孪生素数猜想与弱波林那克猜想的证明

THE PROOFS OF TWIN PRIME CONJECTURE AND WEAKER POLIGNAC’S CONJECTUREWAN-DONG XUSchool of Science,Tianjin University,Tianjin,300072,Chinae-mail:wandongx@Abstract:This paper advanced a new method of appointedly cov-ering prime circles with level colors of black degree,and showed twinprime conjecture and weaker Polignac’s conjecture to be true withthe proof by contradiction.Keywords:prime;distribution of primes;twin primes;Polignac’sconjecture;Goldbach-type problem.2000MSC:11A41,11P32,11N051.Introduction.After C.Goldbach and L.Euler presented their famous conjecture by a letter in1742which states that every number>2is the sum of two primes[1,2p421],A.de Polignac presented yet his famous conjecture in1849[1,2p424,3p11],One named it the conjecture of twin primes,which statesConjecture1:Even number2is the difference of two consecutive primes in an infinitude of ways.Then he advanced extendedly another famous conjecture also in that pa-per,it statesConjecture2:Every even number is the difference of two consecutive primes in an infinitude of ways.One named it Polignac’s conjecture.Conjecture2is a very stronger con-jecture,one couldn’t now not only show it but decide whether it is true as well.And one have advanced a slightly weaker conjecture which is analogous as the former,we name it a weaker Polignac’s conjecture,it states Conjecture3:Every even number is the difference of two primes in an infinitude of ways.Owing to Conjecture3doesn’t need that two primes are in order,that is, it is allowed that there are some other primes between these two primes,its condition is weaker than Conjecture2and it is possible to show it.Although Conjecture3is weaker,yet it still includes Conjecture1as a special one.In this paper we will advance a method of appointedly covering prime circles with the level colors of black degree,and show Conjectures3and1to be true with the proof by contradiction.Some symbols using in the paper is analogous as those in ref.[4,5,6,7].2.Preparative works.Prior to showing this important problem in number theory or in mathematics,we should do some things.1a.A sequence of odd numbers.Firstly,we should write a sequence of odd numbers except1,and denote ordinal number of odd numbers by i and its increment by i and i ,and ordinal number of odd prime numbers by s and its increment byσ.That iso dd number:3,5,7,9,11,13,15,17,19,21,···, o rdinal number i:12345678910···, o rdinal number s:123−45−67−···.b.An additive graph of difference formulae of odd numbers.Sec-ondly,we can draw a graph with two-color circles indicating odd numbers. This graph is analogous as Fig.3in ref.[5],n heading numbers was cut offfrom a sequence of odd numbers and rearranged vertically in right line,then, an ordinal sequence of odd numbers was arranged paralleling with former in left,both the right and left lines construct a column for an even number, and there were imaginatively a minus sign between correspondingly two odd numbers in the same row in this column indicating this even number.we denote ordinal number of odd number vertically laying infirst column by k which is starting at2,and its increment by k .There are two kinds of backed circles colored with black and snow,the black is to prime number,the snow to odd composite number.For all n,here n=1,2,···,we can draw out some analogous columns consisted of odd numbers and get an additive graph of these numbers(Fig.1).In Fig.1we draw a horizontal dotted line for every column which is parting the difference formulae of odd numbers into two parts:one part,laying up on the dotted line,contains n difference formulae of odd numbers for the n th column which includes some formulae consisted of two primes;another, laying down under the dotted line,contains infinitely many of rest difference formulae.c.An important conclusion.According to our showing in ref.[5-7],there must exist some difference formulae of two odd primes above the dotted line for every column,that is,Polignac-Xu’s number is≥1,i.e.,#{D(n)}≥1, and it is oscillatingly increased as even number increases.So we can conclude Conclusion4:For every even number its Polignac-Xu’s number is≥1and is oscillatingly increased as even number increases.d.Some convention of using symbols.In order to conveniently indicate its significant for some mathematical items,we use some symbols to denote them.For an odd number laying at a sit in the n th column,we use a front superscript n to denote it is in the n th column,a front subscript r to denote it is in the right line,a front subscript l to denote it in the left,we use a letter c to denote a composite number,a letter p to denote an odd prime,and the ordinal number i for indicating an odd number which is in the i th row in left line in the n th column sits at a back superscript,the ordinal number k for indicating an odd number which is in the k th row in right line in the n th column sits at a back superscript also,the ordinal number s for displayingan odd prime number sits at a back subscript,that is,n r c k,n r p k s,nl c i,or nlp i s.Furthermore,we use symbol“|=|”to indicate a prime meet another prime in the i th row in the n th column,that is,there is n r p k s+n|=|nlp i s,and use2Figure 1.The appointed couples of two primes after re-arranged vertically the sequence of odd numbers.symbol“ ”to indicate a prime circle is being covered by a level color with black degree,that is,n r p k s+n n r p k s+n.3e.About the levels of black degree.And here the term“black degree”possesses the mathematical significant.We define snow is zero level of black degree and black is infinite level of black degree,there are integer levels between them,so that,the potence of the set of the levels of black degree #{B d}is equal to that of the set of nonnegative integers#{N}.We know that according to the theorem of prime numbers[8,p153-180],the potence of the set of odd prime numbers#{P r}is infinity of lower order comparing to that of natural number#{N∗},we all know there is#{N}=#{N∗},so there is#{B d}>#{P r}and the number of the levels of black degree for covering prime circles is sufficiently.f.Three relationship formulae.And from Fig.1there are some important relationship formulae.For a prime in the n th column,here n≤K−1and K is sufficiently large,there are(1)k=i+n.(2)k =i .(3)n r p k s+n=2n+n l p i s.(4)n r p k s+σ=n l p i+ns+σ.3.The proof of conjecture3.Proof.We use the proof by contradiction.Assume that Conjecture3is not true,thus,it could deduce four cases possible.a.Thefirst cases assumes that every even number is the difference of two primes but not in an infinitude of ways.It is directly in contradiction with Conclusion4.Thefirst case indicates that for all even number its Polignac-Xu’s number isfinitely,but Conclusion4indicates that for a sufficiently larger even number its Polignac-Xu’s number is infinitely.Both is in contradiction each other,and the assumption is wrong.b.The second cases assumes that all sufficiently larger even numbers are the difference of two primes in an infinitude of ways but all not sufficiently larger even numbers aren’t in an infinitude of ways.For this cases possible,all even number<M,here M is a sufficiently larger number,is onlyfinite many of the difference formulae of two primes not in an infinitude of ways.That is,we may assume that when k≥K,every even number<M doesn’t be expressed as the difference of two primes(Note: When an even number approaches to M,n is to approach to m,here m also is a sufficiently larger number).This will introduce some error conclusion.Let we return to Fig.1,for n=1,suppose that there is a last pair of twin primes in the(K−1)th row in1st column,according to the assumption of the proof by contradiction,there weren’t any appointed pair of two primes in all rows while k≥K and n<m.Now let we consider a sufficiently larger number K,and draw a horizontal real line between K−1and K. And if there then appears a prime meet another prime under this real line,1 r p K s+1|=|1lp i s in1st column,we cover1r p K s+1and1lp i+1s+1with the1st level colorof black degree,and there are1r p K s+1 1r p K s+1and1l p i+1s+11lp i+1s+1.And if4there appear some primes meet other primes yet in this column,1r p K+ks+1|=|1 l p i+is,we cover1r p K+ks+1and1lp i+i +1s+1for all pairs of k ,i andσwith thatlevel color,and cover continually again any2r p K s+1and2l p i+1s+1,and all2r p K+ks+σand2l p i+i +1sin2nd column,3r p K s+1and3lp i+1s+1,and all3r p K+ks+σ+1and3lp i+i +1s+σ+1in3rd column,and go on.Thus,there are1r p K+ks+σ+1 1r p K+ks+σ+1,2r p K s+12 r p K s+1,2r p K+ks+σ+12r p K+ks+σ+1,3r p K s+1 3r p K s+1,and3r p K+ks+σ+13r p K+ks+σ+1,···;and1 l p i+i +1s+σ+11r p i+i +1s+σ+1,2lp i+1s+12lp i+1s+1,2lp i+i +1s+σ+12lp i+i +1s+σ+1,3lp i+1s+13lp i s+1,and3 l p i+i +1s+σ+13lp i+i +1s+σ+1,···.Next,for n=2and k≥K,if there appear some primes meet others,2 r p K+ks+σ|=|2lp i+is,in2nd column,we cover2r p K+ks+σ,2lp i+i +is+σ,3r p K+ks+σ,3lp i+i +is+σ,···with2nd level color.Thus,there are2r p K+ks+σ 2r p K+k 2s+σ,2lp i+i +is+σ2 l p i+i +is+σ,3r p K+ks+σ3r p K+ks+σ,3lp i+i +is+σ3lp i+i +is+σ,···,and go on.Then,we do this for ever and ever for all n,here n=1,2,3,···,n−1(Fig.2).For example,let we see in Fig.2,There are some clearly examples.Thereis1r432113|=|1l 412012in1st column,then,we cover1r432113and1l432113with1st levelcolor,and cover n r432113and nl 4320+113for all n>1;and there is1r612917|=|1l592916yet in that column,and we cover1r612917and1l613017with that level color,andcover n r612917and nl 6129+117for all n>1.And there are n r 432113,nl4320+113,n r 612917,and nl 6129+117for all n=1,2,···.There is2r412012|=|2l 371811in2nd column,then,we cover1r412012and2l4118+212with2nd level color,and cover n r412012and nl 4118+212for all n>2,there aren r 412012and nl4118+212for all n=1,2,···.And there is3r532615|=|3l472314in3rd column,and we cover3r532615and3l5323+315with3rd level color,and we cover n r532615and nl 5323+315for all n>3;and thereis3r673317|=|3l 613017also in3rd column,and we cover3r673317and3l6730+317withthat level color,and cover n r673317and nl 6730+317for all n>3.So that there aren r 412012,nl4118+212,n r 532615,nl5323+315,n r 673317,nl6730+317.When n from1to n−1,due to n=K−i now,all black prime circlesbetween1st and k th in left line will be through all black circles between (k+1)th and(2K+1)th in right line,and all these circle were being coveredwith K−1level colors of black degree.And there are n−1r p K+ks+σn−1rp K+ks+σand n−1l p i+is+σn−1lp i+is+σfor all pairs of k andσin right line in the(n−1)thcolumn.While n turns from n−1to n,there will appear n r p K+ks+σ+n |=|nlp i+is+σand turnall nl p i+i +is+σ+nto nlp i+i +is+σ+nfor all pairs of K+k and s+σ,here1≤i+i ≤Kand K+1≤K+k ≤2K+1,And we do not consider that there were probably the primes meet primes in infinitely in the n th column.So we can conclude three error conclusions:(i)There is not any appointed pair between two primes,i.e.,#{D(n)}=0and there is no Polignac-Xu’s number,above dotted line while n≥K,this is contradiction with Conclusion4;(ii)There is no prime in open interval(K,2K)under dotted line also while n≥K, this is contradiction with a prime theorem that there must exist at least one5Figure2.Nearly all prime circles would be covered when aneven number is sufficiently large.prime between K and2K,and(iii)There is no prime in open interval(K,∞) also under dotted line while n→∞,this is contradiction with Euclid’s prime6theorem that there are infinitely many primes in natural numbers.And the assumption is wrong.c.The third cases assumes that all sufficiently larger even numbers and some not sufficiently larger even numbers are the difference of two primes in an infinitude of ways but some not sufficiently larger even numbers aren’t in an infinitude of ways.In this status,we divide all even number in to two sets,A and B,the set A contains the former,and the set B contains the latter.For all even number in B we take analogous method of covering sufficiently larger odd primes in the column indicating these even numbers.After processed all these columns we mayfind that all prime number≥K have been covered.Of cause,all odd primes≥K which lay in all columns indicating all even numbers in Ashould also be covered.And there is n r p K+ks+σ+n |=|nlp i+is+σfor all pairs of K+kand s+σfor all columns in A,so that there is no n r p K+ks+σ+n |=|nlp i+is+σ,and theassumption is contradiction itself in this case and it is wrong.d.In the last case one assumes that every even number except2is the difference of two primes in an infinitude of ways.There is indeed no good method to show this assumption to be wrong.But if it were true,one yet assumes that every even number except4is the difference of two primes in an infinitude of ways;every even number except6is the difference of two primes in an infinitude of ways;and go on.That will introduce a new antinomy.The assumption is wrong.Up to now,we havefinished showing Conjecture3to be true and also Conjecture1to be true.Acknowledge:I am grateful to Prof.Rong-Qing Jia,Prof.Geng-Pu Yu, Prof.Zai-De Wu,Prof.Jing Xu,Eng.Yu-Zhi Li,and Bachelor Heng Xu for beneficial discussion,and my student Yue-Bo Ding for her verifying all of the data in this paper.REFERENCE[1]D.Hilbert,Mathematische Problems,Archiv5.Math.u.Phys.3,Bd.1(1901),44-63;213-237.[2]L.E.Dickson,History of the theory of numbers,Carnegie Institute of Washington,Washington,vol I.1919.[3]T.M.Apostol,Introduction to Analytic Number Theory,Springer-Verlag,NewYork,1976.[4]W.-D.Xu,A new two-dimension sieve method and the proof of Goldbach’s con-jecture,/200606266(unpublished).[5],Every even number is the difference of two odd prime numbers,www./2006006456(unpublished).[6],A C++source program of calculating Polignac-Xu’s number with recur-sion method,/200609373(unpublished).[7],A FORTRAN source program of calculating Polignac-Xu’s number withrecursion method,/200609392(unpublished).[8]P.Ribenboim,The Book of Prime Number Records,(Second Edition),Springer-Verlag,Berlin,1989.7。