Download The Laws of Prime Numbers and

The Laws of Prime Numbers and Odd Composite Numbers
Aixinjiaoluo . Xi Guo-wei
Fed . 2004
Reg No : O7-2005-A-271
By The Copyright Bureau of Jilin Province , CN
The Preface
About 330 ～275 B . C Euclidean put the mathematic law of
expressing prime numbers , and for 2200 years so far there is no results
for them.
The law of holography of numbers of《Theory on Motion》tells us
that the infinity in Math. only is the relative infinity , not absolute
infinity , and it is impossible to seek for the numbers of absolute infinity ,
and it’s wrong idea. The concept of infinity is the concept of motion in
essence , of course , the numbers are produced in motion . The first ones
are irrational numbers , and the integers only are tiny spots among them.
Also《The Theory on Motion》found in Lucas Number the formula of
“the new formula（M）of remainders”, and from this derived the
following :
M = a r/r
a =（5 1/2+1）/ 2
——（M）
“r”is an odd number and leading to the laws of
prime number and odd composite numbers.
The Discriminant of Prime Numbers and Odd Composite Numbers
A． Division and screen analysis :
1. The dividend“b”is divided by“a”and gotten the quotient“q”,
and the relation of which can be expressed in form of fraction , as
follows :
a
q
b
——（q）, and when a、b、q are all positive integers the
“b” can be exactly divided by “a” , and this relation of division can
be thought in an images as“screen”. The“b”can be divided by“a”,
in an analog the“a”be screened and gotten“q”.
2. When“b”can’t be exactly divided by“a”there is a relative
formula :
b=a·q+c
——（b）
1≤ c ＜ a , c is the positive integer in one hand ,“b”can’t be
exactly divided by“a”, or in others ,“b”has no integer’s factors , and an
images drawn that“b”can’t be“screened”by“a”.
3. The binomial expansive co-efficient formula is called“binomial
coefficient screen”:
n
n(n  1)( n  2)n  (k  1)
( )
q
k
1.2.3.4 k
“q”is the positive integer.
The denominators are the factorial numbers increased by
“k”, and the
numerators are the continued multipliers by successive decreasing“k”,
and“n”is the binomial involution number (exponent).
The“k”is the number of term of the binomial expansion and
4. The factors of minimum prime numbers of the positive integer
“N” is not greater than
N.
The“N”divided by the positive integer , less or equal to
N , would
confirm soon that whether“N”has factors of the integers or not. The
learner Eratosthenes earlier set up the “screen”of the prime numbers by
this kind of division.
B. The Definitions of Prime Numbers and Composite numbers
1. One positive integer , which can be exactly divided by“1”, and
itself , is called as a prime number . or one positive integer , which only
has“1”and two factors of its itself , is called as a prime number.
2. One positive integer , not only is divided by“1”and itself , and
other positive integers , is called as a composite number , or one positive
integer , which at least has 4 or more than 4 factors of numbers , is called
as a composite number.
3. All the even numbers (except 2) are composite numbers , and in
fact , the prime numbers only indicate prime numbers in the odd numbers
and are called as odd prime numbers , and the composite numbers in the
odd numbers are called the odd composite numbers , and
4. See from the above that there are only two kinds of numbers in
odd numbers : the prime numbers and the odd composite numbers.
C. The Distributive Law of Multiplication
Here is formula :
m·(a + b + c) = ma+mb+mc or ma + mb + mc = m·(a + b + c)
There are the same factors (m) in the terms of algebraic sum , first
all , take the different factors in all terms for algebraic sum operation ,
then take the sum numbers multiplied by the same factors (m) for
multiplication operation , and the results are same , and vice versa.
D. The Binomial Expansion and its Coefficient and the Special
Formula :
1. The expansive formula of binomial theorem.
n
n
n
n
o
1
2
k
(a+b) n= ( ) an·bo + ( ) an-1·b1 + ( ) an-2·b2 + … + ( ) an-k·bk
k n n
n
(
+ … + ( ) a ·b =
n-n
n
n
k o k
) an-k·bk
————

0
2. The binomial expansive general term formula of coefficients
n
( )=
k
n
n(n  1)( n  2)n  (k  1)
—— ( ) o
k
1  2  3  4 k
(k<n , K

N) take“k”from“n”, and the“k”is the term number
(k=0,1,2,3…≤n)
In the numerator there are “K”factors from“n” by decreasing
successive multiplication and the denominator is the factorial of the
n
factors by increasing “K” in which the leading coefficient ( ) = 1 ，
o
n
and the last term coefficient ( ) = 1 ，
n
n
n
1
when k=1 , ( ) = =n
1
n
when k=2 , ( )=
2
n
when k=3 , ( )=
3
n( n  1)
1 2
n
n(n  1)( n  2)
et , till ( ) .
n
1 2  3
n
It can be seen from ( ) o:
k
n
n
1
n 1
(1) The“n”exists in the terms from ( ) to (
) , and is the first
main factor.
(2) The“n”in each term , and screened by“k!”(the factorial of the
K) , and the“k！”here is a divisor.
(3) As the numbers of terms increase (K enlarges)
n number must
appear in K ! , and when“n”is an odd composite number , and it will be
screened , and when the“n”is a prime number , it will be left in the terms
to meet the definitions of prime numbers or odd composite numbers.
3. when a=b=1 the binomial becomes the theorem of the sum of the
binomial coefficients :
(a+b) n=(1+1) n
n
n
n
n
n
o
1
2
k
n
2n = ( ) + ( ) + ( ) + …+ ( ) + …+ ( )
—— (2n) and
certified that :
The sum of the coefficients of each terms of the binomial expansive
n
n
n
o
n
k  n 1
n
formulas are equal to 2 , for ( ) = 1 , ( ) = 1 , so : 2 -2=
n
 (k )
—
k 1
— 1
4. The recursion relation of the formula of the binomial expansive
n
coefficients decides that ( ) always is a positive integer , because .
k
1. (
n0
k 0
) =1
2. (
3.(
n 1
k 0
n
k 0
) =1 , (
) =1 , (
n 1
k 1 n
n
k n
) =1
) =1
n
n 1
k
k
the recursion relation : ( )=(
)+(
n 1
k 1
) are always the positive
integers , here are the“Yanghui Triangle”constructive formulas.
E . How can one odd number be known that it is a prime number or
an odd composite number ?
The steps follow as :
1. Take ( 1 ) formula and the“n”is expressed by an odd number
“r”.
2. Write down all the coefficients in the terms from K=1 to K= r-1:
r
( )=
1
r
( )=
2
r
( )=
3
r r
 r
k 1
r (r  1) r (r  1)

1 2
2!
r (r  1)( r  2) r (r  1)( r  2)

1 2  3
3!
……
r
( )=
k
r (r  1)( r  2)r  (k  1)
1  2  3  4 k
3. Constitute the above coefficients of all terms gotten by ( 1 ) 。
4. Observe and compare all the coefficients of the terms :
n
n
1
n 1
(1) From ( ) to (
) only numbers “r” are the common factors.
(2) When “r” is an odd composite number , there must be a factor
“r” in the denominator of all terms and it can be screened (divided) .
(3) When “r” is a prime number the numbers “r” remaining in
n
n
1
n 1
the coefficients of all terms from ( ) to (
) can’t be screened.
5. From the above known that when only “r” is a prime number ,
( 1 ) meets the integer relation of distributive law of multiplication ,
then

1
= r·m
k  r 1
m=

k 1
r
( )
k
r
—— (r·m)
—— (m)
in which “m”is a positive integer.
6. From the above known that when the “r” is an odd composite
number , ( 1 ) the formula can’t meet the integer relation of the
distributive law of multiplication , and for that “m”is a hybrid decimal
(that is a decimal of which the integer part is not equal to “o”) , and seen
from the formula (m).
r
(
)
2 r  2 k r 1 k
 
m
7. The formula here :
r
r
k 1
—— (2r-2)
for ex. : r =7
The first step : is that to set up the sum of the screens of the
coefficients :
7
76
765
7 65 4
7 65 43
7  6 5 4 3 2
7  6  5  4  3  2 1
27  1 + ( ) + (
)+(
)(
)(
)(
)(
)
1
1 2
1 2  3
1 2  3  4
1 2  3  4  5
1 2  3  4  5  6
1 2  3  4  5  6  7
The second step is that :
By screening
27=1+7+(7×3)+(7×5)+(7×5)+(7×3)+7+1
and the third step : Take the formula of (2r-2) :
27-2=7×(1+3+5+5+3+1)=7×18
7
(
)
k 7 1
k
1  3  5  5  3  1  18
Here m  
k 1 7
The “ m ” meets the integer relation of distributive law of
multiplication , so “r=7”is a prime number .
Ex. : r = 9
The first step is that : To set up the sum of the screens of the multi –
coefficients :
9
9 8
9 8 7
9 8 7  6
9 8 7  6 5
9 8 7  6 5 4
9 8 7  6 5 4 3
29  1  ( )  (
)(
)(
)(
)(
)(
)
1
1 2
1 2  3
1 2  3  4
1 2  3  4  5
1 2  3  4  5  6
1 2  3  4  5  6  7
9 8 7  6 5 4 3 2
9  8  7  6  5  4  3  2 1
(
)(
)
1 2  3  4  5  6  7  8
1 2  3  4  5  6  7  8  9
The second step is that : by screens :
29=1+9+(9×4)+ (3×4×7)+(9×7×2)+( 9×7×2)+(3×4×7) +(9×4)+9+1
and we have paid the attention to that the factor “9”in the forth ﹠
seventh terms had been screened.
The third step is that : Take the formula of (29-2) :
29-2=9×[1+4+(
3 4  7
3 4  7
)+(7×2)+( 7×2) +(
)+4+1]
9
9
29 - 2
 m  56.6 is a hybrid decimal , in which the “m”is a hybrid
9
decimal which does’nt meet the integer relation of the distributive law of
multiplication , so , r = 9 is an odd composite number .
F. The conclusion :
1. When “m”is the positive integer and the odd number “r”meets
the definition of the prime numbers (remained by screening) , and meets
the integer relation of distributive law of multiplication of the formula
(2r-2) so , “r”is a prime number .
2. When “m”is a hybrid number and the odd number “r”meets the
definition of a composite number (in some terms are screened , not left)
and does’nt meet the integer relation of multiplication , so “r”is an odd
composite number.
3. Take the formula (b) , and let :
b=2r
c=2
a=r
q=m
then : b=a·q+c
——(b)
2r=r·m+2
Result is :
2r-2= r·m
——(2r-2)
So , we can see that :
The formula (2r-2) is exactly the special example of the formula (b) ,
and has certified that :
This is the dialectical relation of existence between the special
phenomenon and general phenomenon.
The Laws of Prime Numbers ﹠Odd Composite Numbers
A. The Evolution of “The Theorem of Prime Numbers.”
“The theorem of prime number ”is the “quantity of prime
numbers π (x) < x (not over than x) meet the following formula :
π(x)～
x
ln x
(x→∞) ”
The theorem was gotten by Hadamard and De La Vallee poussin in
1896. If we have thought the formula is equal and the signs changed :
“x” changed for “r”, and “π (x)”changed for n/ lna , and “n” is
1
2
the natural numbers here , a= (1+ 5 ) , thereby
n
r

——(n)
ln a ln r
lna is a constant , and when n=2 its form of exponential function is :
ar =r2
r=
ar
r
and take the last equality as the new function , and “r” on the left
is named as another “M”. then the formula (M) o is :
M=
ar
r
——(M)o
The exponent “r”of “ar”and the “r”of the denominator are the
same numbers , which have two inverse functions expressions :
a r
r′ =
M
r=
ln( Mr )
ln a
—— φ(M)
—— f (M)
They together become one curve by f (M) and φ (M) in the rOM
coordinate and r’OM coordinate , and the curve is symmetrical about the
straight line M/r=1 with the (M)o curve .
The formula of (M)o is just one of the main objects we are to study ,
and is sure to take the following formula (M)o :
rn = ar
r(n-1)·r = ar
Let the equality r(n-1) = M , then
M=
ar
r
—— (M)o
B. The Lucas Number Sequence in the Fibonacci Number Sequence
and Their Features.
The Lucas Number Sequence is a generalized number sequence of
the Fibonacci . Lucas found in 1876 that “The linear combination of any
1
2
1
2
orderal equation powers of the two roots a  (1  5 ) and   (1  5 )
in the equation of x2 - x - 1= 0 would meet the following relation :
Fn+1= Fn+Fn-1 . ”
Here : Fn =

1 n
a n
5

, which is called as the closed general term
expression by Fibonacci , and the result is the Lucas Number Sequence
Ln = an + (-1)n a-n
When “n” is the odd number “r” , it means that :
Lr = ar - a-r
—— (Lr)
in which (Lr) is always a positive integer , and the decimal part “a-r”
is exactly the reciprocal of the whole number “ar”，and the three
numbers combine the special close relation
(1) the relation between
(Lr) , and more important :
(Lr) and (M)o
(2) in the (M)o , ar can’t be exactly divided by “r” forever, that is
r
αr
See 《Fibonacci Quarterly》for reference.
C. Fomulas (M)o is the new “remainder formula”
The old “remainder formula” is expressed by (b):
b = a·q + c
(0 ≤ c < a) ——(b)
in which the “b” divided by “a” and the quotient q , and the
remainder is “c”. The numbers of “b、a、q and c” are positive integers ,
and this “Theory on Numbers” is regulated by man and the new
“Formula of remainder” is expressed by (ar) ,
Let
ar
r
M=m+ 
=m+ 
—— (ar)
Here ,“m”may be a positive integer , also may be a mixed decimal ,
 is a pure decimal , and there are two handwritings :
ar = rm + r 
1=
m 

M M
——(1)M
The old “remainder formula”(b) when can’t by exactly divided by
“c” , and the integer “c” remainded , and the division operation stops
at once , and plus “c” become a “remainder formula” by addition ,
so the division operation here failed , without actions. It consists of two
concepts of division and addition operations .
The new “remainder formula” (ar) is different with (b) , it can be
exactly divided without any whole remainder , in which “ar”divided by
“r” , and the quotient always includes the sum of the integer part
‘m’ and the decimal part ‘  ’.
D . The binomial expansion of “ar”and its coefficient features
(1) The binomial theorem
k n n
n
n
n
n
(a  b) n  ( )a n  ( )a n 1  b  ( )a n2  b 2    ( )b n   ( )a nk  b k
o
1
2
n
k 0 k
（2）lt’s coefficient formula
n n(n  1)( n  2)n  (k  1)
( )
k
1  2  3 k
n
—— ( )
k
Take k (k < n , K ∈ N) from n
There are K number decreasing factors in numerator while in
denominator there are K number increasing factors .
n
n
o
n
Here is : ( )  1, ( )  1
(3). Expanse it :
1

 a  2 (1  5 )
a  (1  a ) , 
1
a 1  ( 5  1)
2

n
1 n
n
n
n
n
 n(n  1)  2
(1  a 1 ) n  ( ) 1  ( )1n1  a 1  ( )1n2  a 2    ( )  a n  1  na 1  
a 
o
1
2
n
 2! 
k  n 1 n
 n(n  1)( n  2)  3
 n(n  1)n  (k  1) ( n 1) n
n

a




a

(
)

a

1

( )  a ( k )  a  n





3!
k!
n
k 1 k
That is : a n  a n  1 
k  n 1
n
 (k )  a
k
k 1
 Ln  1
and this is exactly the sum of expansion formula of the Lucas
Number “Ln-1”, for that :
Ln  a n  (1) n  a  n , and when n = r , the odd number , and
Lr  a r  a  r , then Lr  1 
k  r 1
r
  (k )  a
—— Lr 1
k
k 1
r
(4) The features of the coefficients ( ) observed .
k
a. The formula Lr 1 is the expansive formula of the number Lr ,
and Lr is the retracted number of the formula of Lr 1, they have the
sum of terms of K= r-1 .
r
b. In the every coefficients ( ) from all terms of K=1 to K=r-1
k
there are the public factors“r”.
c. (K<r , K∈N) take the “k”from “r” are of the natural numbers .
r
In ( ) formulas the denominator is the factorial of k , and the
k
numerator is the continuous multiplication from r to [r-(k-1)] . So , in the
r
( ) quotients the numbers , which can be cancelled , of the denominators
k
and numerators have been cancelled , and only the two coefficients in the
r
r
1
r
second term ( ) and ( ) in the last term are equal to r and 1 , and the
prime numbers in numerators , which can’t be divided , are remained .
The above is that if the public factors in all terms “r” are odd
composite numbers , then that must be divided by the factors (K!) in the
denominators , and if “r” is a prime number , then only the second term
and the last term can be divided , for in the two terms , the “r” can be
divided by “1”, and the “r” also be divided by “r” , and this is exactly
the main meaning of the definition of odd composite numbers and prime
numbers .
r
d. So “r” is remained in all terms of all equation expansions ( ) ,
k
(that is the “r” is not divided by the factors in the denominator) , and
then “r” is prime numbers , or in another way the prime numbers must
r
by remained in ( ) .
k
r
e. The “r” in the all equations of the expansion formula ( ) is
k
divided , if in one term or some terms , except the second and the last
r
terms , then the “r” is the odd composite number and the feature of ( )
k
is called as “ the binomial coefficient screen ” .
r
r
r
0
1
r
f. For ( )  ( )    ( )  2 r , and the denominators and numerators
are all continuous multiplications of the odd even numbers , and can be
r
exactly divided so , all terms ( ) are positive integers.
k
(5) The retracted number Lr in the formula of [Lr-1]
The known formula :
——(5)1
Lr  2 sinh(lnar)
a e
r
Arc sinh(
Lr
)
2
 (1 a -1 ) r
a -n  a  ( n1)  a  ( n 2)
——(5)2
——(5)3
(recursional formula)


Fn  a n  F( n1)  a ( n1)  a 1 
m3
Fn  a m  F( n1)  a ( m1)  a ( mn1) 
n 1
Fn a ( n1)  F( n1)  a ( n 2 )  a 2
n
1 1  5 
1 1  5 
Here , Fn  
 


5 2 
5 2 
——(5)4
n
The Fibonacci Numbers , n∈N
Using (5)3 and (5)4 retract the expansion formula of a r  (1  a 1 ) r
into Lr-1 number , and checked by (5)1 , and (5)2
In the expansion formula the absolute values of the involution
numbers are increasing from the second term of “a-1” to the last term of
“a-n” and these meet the recursion relation.
(6). The prime numbers and odd composite numbers in the binomial
coefficients :
The known coefficient sum of the binomial formulas are
r
r
r
r
r
2r  ( )  ( )  ( )    ( )    ( )
0
1
2
k
r
r
r
n!
( )  1, ( ) 
r
k
k!(n  k )!
so , seen :
here
,
r
( ) 1
0
,
and
2r  2 
k  r 1
r
 (k )  r  m
——— (r·m)
k 1
When the odd numbers“r”existing in all coefficients from
(
r
( ) to
1
r
) can be recomposed into (r·m) formula , in which “m” number is
r 1
the positive integer , then “r” is a prime number , or that “r”is an odd
composite number.
as same as Lr  1  r  m 
k  r 1
r
 (k )  a
k
—— [Lr-1]
k 1
The number “r” can be recomposed from [Lr-1] , and becomes a
formula [Lr-1]=r·m ,
then “r” is the prime number , in which “m” is a positive integer.
E. when the Lucas Number meets the new remainder formula and
binomial expansive formula , the “r” must be prime numbers
Let :
rm=Lr-1
——(rm)
r  =1+a-r
——(r  )
and mean that :
The (rm) and (r  ) formulas hold water :
(1) for r 
then
ar
, Lr  a r  a r
M
ar
 m  (a r  a r )  1
M
m
 1  a 2 r  a r
M
ar
  1  a r
(2) here is M

 a r  a 2r
M
(3)
m 

 1  a 2 r  a r  a 2 r  a r  1
M M
and be certified .
Analysis :
(1) r =
ar
M
—— (r) M
This is a new concept and “ar”is the relative factor of “r”.
The odd number “r” has two factors “ar” and “
1
”, for
M
M  m   , so here “r”number becomes a new concept of share of “ar”
over “M” , and meanwhile “ar”is the whole number of Lucas Number.
See (Lr) formula for reference.
Obviously , “ar”>M , for the “r” is the positive integer (odd
number) , indicating “ar” can be exactly divided by “M”.
As clear as the above :
M
ar
r
always has
r
ar but here r 
ar
, M/ar
M
(2) To substitute the equation (r)M into (rm) , and result
gotten : rm  a r 
m
m
and is named “the share of integer
M
M
occupies the
proportion of the whole number of ar ”.
(3) Observe the relation between the shares of the whole numbers
and decimal shares . The equation (rm) divided by “ar”, relative factor
of the “r”, gets the integer shares :
m a r  a r  1

 1  a 2 r  a r
r
M
a
The equation (r  ) divided by “ar”, the relative factor of “r” , and
gets the decimal shares :
 1  a r

 a r  a 2 r which are certified that :
M
ar
(a). The shares of the integer :

M
m
M
and the shares of the decimal :
can be completely separated at the time for that the share of integer is
“1” (max.probability) minus the share of the decimal , which is not
straight expressional numbers , and just the decimal share is the straight
expansive numbers . From this surely confirmed that the integer’s part of
M is the “m”, and the decimal part of “M” number is just “  ”
number.
(b). Because the results of integer “m” and the decimal “  ”, that
is to say , the integer part (rm) has been divided exactly by the odd
number “ r ” and in the binomial expansive equations , seemly to say , the
r
“ r ” number can’t be divided in all coefficients ( ) of the whole
k
equation , then “ r ”must be the prime number , Or in another way , as
“r” can be recomposed from [ Lr -1 ] formula , the number “ r ” is just
a prime number , and “ m ” is a positive integer .
(4) The integer “m” is the max . relative factor of (Lr-1) ,
including all the factors except “r” ; and the decimal △ is the min .
relative factor of (1+a-r) , for the comparision :
(Lr-1)﹥(Lr- )
1﹤ ﹤r
(1+a-r) ﹤( +a-r)
When  =1, it’s :
(Lr-1)=(Lr-1)max
(1+a-r)= (1+a-r)min
r

(
)

k  r 1
k
k
m  mmax  
a 

r
k 1
r

1 a

   min 
r

—— (m  ) max
min
when (Lr- ) , (  +a-r) taken , 1< <r
then, the equations
 rm  Lr  

r
r     a
__( rm )
__( r )
Obvisously :
——(solution)1
So , here is
m′﹤m=mmax
△ ′﹤△=△min
The above indicates that in the binomial expansion equations if the
“r” can be divided by the factors ( K !)of denominator of some
coefficients , then the number “r”is an odd composite number , and at
the time the “r”only can be matched to the equation “ Lr- ”with the
number “m”.
F. when 1≤  <r meet the new remainder formula , and the odd
number factor “r” is not the prime number , and it must not in the field
of “ =1”,and exists in the field of “1﹤ ﹤r”, and the kind of odd
number is just an odd composite number , because in the odd numbers
there are only 2 kinds of numbers : the prime numbers and odd composite
numbers .So here is the formula (r ) :
r
ar
  a r 
 Lr   



 r 
m
 r 
—— (r )
The new definition of prime numbers and odd composite numbers :
“The odd number“r” to meet the formula of “ (r ) ”,and that
(1) when  =1 , then the “r” is a Prime number , and
(2) when 1﹤ ﹤r , then the number “r” is an odd composite
number . ”
G. In practice how to check a given odd number “r” is a prime
number or an odd composite number ? The steps are :
(1)To
seek
for
evaluation
the
r-order
power
“ar”
of
1
2
“ a  (1  5 ) ”, and the integer part of the “ar”is the “Lr”, and the
decimal part is the “a-r”.
(2)The “ar”divided by “r”and the quotient is the “m”of the
integer part of “M”.
(3)The “m”multiplied by “r”and the product is (rm), and
(4)From Lr minus (rm) , and if the difference is equal to “1”,the
odd number “r”is a prime number , and if the difference , greater than
“1”and equal to  (1﹤ ﹤r) , then the odd number “r”is an odd
composite number , that is the following:
Lr-rm=1(= )
—— (rm)
As the same theorem:
(1)To seek for evaluation the r-order power “ar” of “a”.
(2)The “ar”divided by “r”and the quotient is the decimal part
△of M.
(3)The “△”multiplied by “r”and the product is (r△), and
(4). From (r△) minus a-r if the difference is equal to “1”,the odd
number “r”is a prime number , and if the difference , greater than
“1”,and equal to “ (1﹤ ﹤r)”,then the odd number “r” is the
odd composite number , as the following:
r△-a-r=1(= )
—— (r)
In the equation of (r△)  formula , for the △ and a-r are all
irrational numbers and the result may be affected by the accurate
precision for evaluation
H. The analysis of “incremental method” gives a clear margin
between the prime numbers and an odd composite numbers for odd
numbers “r”.
(1)The ar type function is an analytic function and the result is of the
common ones and suitable to the “incremental method”.
Suppose that :
r = r1 · r2 , the simplest odd composite numbers , r1 and r2 are two
prime numbers.
So each has its new remainder equation expressing “M”
r
a r ar 1  r 2

M
M
a r1 

a ( r1  r2 )
M1 
r

r


ar2  1 2 M1  M 2
r2 
M 2 
r1 
Here is the equation :
ar
a ( r1  r2 )

M M1  M 2
a r1r2
M  ( r1  r2 )  M 1  M 2
a
——（M）12
Except r1=r2=2 ,
There’s r1·r2﹥r1+ r2
So there are the independent variable increments
r  r1r2  (r1  r2 )
——( r )
At meanwhile:
a r  a r1r2  a ( r1  r2 )
——( a r )
there is the functional increment.
That is to say that:
The ( r )and ( a r ) , the numbers , are the additional conditions rising
from the odd composite number “r” to meet the (M)12 , and r
cancelled:
a r  a r1r2  ( r1  r2 )
——( a r )1
The solutions are :
a ( r1  r2 ) 
a r 
a r1r2 1 2 r1r2

a
 4a r1r2
2
2
a r1r2 1 2 r1r2

a
 4a r1r2
2
2
——（ a r  r ）
1
2
——( a r )2
(2) Always known the following :
a. ( Lr   ) 
ar 1

2 2
=
=
( Lr  )  (  a )
ar 1

2 2
2
r
a
r
 2(  a  r )

2
ar 1 2r

a  4a r  4( 2  1)  4(a  2 r  2a  r )
2 2
b. When   1 , then
(Lr-1)=
ar 1 2r

a  4a r  4(a  2 r  2a  r )
2 2
ar 1

2 2
( Lr  )  (  a )
=
ar 1

2 2
a
=
a r 1 2r

a  4a r  4( 2  1)  4(a 2 r  2a r )
2 2
c. (  a  r ) 
r
r
 2(  a r )
2

2
d. When   1 ,then
(1  a  r ) 
ar 1 2r

a  4a r  4(a  2 r  2a  r )
2 2
The four equations of a .b.c.d. are all identities , and （ a r  r ） and
1
2
( a r )2 are analytic equations .
Comparing with the equations of b and d ,and in the radical signs of
the two equations of （ a r  r ）and ( a r )2 there is no the term:
1
2
+4(a-2r+2a-r) ——(O)
be surely known.
When “r1=r2=3” is the min. odd composite numbers of
“r=3×3=9”,(O) equality is the max . value (O)max:
(O)max=O·105937221……
and here are the equations seen .
rm′=
——（solution）2
r△′=
The odd number “r”to meet the equation on the right in the
(solution)2 equation is a prime number. And the odd number “r” to
meet the equation on the left must be an odd composite number . The
（ a r  r ）and ( a r )2 are the margin functions of the quantitative analysis of
1
2
the equations on the left and the right , and conforming to the (solution)1 ,
and certified the (solution)1 actual.
J. The general relation between Fibonacci Number and Lucas
Number :
———（ L(i  j 1) ）
L(i  j 1)  Li  F j  L(i 1)  F( j 1)
Here , the “i”and“j”are two different arbitrary positive
integers .
i
1  5  1  5 
Li  
 

 2   2 
j
i
j
1 1  5  1  5 
1
Fj 

 
 
5 2   2 
5
（ L(i  j 1) ）is calculated by the recursion relation of Li and Fj , and its
functions are that to seek for the evaluation of greater numbers of L(i  j 1)
by the known smaller numbers of Li、L(i+1)、Fj and F(j+1).
K. The expression of hyperbola by Lucas Numbers
——（Ln）
Ln  a n  (1) n a  n
When “n”is an even number
n=e
——(Le)
Le  a e  a e  2 cosh(ln a e )
When “n”is an odd number
n=r
——(Lr)
Lr  a r  a  r  2 sinh(ln a r )
So here are :
a r  cosh(ln a r )  sinh(ln a r ) 

a r  cosh(ln a r )  sinh(ln a r )
——（ a r  a  r ）


r  cosh(ln a )  sinh(ln a )  1(  )
——

r  
2 sinh(ln a r )  rm  1(  )
r
Arc sinh(
Lr
)
2
r
 1(  )


——（r△）
1
e
e = Napierian Numbers
L. Complete square numbers
for
m 

1
M M
m
 1  (a  r  a  2r )
M

 a r  a 2r
M
The solutions are :
1
1 
a r   

2
4 M
——(1)M
(rm)
( r )
1
5 m
 

2
4 M
Let a  
2ab 
1 
1
, b  (  )
2
4 M
1 

4 M
So , x  a  2ab  a  r
1
 a 2r
4
1
z  (a  b)  2ab  (  a  2r )
4
y  b  2ab 
The triangle function relations are :
cos  
x
1
 r
z
a
( )  a r
4
ar
)  a r
y
sin    4r
z
a
( )  a r
4
(
tg 
y ar

 a r
x
4
sec   tg  2a  r
sec   tg 
ar
2
1
r  1  (sec   tg )
2
rm  Lr  1
1
 (3 sec   5tg )  1
2
a r  rm  r
 2(sec   tg )
The old question of “Number Theory” turned to the old relation
of Pythagoras and becomes very simple and practical , and the substantial
results of the number theory have been witnessed.
Here are :
x
cos 
2(1  sin  )
y
sin 
2(1  sin  )
z
1
2(1  sin  )
M. The geometric implications of x、y and z
(1) Confirm the coordinate system of yox,
(2) Cross a point “O” leading in a straight line “T” of an
arbitrary angles of “  ”with x axis .
1
2
1
4
(3) To take lengths of “1”,“ ”and“ ”at the line “T”,
(4) To draw two circles by radius of
1
1
and
, and the circle of
2
4
1
intersects with the straight line “T” at the point of intersection “h”.
2
(5) Cross the point “h” , to draw a perpendicular line “T”, and
meet at the point “e” with the axis “x” and
(6)To make subangles line of∠heo and meet at the point “f”with
the line “T”.
Obviously , for that :
z y
1
2
fg  fh  y
oh  fh  of  z
og  x  a  r
(7) To make an equation ok  y on the line of “T” , then
z  y  kf  2a 2 r
obviously , the length from the midpoint to the “zero-point” of
kf is exact
1
.
4
(8)To make T  of infinite numerous  and try to find many points
K  and f  , and connect all the points K  and become a curve K  ,
and connect all the points f  and become a curve F  , in which
1
kf  a 2 r  x 2
2
1
 y
4
So be seen the following :
The circle by
1
radius is always the bisector (line) of the line
4
segment kf and here are :
1
 a r  a 2 r
4
1  1
m
 
  (1  )
4 M 4
M
1
  x  x2
4
1
z   x2
4
xz 
The x . y and z and their relations can be expressed by the geometric
relations of the line segments (diagram).
The documents for reference :
1.《The Theory on Motion》by Xi Guo-wei published by The
Haiyang Publishing House in 1993
2.《Mathematical Formula》Compiled by H.Nyetz , Head compiler ,
published by The Haiyang Publishing House , Beijing , 1983