端点左边的第一个奇数点全部是RPLS
2 m-ỹ-2
的左端点.
所以，每一条RPL
2m-ỹ-2
都能反向延长 成为至少一条RPL
2m-y+2
. 那就是
说，RPLS
2m-ỹ-2的每个左端点都一定是RPLS
2m-y+2
的至少一条的左端点，这
里 y是≦ỹ+2的正奇数.

因为RPLS
2m-ỹ-2
的左端点一对 一地全部在RPLS
2m-ỹ
左端点（包括特有
奇素数点）左边的第一个奇数点上，于 是，在特有奇素数点左边的第一
个奇数点一定是RPLS
2m-y+2
的左端点，这里 y是≦ỹ+2的正奇数. ---{2}

所以,在既要满足结论{1}、又要满足结论{2}的情况下，我们只能够
得出: 在特有奇素数点左边的第一个奇数点是RPLS
2m+1
的左端点.

既然如此，让我们向右移动任意一条这样的PLS
2m+1

到相邻的奇数< br>点，那么,这条PLS
2m+1
被移动成了一条PLS
2m+3
. 显 然，这条PLS
2m+3
的左
端点重合这个特有奇素数点，而它的右端点是奇数点2m +3.
22

这样，OD 3（2m+3）可以被表为两条素长之和，这两条素长的共同
端点就是这个特有奇素数点.

到此为止，我们已经证明了: 当OD P
m
（2m+3）是一条合长时，OD 3
（2m+3）也能表为两条素长之和.
那么，根据定理3，3加上第m+1个奇数所得的偶数,即 3+(2m+3)可以
表示成两个奇素数之和.

每次都从已证得的结论出发，就可以推出: 当k等于每一个自然数时，
每一个偶数3+(2k+1)都可以表示成两个奇素数之和. 即 当N>2时，每一
个偶数2N都可以表示成两个奇素数之和.

又 当N=2时,有2N=4=偶素数2+偶素数2.
所以,每一个偶数2N都可以表示成两个素数之和，这里 N
≧
2.
因为每 一个奇数2N+3等于偶数2N加上奇素数3，因此，每一个奇数
2N+3都可以表示成三个素数之和， 这里 N
≧
2.

综上所述，我们已经证明了哥德巴赫猜想的两个命题都是站得住脚
的，因此，哥德巴赫猜想成立.
23