Problems & Puzzles:
Conjectures
Conjecture 3. Twin Prime's Conjecture
If we define d_{n} as : d_{n} = p_{n+1}  p_{n},
is easy to see that d_{1}=1 and d_{n}=even for n>1.
Now, it’s believed that "for n>1, dn=2 infinitely often"
(Ref. 2, p. 19). This is the "Twin prime Conjecture", which can be paraphrased
this way : "There are infinite consecutive primes differing by 2".
SOLUTION
Mr Liu Fengsui has sent (3/9/01) an argument that
proves  according to him  the well known and named "ktuple
conjecture"
This conjecture can be expressed the following way (see
1,
2
& 3):
 "Any admissible constellation
of primes occurs infinitely often".
Therefore, if this the Mr Liu's argument is
correct then also the Twin Primes conjecture has been proved.
As you soon will discover this argument is close related
to the Liu's approach to the prime numbers definition, approach
that has been exposed in detail in the Problem 37
of these pages.
What follows is Mr Liu's argument. I
should strongly point out that the most that Mr. Liu wants is to
receive some criticism about this his work, through these pages or direct
via his email below.
***
On the prime ktuple conjecture
China, Liu Fengsui
Email: fengsuil@yahoo.com.cn
In this paper we constructed a second order arithmetic P(N), Given a recursive formula
Tn which approximates the set of prime ktuples and show its pattern occurs
infinitely
often.
After all we 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[n1]+bm, an+bm>,
A * B = <a1*b1,a2*b1,...,ai*bj,...,a[n1]*bm, an*bm>.
A \ B be the subtraction of sets.
We 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[n1,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 one element 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.
We shall show the prime ktuple conjecture of the first order
arithmetic <N,+,*, <,0,1> In the second order arithmetic <P(N),+,*,<,0,1>.
The prime ktuple conjecture states that every admissible pattern for a prime
constellation occurs infinitely often. ( the prime glossary).
To describe easily, we discuss the conjecture:
There are infinity a such that
x^2  x + a
all are prime for x = 0,1,2,...,k.
We use R(k,a) denote that x^2 x + a all are prime for x = 0,1,2,...k.
and say this a be a solution.
Let pn be nth prime , P0 = 2.
For random 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,...,pn1>) \ 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 random 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]  = (p11)/2 * (p21)/2 *...*(p[s+1]k) *...*(pnk).
From the recursive formula Tn ,obviously the criterion of prime ktuples is
R(k,a) iff a = pn = min Tn > k1.
Where min Tn is the smallest number of the set Tn.
This criterion R(k ,a) recursively enumerate all prime ktuples.
This algorithm is a second order version of the Eratosthenes sieve. At nstep, we had
computed all solutions R(k,a) , a < pn. In set Tn, if t belongs Tn and k^2  k
+ t < pn^2 then t is a solution R(k,t). As n larger enough , almost all
elements of set Tn are solutions. If a > pn is a solution then a belongs the
set Tn mod mn. So that the recursive formula Tn approximates all solutions.
As an elementary result of recursive formula Tn ,we obtain another solution
for the Conjecture 17. ( http://www.primepuzzles.net/conjectures/conj_017.htm
)
Let (pn+1)/2 <= k, let A = min Dn,  Dn  =  Tn *(pn+1), then
SPD(T,A) = pn for x = b. Where min Dn is the smallest number of the set Dn.
We can design a program and compute some exactly solutions of R(k,a). But with our
computer we can never 'touch' entire set of R(k,a) and can never know if they are
infinite.
Now let us see the result without actually "running the program" from theory.
We consider the limit of the sequences of sets T1,T2,...,Tn,......, as the
classes of
residues, there is include relation N > T1 > T2 >......>
Tn>......., thus there is the Lim Tn.
As n goes to infinity, we canceled all classes B1,B2,...,Bi,......, we
will obtain what
result? It is easy to prove lim Tn is empty ,since that when
we canceled all classes
B1,B2,...,Bi,... to sift the solution R(k,a) , we removed the solution R(k,a) also by
pn = a and pn  a. To show the infinity first we consider the non empty set. We modified
the set Bi to be Bj:
Bi = {a: pi  x^2  x + a for some 0<=x<=k but pi =/= k^2  k +a}
= {a:x^2 x + a = 0 mod pi for some 0<=x<=k but pi =/= k^2 k +a}.
As n goes to infinity we canceled all classes B1,B2,...,Bj,......, since lim  Tn  is infinite
then lim Tn is not empty .
Now we try prove the pattern R(k,a) , k < 41 occurs infinitely often.
Proof: Suppose that the number of patterns R(k,a) is finite, then there is a maximum
number a0 such that for all a > a0 R(k,a) is false. From the set of all
natural numbers we
cancel the classes B1,B3,...,Bi,...,Br successively, and obtain the class Tr such that a0 =
mim Tr = pr, then for all j>=r in the class Tj there is not any a such that R(k,a) ,and for
all 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, lim  Tn  is infinite and lim Tn is not empty . Take
number e belongs lim Tn 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) is true 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 pi, except pi  a, let Bi = {x: ax+b = 0 mod pi}, then
 Tn  =( p11)*...*(pn1), but pi  a. Iterate above proof ,we obtained the
Dirichlets Theorem again.
Now we find the admissible or necessary condition of infinity by above method.
1). lim  Tn  is infinite, so that the k is a constant.
2).The set of solutions is admissible nonempty, or at least there is one non
trivial solution.
It is no meaning to discuss if infinite for empty, to extend above method to
empty, we
discuss the situation that the set of solutions is empty. Assume the
set of solutions
R(k,a) is empty, we select a disjunctive proposition:
There is no any solution a > a0 or there is no solution a <= a0.
Assume the disjoint "there is no solution a > a0" , run a proof, if we do not obtain
contradiction, we can not prove anything, if we obtain a contradiction, then we have
proved the disjoint " there is no solution a <= a0" is true, namely we have proved the
set of solutions is empty.
There is an Euler's formula x ^ 2  x + 41 all are primes for x = 0,1,2,...,40. So that, for k
< 41 , at least there is one non trivial solution
R(k,a), 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 find one solution, then they are infinite.
This is incredible! In less than one week this noble
Conjecture has received an attack by two flanks. Now (12/9/01) it is the Alan
Tyte's turn.
Alan has constructed an analytical proof that
shows that there are more twins in the range (q, q^2) than in the range
(p, p^2) being q the following prime to p. But if this is rigorously true
then it's also true than in each new range we have at least one more twin
primes. Being the quantity of primes infinite, this also means that the
number of twin primes is infinite.
The most surprising fact is that  as you will discover
soon  both attacks uses as background for the reasoning, the Eratosthenes
sieving machine...
This is the Tyte's proof in detail:
A proof that there are an infinite number of twin
primes
by Alan Tyte
Notation 1
 Let P be the set of odd positive integers > 1, i.e. P
= {3, 5,7,9, . . . . . }
 Let m represent a composite integer.
 Let u
represent an integer whose status is currently undetermined, that is an
integer which has not yet been shown to be composite by casting out a
prime of which it is a multiple and which thus may, or may not, be a
prime.
 Let a triple be a subset of P consisting of three consecutive odd
integers, starting with a multiple of 3 ( e.g. triples are {3 5 7} ; {9 11
13}, etc.)
Lemma 1
After casting out all multiples of 3 , each triple in P
, subsequent to the triple containing 3 , has format muu (the triple {3 5
7} has format uuu ).
Proof : follows from definitions.
After casting out multiples of 3 , P consists of an
infinite set of triples beyond {3 5 7}, each of which contains a possible
twin prime. As each further prime is cast out, a proportion of the u
integers become m integers and the number of possible twin primes
decreases.
Lemma 2
As further primes are cast out, a triple may have one of
four formats : mmm, mum , mmu or muu . Proof : follows from definitions.
Examples:
 After casting out 3 and 5, the triple {21 23
25} has become mum.
 After casting out 3, 5, 7 and 11, the triple {117 119
121} has become mmm .
Notation 2
 Let A represent a triple of format mmm, B of format mum,
C of format mmu and D of format muu .
 Let t be the number of triples in a
pattern repeat (see Lemma 3 below).
 Let a be the count of A triples in a
pattern repeat; similarly b for B , c for C and d for D so that t = a + b
+ c + d.
 Let p be any prime > 3 Let q be the next greater prime to p (
e.g. if p = 7 then q = 11 or if p = 59 then q = 61).
Lemma 3
After multiples of p are cast out, starting with the
first triple beyond that containing p , there will be a pattern of As, Bs,
Cs and Ds which repeats infinitely. The number of triples in the pattern,
t , will be the product of all the primes from 5 up to and including p .
After multiples of q are cast out, the pattern length is t . q triples.
Proof:
Initially after casting out multiples of 3 there is one triple in
the pattern repeat, i.e. a D . There are 6 integers effectively spanned by
each triple if even numbers are included. Since 5 is prime to 6, multiples
of 5 will only occur at the same position within those 6 integers every
5th triple. Hence there will be 5 triples in the pattern repeat after
casting out multiples of 5. Similarly, since 7 is prime to the new
effective pattern span of 30 integers, there will be 5 x 7 triples in the
pattern repeat after casting out multiples of 7 . As each new prime is
prime to the existing pattern length, the lemma follows.
Example: After casting out 5 and 7, Lemma 3 predicts a
pattern repeat of 35 triples as shown below :
9 11 13, D
15 17 19, D
21 23 25, B
27 29 31, D
33 35 37, C
39 41 43, D
45 47 49, B
51 53 55, B
57 59 61, D
63 65 67, C
69 71 73, D
75 77 79, C
81 83 85, B
87 89 91, B
93 95 97, C
99 101 103, D
105 107 109, D
111 113 115, B
117 119 121, C
123
125 127, C
129 131 133, B
135 137 139, D
141 143 145, B
147 149 151, D
153 155 157, C
159 161 163, C
165 167 169, D
171 173 175, B
177 179 181, D
183 185 187, C
189 191
193, D
195 197 199, D
201 203 205, A
207 209 211, D
213 215 217, A
then pattern restarts:
219 221 223, D
225 227 229, D
231 233 235, B
237 239 241, D
243
245 247, C
249 251 253, D
255 257 259, B
261 263 265, B
etcetera
Lemma 4
When multiples of q are cast out:
1 / q of what were Ds
in the t . q triples of the new pattern repeat become Bs
1 / q of the Ds
become Cs
1 / q of the Bs become As and
1 / q of the Cs become As.
The
counts in each pattern repeat after casting out multiples of q is
thus:
for Ds is d.( q  2 )
for Cs it is c . ( q  1 ) + d
for Bs it is b . ( q
 1 ) + d
for As it is a . q + b + c
And
d.( q  2 ) + c.( q  1
) + d + b.( q  1 ) + d + a.q + b + c = (a + b + c + d).q = t.q
Proof:
This is similar to that for Lemma 3 . The new
pattern consists of q sets of pattern length t . If, for example, the nth
integer in the first of set of length t is a multiple of q , then in none
of the remaining q  1 sets can the nth integer be a multiple of q , since
q is prime to t ; i.e. 1 / q of the nth integers in each set are multiples
of q . If a multiple of q is already known to be a multiple of a smaller
prime, then the triple's format is unchanged ; e.g. once an A, always an A
. Only one u in any triple can change to an m as odd multiple of q are
separated by 2q . Hence the only changes possible are from D to B or C ,
and, from B or C to A . If the nth triple in the first set is changed from
a D to a C , an identical change cannot occur to the nth triple in any
other set of the q sets; in one other set the nth triple will be changed
from a D to a B .
Examples:
After multiples of 3 are cast out, then a = 0
, b = 0, c = 0, d = 1 and t = 1
9 11 13 pattern restarts 15 17 19 pattern
restarts 21 23 25 etc. (D) (D) (D)
When casting out multiples of 5, a => 0 x
5 + 0 + 0 = 0 b => 0 x (5  1) + 1 = 1 c => 0 x (5  1) + 1 = 1 d
=> 1 x (5  2) = 3 t => 1 x 5 = 5.
9 11 13, 15 17 19, 21 23 25, 27 29 31,
33 35 37, pattern restarts 39 41 43, 45 47 49, 51 53 55, etc. (D D B D C)
(D D B...)
When casting out multiples of 7, a => 0 x 5 + 1 + 1 = 2 b => 1 x (7
 1) + 3 = 9 c => 1 x (7  1) + 3 = 9 d => 3 x (7  2) = 15 t =>
5 x 7 = 35. See list of triples in Lemma 3
More examples:
Repeating these iterations gives : after
casting out multiples of 19 a = 336410 b = 450765 c = 450765 d = 378675 t
= 1616615 ( these figures were verified by an actual count) after casting
out multiples of 29 a = 271120850 b = 296226315 c = b d = 214708725 t =
1078282205 Although the counts rapidly reach huge numbers as further
primes are cast out, they are finite and calculable.
Corollaries to Lemma 4
After multiples of p are cast out, the proportion of Ds
in the infinite set of triples beyond the triple containing p is d / t .
The ratio d / t applies exactly over any set of t consecutive triples.
Over sets of less than t triples it is the average proportion.
After
multiples of q are cast out, the proportion of Ds in the infinite set of
triples beyond the triple containing q is d . (q  2) / t . q
Lemma 5
When casting out all multiples of p , the format of all
triples less than the triple containing p ^2 cannot be affected. Any
Ds between p and its square are triples containing twin primes.
Proof:
When casting out multiples of p , all multiples
of p less than p ^2 will have been found when casting out smaller primes.
Example:
All odd multiples of 7 less than 49 are also multiples of 3 or
5.
Proof that there are an infinite number of twin
primes:
It will be sufficient, by Lemma 5, to show that there
are always Ds between a prime and its square. We have shown that there are
Ds between 5 and 25 and between 7 and 49. We now
show that there are more Ds between q and q ^2 than there are between p
and p ^2 , in fact as primes get larger there are more and more
twin primes between a prime and its square.
Let Np be the number of Ds
between p and p ^2.
Let Nq be the number of Ds between q and q ^ 2.
From
Lemma 4, the proportion of Ds in the pattern repeat after casting out
multiples of p is d / t .
The number of triples between p and p ^2 is the
integral part of ( p ^2  p) / 6 = p . (p 1 ) / 6 .
Hence, on average:
Np = [ p.(p  1 ) / 6 ] x [ d / t ] = d.p.(p 1) / ( 6 . t )
From
Lemma 4 , the proportion of Ds in the pattern repeat after casting out
multiples of q is ( d . (q  2) ) / ( t .q ) The number of triples between
q and q ^2 is the integral part of ( q ^2  q) / 6 = q .(q 1 ) / 6
Hence,
on average:
Nq = [ q.(q 1 ) / 6 ] x [ ( d.(q  2) ) / ( t.q )] = d.(q 1).( q  2) / ( 6.t )
Now q >= p + 2 , therefore (q 1) . ( q 
2) > p . (p 1)
Hence Nq > Np .
Conclusion:
Since this (Nq>Np) means that there are always new
twin primes between a prime and its square, and since there are an
infinite number of primes, there must be an infinite number of twin
primes.
Examples of counts of twin primes between a prime and
its square obtained from actual counts:
Between 5 and 25 there are 2 twin primes, i.e. 11 13 and
17 19 . Between 101 and 101 x 101, there are 201 twin primes. Between 199
and 199 x 199 , there are 574 twin primes. Between 293 and 293 x 293 ,
there are 1056 twin primes.
***
And my question: Is this proof correct or
is it flawed?
***
Regarding the Tyte's proof I have received three
enthusiastic commentscontributions, pointing out that while the
averaging step made by Alan is questionable, maybe this approach shows
where to look for in the solution of this Conjecture. These first
three comments are from Chris Nash, Fabrice Marchant and Leadhyena
Inrandomtan:
"About Alan Tyte's proof : all the
beginning up to "Lemma 5" is right but there are 2 errors in
the end of the proof, after each "Hence, on the average :"
because we do not know the way our beloved Ds are spanned : no reason to
be sure they are put at the same rate between x and x^2 than between a
whole pattern. However, I think the idea of the proof with As, Bs ... is
great and I'll try to work in the way of Alan." (F. Marchant)
***
"I would like to bring this up in a forum... I
feel that his proof is almost correct, I have but one unfortunate
problem with it. I do feel that the first person to reconcile this
slight over assumption will clench this problem once and for all...
I verify all of Alan Tyte's proof up to and including lemma
4. Keep in mind however that up to this point the lemmas only deal with
the unidentified letters in the sequence of triples produced. It is in
lemma 5 that he attempts to make the jump to prime numbers. And he does
correctly state that after casting out the p's that all 'unidentified's
that represent numbers under p^2 in the pattern repeat represent primes.
It is the distribution that concerns me.
Alan Tyte states that the proportion of Ds in the pattern
repeat after casting out multiples of p is d/t. This is where I
disagree. If the pattern of letters follows with the pattern of the
primes, they would share the same distribution, IE pi(x) approaches n/log(n)
as n approaches infinity, as stated in the prime number theorem. Since
the pattern repeat models a pattern that does not follow a UNIFORM
distribution, assuming that a segment of the pattern has the same
proportion is illegal under the laws of statistics.
It may seem like a technical point; however the problem becomes
paramount. If we took the proportion of primes to the natural numbers as
the size of the subset extended to the size of the natural numbers, the
proportion would approach lim n>inf of 1/log(n) or 0. If this type
of previously mentioned assumption were allowed, we could say that for
any segment of the natural numbers that it contained no primes, because
it would have the same proportion of primes to naturals, IE 0.
As I said before, this proof can be fixed, but I'm not sure how. I
would agree with his numbers; I do think that the proportion of twin
primes to all primes increases with the size of the segment from n to
n^2. I don't have the statistical expertise to handle the manipulation
of a logarithmic distribution, which I believe that the twin primes
follows. I think that Tyte is on a goldmine of analysis, and I
would even conjecture that these patterns could be analyzed for not just
p,p+2 but also p,p+4 and other possibilities... (Leadhyena
Inrandomtan)
***
Mr. John Washburn as a critical note (8/3/2002) to
the Liu Fengsui's proof, that ends with a happy ending phrase "I
thinks that this problem is repairable":
The proof of Liu Fengsui has a flaw. His approach is to create
sets where the members of the set are the integers which survive an
nstranded sieve. The basis for the stands of the sives are the initial n
primes; 2,3,5,..Pn.
As an aside, the idea contructing the set of sieve surviors from the
previous set of surviviors is brilliant. The Sieve of Eratothenes
(and variants) is essentially a subtractive process. The method of Fengsui
is essentially additive.
But back to the proof. For example T4 is the set of integers which
survive a 4stranded sieve. The basis of the strands are 2, 3, 5, and 7.
The set constructed is T4 = <11, 17, 29, 41, 59, 71, 101, 107, 137,
149, 167, 179, 191, 197, 209>.
The essence of Liu Fengsui's proof is that Tn never becomes the
empty set as n approaches infinity. The problem is that this in not
sufficient. Liu Fengsui must also prove the minimal member of Tn is
less than (Pn+1)^2  2. Otherwise the set of survivors nonempty, but all
be composite. The composites would have no prime factor < P(n+1).
In the example of Tn above, what if the minimal member was 119, 141 or
167 and not 11? (121= 11*11, 143=11*13, 169=13*13).
I think this problem is repairable.
***
This is the Mr. Liu's answer (28/302) to Mr.
Washburn:
I thank Mr. John Washburn very much for his kind
comment and points the flaw.
I welcome anyone comment my proof or point any flaw.
New I post my opinion,please Mr. John Washburn
bear with it.
The essence of the flaw pointed by Mr. John Washburn
is sentence:
Liu Fengsui must also prove the minimal member of
Tn is less than (Pn+1)^2  2.
I think, perhaps, Mr. John Washburn
misunderstanded my criterion of prime ktuples
R(k,a)
iff a = pn = min Tn > k1.
to be
R(k,a) iff a = min Tn > k1.
Perhaps,Mr. John Washburn misunderstanded my lim
of set sequences Tn Lim Tn to be every Tn.
Perhaps, it is Mr. John Washburn own some
difficults or approach.
If the sentence is repaired to be an existence sentence:
exist a Tn such that minimal member of Tn is less than (Pn+1)^2  2.
It is right, and Liu Fengsui has proved
it.Otherwise,obviously, it is false.
I wish Mr. John Washburn can articulate the flaw
better.If I misunderstanded the idea
of Mr. John Washburn , please
criticize .
***
Mr Liu Fengshui wrote (4/7/04):
"
When 6th WSEAS International Conference on APPLIED
MATHEMATICS
(August 1719, 2004, Corfu, GREECE) call paper, I tried submit my paper
"On the prime ktuple conjecture".
May 13, 2004, I received the
reply:"We have now
the first results from our reviewers. Based on the review of at least 2
independent reviewers, we are glad to inform you that your paper has
been accepted. Also, your paper has been
selected in the CATEGORY A' of the papers.
That means: publication in the Conference Proceedings as well as in
WSEAS Journals WSEAS Transactions on mathematics."
The main reviews: this paper is an original work,
we do not find error
in the proof twin prime conjecture and so on,
publish it."
