- View SourceHi all:

It is seem, I found a proof of prime k-tuple conjecture. Because this conjecture is very difficult, I am not sure my proof is correct.Please help me find the error.I appreciate any comment,improvement or refution.To read this paper need some knowledges of set theory.

Best wishes

China Liu Fengsui.

On the prime k-tuple conjecture

In this paper we constructed a second order arithmetic P(N), Given a recursive formula Tn which

approximates the set of prime k-tuples and show this pattern occurs infinitely often.

After all give some operations.

Let A = <a1, a2,...,ai,...,an> be the finite set of natural numbers.

If A = <a1,a2,...,ai,...,an> , B = <b1,b2,...,bj,...bm>,

we define A + B = <a1+b1,a2+b1,...,ai+bj,...,a[n-1]+bm, an+bm>,

A * B = <a1*b1,a2*b1,...,ai*bj,...,a[n-1]*bm, an*bm>.

A \ B be the subtraction

Define the solution of the system of congruencies

X = <a1,a2,...,ai,...,an> mod a,

X = <b1,b2,...,bj,...bm> mod b

be X = D = <d[1,1],d[2,1],...,d[i,j],...,d[n-1,m],d[n,m]> mod ab.

where x = d[i, j] mod ab is the solution of the system of congruencies

x = ai mod a , x = bj mod b.

We identify the set <a> which has only single element a with the number a

<a> = a.

So, we had defined a second order arithmetic

< P(N),+,*,<, 0,1>,

where N is the set of natural numbers and P(N) is the power set of N.

In the second order arithmetic <P(N),+,*,<,0,1>, we shall show the prime k-tuple

conjecture. Perhaps, it is undecidable in the first order Peano theory.

The prime k-tuple conjecture states that every admissible pattern for a prime

constellation occurs infinitely often.

To describe easily, we discuss the conjecture:

There are infinity a such that all

x^2 - x + a

are prime for x = 0,1,2,...,k.

Example: let k = 2, this is the Twin Prime's Conjecture.

We use R(k, a) denote that all x^2 -x + a are prime for x = 0,1,2,...k.

and say this a be a prime k-tuple.

Let pn be n-th prime , P0 = 2.

For any prime pi > 2, we consider the set

Bi = {a: pi | x^2 - x + a for some 0<=x<=k }

= {a: x^2 -x + a = 0 mod pi for some 0<=x<=k }.

By (pi +1 - x)^2 - (pi +1 -x ) = (pi + 1 -x)*(pi - x) = x^2 - x mod pi, it

is easy to prove that, in the congruence x^2 -x + a = 0 mod pi , when x

runs through the complete system mod pi , a runs through the classes of

residues <0,-2,-6,...,(pi+1)/2 - ((pi+1)^2)/4> mod pi. Thus the set Bi

are the classes of residues:

Bi = <0,-2,-6,..., (pi+1)/2 - ((pi+1)^2)/4> mod pi , (pi+1)/2<= k.

Bi = <0,-2,-6,..., k - k^2> mod pi, (pi+1)/2 > k.

For example, let k = 4, the first few terms of the sets Bi are

B1 = <0,-2> = <0,1> mod 3,

B2 =<0,-2,-6> =<0,3,4> mod 5,

B3 = <0,-2,-6,-12> = <0,5,1,2> mod 7,

B4 = < 0,-2,-6,-12> = <0,9,5,10> mod 11.

Let m[n+1] = p0*p1*...*pn, from the set of all odd numbers X = <1> mod 2 we

cancel the classes B1,B2,...,Bn successively, and obtain the class T[n+1] mod

m[n+1] such that

T[n+1] = {a: x^2 - x + a =/= 0 mod pi, for x = 0,1,2,...,k and

pi = p1,p2,...,pn}.

Then the recursive formula of T[n+1], which is the set of non negative

representatives mod m{n+1] is as follows:

T1 = <1>,

T[n+1] = (<Tn +<mn>*<0,1,2,...,pn-1>) \ Dn

where X = Dn mod m[n+1] is the solution of the system of

congruencies

X = Tn mod mn,

X = Bn mod pn.

For example, let k = 4 , the first few terms of T[n+1] are

T1 = <1>,

T2 = (<1>+<0,2,4>) \ <1,3> = <5>,

T3 = (<5> + <0,6,12,18,24>) \ <5,23,25> = <11,17>,

T4 = (<11,17>+,0,30,60,...,180>) \ <47,71,77,107,131,,161,191>

= <11,17,41,101,137,167>.

Given any integer k>1, take an integer s such that

(ps+1)/2<= k < (p[s+1]+1)/2,

then the number of all elements of the set T[n+1] is

| T[n+1] | = (p1-1)/2 * (p2-1)/2 *...(ps-1)/2*(p[s+1]-k) *...*(pn-k).

Obviously the criterion of prime k-tuple is

R(k,a) iff a = pn = min Tn > k-1.

Where min Tn is the smallest number of the set Tn. This criterion R(k ,a) recursively enumerate

all prime k-tuples.

If we canceled all classes B1,B2,...,Bi,......, will obtain what result? In other word, we

consider the limit Tn of the sequences of sets T1,T2,...,Tn,....…. As the classes of residues,

there is include relation N > T1 > T2 >......> Tn >......., thus the sequences of sets T1,T2,...,Tn,

...…. have the Lim Tn.

It is easy to prove lim Tn is empty ,since that when we canceled classes B1,B2,...,Bi,...

to sift a prime k-tuple R(k,a) , we removed the prime k-tuple R(k,a) also by

pn = a and pn | a.

We modified the set Bi to be Bj:

Bj = {a: pj | x^2 - x + a for some 0<=x<=k but pj =/= k^2 - k +a}

= {a:x^2 -x + a = 0 mod pj for some 0<=x<=k but pj =/= k^2 -k +a}.

Except retainment the prime k-tuples, Ti and Tj are same.

Now the recursive formula Tj will approximate the set of all prime k-tuples Lim Tj.

When the set of prime k-tuples is not empty, example k < 41, We try prove the pattern R(k,a)

occurs infinitely often.

Proof:

Suppose that the number of patterns R(k,a) is finite and not 0, then there is a

maximum number a0, such that for every a > a0, R(k,a) is false. From the set of all odd

numbers we cancel the classes B1,B3,...,Bi,...,Br successively, and obtain the class Tr

such that

a0 = mim Tr = pr.

then for every j>=r , in the class Tj there is not any a such that R(k,a) ,and for every a > a0 there

is not any a such that R(k,a).

From the class Tr we continue cancel the set B[r+1],...,Bj,......Now as n goes to infinity,

the recursive formula Tj approximated all prime k-tuple a > a0. Because lim | Tj | is infinite so that

lim Tj is not empty . Take number e belongs lim Tj , then

x^2 - x + e

does not contain any prime pi or pj as factor except itself for x = 0,1,2,......,k thus

R(k,e) hold by definition and

e > a0.

This is a contradiction so we have proved that

The pattern R(k,a) , k < 41 occurs infinitely often. #

Change the sets Bi, we can prove the infinity of various patterns with above

method. Example: the primes in arithmetic sequence ax+b, (a,b)=1.

Give any prime pi, except pi | a, let

Bi = {x: ax+b = 0 mod pi},

then

| Tn | =( p1-1)*...*(pn-1), except pi | a.

Iterate above proof, we obtained the DirichletsTheorem again.

If the set of prime k-tuples is empty, It is no meaning to discuss if they are infinite,

to extend above method to empty, we must describe the empty set with a suitable universal

sentence

There is no prime k-tuple a > a0 or there is no prime k-tuple a <= a0.

Assume the disjunct "there is no prime k-tuple a > a0", run above proof, if we do not

obtain contradiction, we can not prove anything, if we obtain a contradiction ,then we have

proved the disjunct " there is no prime k-tuple a <= a0" is true, namely we have proved the set of

prime k-tuples is empty again.

There is an Euler's formula all

x ^ 2 - x + 41

are primes for x = 0,1,2,...,40. So that, for k < 41,the set of prime k-tuples is not empty,

by above proof, we have known that the pattern R(k,a) , k < 41 occurs infinitely often. For k >=

41, we do not know if there is a number a such that

x ^ 2 - x + a

all are primes for x = 0,1,2,...,k, we do not know if they are infinite. If we found one a, then

they would infinite.

The first manuscript of this paper was posted in

http://www.primepuzzles.net/conjectures/conj_003.htm

[Non-text portions of this message have been removed] - View SourceI am afraid I can not answer your question well.

Yes, the number of prime quadruples is infinite.Example <5,7,11,17>.

Every prime k-tuple have its gap between conscutive prime k-tuple.

This gap is computable.

But julienbenny, I use the discrete approach, I do not evaluate the gap between conscutive prime k-tuple. I think It is a job of analytic number theory.

Liu Fengsui

julienbenney <jpbenney@...> wrote:

I am afraid I do not have the mathematical skill (because of my

aversiveness to work, I failed second-year mathematics completely),

but, Liu, this is most interesting dveloment of an early method of

working to prove the prime k-tuple conjecture.

However, what does this say about how large the gap between conscutive

prime k-tuples should be. Even if the nmber of prime quadruples in

infinite, is is easy to see how quickly they become very rare even when

one gets to 10,000.

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