, THERE IS AT LEAST ONE PAIR OF DOUBLE PRIMES FOR ANY EVEN NUMBER arxiv:ath/0502v athgm 6 Oct 2005 SHOUYU DU AND ZHANLE DU This aer is dedicated to Prof Wei WANG Abstract We roved that any even nuber not less than 6 can be exressed as the su of two old ries, 2n Introduction The Goldbach conjecture was one of the oldest unsolved robles in nuber theory, 4 It states that for any even nuber 2n there exists a air of double ries (, ) such that 2n, and usually reresented as The best result is the Chen s Theore, 2 that every large even nuber ay be written as the su of a rie and a seirie (2-alost ries) Our result is as theore Theore There is at least one air of double ries for any even nuber 2n 6 Let P{, 2,, v } {2,,, v } be the ries not exceeding 2n, then the nuber of ries not exceeding 2n 5 is, (π( ( ) 2n 2n 2n 2n) )2n () π(2n) 2 ( v 2n 2n 2n 2 v v ), For silicity, we can write it as, π(2n) (π( 2n) )2n 2 (2) v (π( 2n) )2n v i i, where 2n i 2n 2n, 2n i j 2n 2n 2n The oerator i will leave the ites which are not ultiles of 2n In this aer, x xis the floorfunction, x xthe ceiling function ofx The integral oerator, which is not the floor function in this aer, has eanings only oerating on (real) nuber: Date: Oct, 2005 2000 Matheatics Subject Classification A4; P2; N6 Key words and hrases rie, floor function, ceiling function, integral oerator, Goldbach s conjecture All rights reserved This is one of a serial works, and was suorted financially by Prof Wei WANG and on loan in art Corresonding author: Zhanle Du
2 SHOUYU DU AND ZHANLE DU 2 The nuber of double rie airs in 2n Let Z {,2,,}( 2n ) be a natural arithetic rogression, Z 2n Z {2n,2n 2,,2n } be its accoanying arithetic rogression, so that 2n Z k Z k,k,2,, There are such airs { Z {, 2,, } (2) Z { 2n, 2n 2,, 2n } After deleted all the airs in which one or both ites Z k and Z k are ultiles of the ries 2n, then the airs left are all rie airs or,2n- For a certain, we first delete the ultiles of in set Z, or the ites of Z k od 0, (22) y( ) Note the oerator If has eaning only oerated on integer () (2) λ i (2n)od is not zero, we should delete the ultiles of in Z, ie the ites of (2n Z k )od 0 or Z k od (2n)od in set Z, (24) y (i ) (2n )od ( ), def where0 (2n )od ( ) is the reainderodulo 0 θ i ( ) (2n )od ( λ i ) ( ) is the nuber of eleents which, when added before Z, will ake a new rogression(z {Z a,z },Z a {, 2,,λ i }) having ites of Z k od 0 in the ositions of kod 0 If λ i 0, then it is the osition of the first ite with Z k od 0, and then θ i λ i Eq (22) will delete all airs with Z k od 0 in set Z, and Eq (24) will delete all airs with Z k od λ i in set Z If (2n)od 0, then for soe Z k od 0, we have Z k od (2n Z k )od 0 These two ites are in the sae air and should be deleted only once, thus, (25) y ( ) 0 if (2n)od 0 When (2n)od 0, θ i, (26) y ( ) δ, where 0 δ with, { 0 : 0 odi θ (27) δ i < : else After deleted all the ultiles of in both Z and Z, the airs will leave, (28) M( ) y( ) y ( )
THERE IS AT LEAST ONE PAIR OF DOUBLE PRIMES FOR ANY EVEN NUMBER The oerator, when oerating on, will leaves the ites having no ultiles of After deleted the ultiles of the ries v in both Z and Z, the airs left will be rie airs and have, D 0 () (29) 2 2 v v (2n)od0 (2n)od 0, the eaning is as follows, (20) (2), (i ) (2n )od θi i λ i (i j ) (2n )od θi,j i j λ i,j i i λ ij λ i j def def Let θ ij (2n )od λ ij ( ) λ ij (2n)od istheositionofthefirstiteinz withz k od( ) 0, if it does not exist then λ ij >,θ ij < and this ite will be zero Note that λ ij 0 because (2n) and (2n) def In Eq (2), θ i,j λ i,j ( ), λ i,j is the osition of the first ite with Z λi,j and Z λ i,j (22) { Zλi,jod 0 Z λi,j od (2n)od Let X {λ }, X 2n X {2n λ }, λ,2,,, then (2) { λi j od (2n)od λ i,j λ i j If it exists and (2n)od 0, then λ i j, and θ i,j λ i,j ( ) If there is no such λ i,j in set Z, ie, λ i,j (), then the last ite in Eq (2) will equal zero Let 2n,thenZ {,2,,2n }andz 2n Z {2n,2n 2,,} For each ais, there is another sae air: 2n Z k Z k Z 2n k Z 2n k When k n is rie, it ust be the sae air 2n Z n Z n nn Therefore, fro
4 SHOUYU DU AND ZHANLE DU Eq (29), the actual nuber of rie airs in 2n is, D(2n) D0(2n ) 2 D( 2n) D (24) 2 (2n ) 2 v i2 D( 2n) D, where D( 2n) 0 is the nuber of rie airs 2n (2n ) when v < 2n, 2n are both ries, and { 0 2n rie (25) D 2n rie Reeber that when (2n)od 0, 0 Exale 2 Let 2n 46, then 2,,5, n,5 n, 2n 45, (26) Fro Eq (2), (27) { Z { 2 2 44 45} Z { 45 44 4 2 2 } { i : 5 5 λ i : Fro Eq (2), { (i, (28) j ) : (2,) (5,) (0,) (2,5) (,5) (6,5) (2,5) λ i,j : 4 0 0 6 6 6 6) Fro Eq (29), (2), D 0 (2n ) (2n ) 2 45 2 6 2 5 2 5 5 6 5 5 2 5 45 45 2 45 5 45 45 0 455 5 455 5 45 6 45 5 450 6 0 5 5 5 0 5 45 45 0 450 6 0 45 22 57 57 94 2 94 2 455 0 5 455 6 5 456 4 6 0 5 0 450 0 0 450 6 0 Wecancheckthisresultdirectly AfterdeletedtheitesofZ k od2 0;Z k od 0,;Z k od5 0, fro set Z, it is left Z {7,2,29} D 0 (2n ), it is the sae as before Fro D( 2n) 2(46 4 5 4), D 0(46-45 is not a rie) Fro Eq (24), D(2n) 2 2 0 4 The set left is Z {,5,7,2},Z
THERE IS AT LEAST ONE PAIR OF DOUBLE PRIMES FOR ANY EVEN NUMBER 5 44 Z {4,4,29,2} Thus there are four double rie airs of 46 : (,4),(5,4),(7,29),(2,2) Definition 22 The ites of Z k od 0, or the ultiles of, have, (29) S(, 0) : Z k od0 So the ites of Z k od 0,λ j have, (220) S(, 0,λ j ) : Z k od 0,λ j Definition 2 Let X {X,X 2,,X t } be an (any) integer set, then (22) t tφ : t i tφ i 0 X k od 0,λ i is the nuber of ites left after deleted the ites of X k od 0,λ i ( ) Usually, We can exress it as S(, 0,λ i ; 0,λ j ) (222) i ( ), is the nuber of ites left when we first delete those Z k od 0,λ i fro set Z, and then delete those X k od 0,λ j fro set X {Z,X k od 0,λ i,k,2,, }, where set X is no longer an arithetic sequence In general, (22) i ( i ) i i Soe roerty () (2) i j j i 2 2 od 2od () i 0 ( (4) a 2 )b for a b (5) 2 ( 2 ) 2
6 SHOUYU DU AND ZHANLE DU Proof Let α od,β 2 od, δ 2 ( 2 ) 2 2 2 2 αβ 2θ i αβθi 2θ i αθi βθi 2 θ i The iniu: in(δ 2 ) 2, when αβ, αθ i <, βθ i < The axiu: ax(δ 2 ) 0, when αβ <, αθ i, β θ i We can reresent Eq (5) as ( 2 ) (6) 2 2 2 2 (2n )od (2n )od 2 od 2 2 (2n )od) 2 will delete the ites of X k od 0 in the sequence of X {, 2,, 2 }, and 2 will delete the ites of X k od 0 in the sequence of X {2n X} or the ites of X k od λ i in set X For any 2, 0 2 (7) 2, so, for Lea 4 For, (4) Proof left 4 Soe lea ) ( i θi 2 right Lea 42 For 2 j, > 2, ) (42) ( j Proof Let s t, t a b, 0 a ( ), 0 b ( ), Suose λ i (2n)od 0 else 0, and λ j (2n)od 0 else 0, so θ i,θ j > 0 and θ i,j,θ j,i > 0,θ ij > 0 Because 2 j, >, so s
THERE IS AT LEAST ONE PAIR OF DOUBLE PRIMES FOR ANY EVEN NUMBER 7 Because (v) 2 v Let (j ) (j)( /), then (j ) (j)( / ) 2 j ( /) ( ) ( 2)( ) 2 j 2 2 j So for any i (j ), we have (i) 2 i Fro Eq (2), (4), ) ε ( j ( )( ) ( )( ) s 2 2 s 2 t t tθi t t tθj,i tθj where tθij ( t s( i 2) t tθij s( 2) a ajbθ ij ) tθj a a 2 t ε a s( 2)a ε ε 2 2ε ε 4, ε ε 2 ε bθj t 0 tθ i tθj,i tθi,j ε 2 ε ε 2 ajbθ j,i ajbθ j,i ajbθ i,j ( od i t )od j ( od i tθ i a 2 )od ajbθ i,j tθi,j ajbθ ij ε 4 bθj, ε ε ε 2 2ε ε 4 7 () If, then ε s( 2) ε 9 7 > 0 (2) If 7, then (a) if a 2, then ε s( 2)a ε 52 7 0 (b) if a, then ε a 2 7 0, ε, so ε s(i 2) ε 5 > 0 () If 5, (a) if a 4: ε s( 2)a ε 4 7 0 (b) if a 0: ε 0, ε, so ε s( 2) ε 0
8 SHOUYU DU AND ZHANLE DU (c) if a : if then ε, ε 5, so ε s( 2)a 0 then ε 0, ε, so ε 5 0; else for ε s( 2)a ε > 0 (d) if a 2: ε, ε 5, so ε s( 2)a ε 2 5 0 (e) if a : if 7 0; else for then ε s( 2) a a 0 then ε ε s( 2)a ε 5 > 0 (4) If, then 0 a 2, a 2ε a 2 ε ε 2 ε s a ε ε 2 ε 4 (a) if a a 2a a ( ( )od a j jb )od j ( ( )od a j jbθ i )od j 0, then ε ε 2 0 a 0: ε s ε 4 0 if a : ε s ajbθ j,i ajbθ i,j ajbθ i,j ε 4 Proof If s 2 then ε If s ie, s 2 j ε, ε 5, so, ajbθ ij 5 2 52 5, we have a 2 Fro, so 5 Fro 0, we have b Because θ j, so bθ j If bθ j, ie, b 0 or θ j 2, then ε 4 0 Else for bθ j, ie, b and θ j, ε 4 But θ j (2n )od, so (2n )od 0, or (2n)od Let us consider θ i,j λ i,j Fro Eq (2), λ i,j λ λ, with the condition (2n λ)od 0 (λ)od (2n)od5, so λ 2 and θ i,j ajbθ i,j λ 5 2 9 Therefore, ajbθ i,j 2 59 5 ε 4 0 if s or ε s ajbθ i,j ε 4 ε ε ajbθ a j,i ajbθ ij
THERE IS AT LEAST ONE PAIR OF DOUBLE PRIMES FOR ANY EVEN NUMBER 9 (b) if If a, then ε a 0 Else if a 2(< ), fro Eq (2), θ j,i λ ( ajbθ λ ), j,i 2j j So ε a 0, then ε Proof if ε ε 2 0 if 2, then ε 0, because θ i, ε ε 2 (ajb )od 0 if (a b)od 0 0 if (a b)od 0 if (a b)od 2 So ε ε 2 and ε (ajbθ i )od ε ε 2 0 Besides, ε ajbθ s i,j ε 4 Proof If s 2 then ε If s ie, s 2 j 52 5,we have a 2 Fro, so 5 Fro 5 2,2, we have b 2,,4 Let us consider θ i,j λ i,j Fro Eq (2), λ i,j λ λ, with the condition (λ)od (2n)od5 Because (2n)od5 0, so λ 4 and θ i,j λ (5 λ) ajbθ i,j j ajbθ i,j So, Therefore, or ε s 2 52 5 ε 4 0 if s, ajbθ i,j ε 4 ε ε aε a ajbθ j,i ajbθ ij ajbθ j,i If a then ε a 0 Else for a 2(< ), because θ j,i λ ( λ ), ajbθ j,i 2j j So ε a 0 In suary, ε 0 for all < the Lea is roved If (2n)od 0,
0 SHOUYU DU AND ZHANLE DU ε i ( ) j i ( )( ) ( )( ) s 2 s t t t ( t t ) s( ) s( )a s( )a where t t tθj tθj a a ε ε ε 4 tθi,j t tθi,j bθj ajbθ i,j ajbθ i,j ε ε ε 4, ( od i t )od j a bθj, 2 ε a ε ε ε ε 4 4 We can follow the sae ethod above (it is a little easier), i (4) j ( ) j i ) ( j ε Eq (4) and (42) ean that, after deleted the ites of Z k od 0,λ j fro Z, the ites of Z k od 0,λ i, will have, (44) S(, 0,λ i ; 0,λ j ) S(( / ), 0,λ i ) ( 0,λ i ), where ( 0,λ i ) 0 is the extra ites of Z k od 0,λ i and Z k od 0,λ j Thus the effect of is that it constructs a new effective nature sequence with at least ( ) ites which satisfy the condition Z k od 0,λ j This lea eans that, fro equation (4) and (6), we can let the ites which have no ultiles of and (45) ( j ) t,t 0, and oerated by S(, 0,λ i ; 0,λ j ) ( ) ) ( j t S(( / ) j ), 0,λ i ) ( i 0,λ i ) ( j t ) ( j to attain
THERE IS AT LEAST ONE PAIR OF DOUBLE PRIMES FOR ANY EVEN NUMBER Lea 4 (46) 2 6 Proof For 2,, 6st, fro Eq (4), 2 6s 2 6s ( 2)( 2 ) s 6 6 t 2 For the residual class of odulo 6, X {6s,6s2,,6s6}, X k od6 {,2,,4,5,6}, there are 4 eleents (2,,4,6) of ultiles of 2 or For the other eleents X k od6 {,5}, there is at least one ite with (2n X k ) or X k od (2n)od So will have at least 6 6 double rie airs Lea 44 For i,2,,i,j, (47) i i ( ) i i Proof Fro equation (220), (4), (42), (44), (45), For i k, we have S(, 0,λ i ; 0,λ j ) S(( / ), 0,λ i ) and S(, k 0,λ k ; 0,λ j ) S(( / ), k 0,λ k ), so that S(, 0,λ i ; k 0,λ k ; 0,λ j ) S(( / ), 0,λ i ; k 0,λ k ) ( 0,λ i ; k 0,λ k ) S(( / ), 0,λ i ; k 0,λ k ) Thus S(, 0,λ i ;2 0,λ i2 ; 0,λ j ) S(( / ), 0,λ i ;2 0,λ i2 ) S(, 0,λ i ;2 0,λ i2 ;, 0,λ i ; 0,λ j ) S(( / ), 0,λ i ;2 0,λ i2 ;, 0,λ i ), i i S(, 0,λ ; 2 0,λ 2 ; ; 0,λ i ; 0,λ j ) S(( / ), 0,λ ; 2 0,λ 2 ; ; 0,λ i ) ) i ( j Suose that for < r i, i (48) i ) i ( j t, wheret 0 Iteansthattheeffectofoerator whenoeratingon ) isthat dividingz {,2,,}intotwosetsX {,2,, ( j } and X {X,X 2,,X t,t 0} Fro equation (222), (22), (4),
2 SHOUYU DU AND ZHANLE DU (42), and (48), we have i i ( ) i ) ( j t ) i ( j ) i ( j or i i i i ) i ( j t ) i ( j t ) i ( j t ) i ( j t ) i ( j If (2n)od 0, then i (49) ( ) i j In fact, if i i < ( ) i i, then for any,, before deleting the ultiles of other ries, it ust have ) < ( j which contradicts Lea 42 So this lea is true
THERE IS AT LEAST ONE PAIR OF DOUBLE PRIMES FOR ANY EVEN NUMBER ) With Lea 42, the oerator can be reresented by ( j, and other oerator can oerate on this inequality unchanged 5 Exlantation Let s a b 2 j, then for all <, the effect of is a nature sequence X {f(z)} whose nuber is not less than ( / ) The reason is as follows When a nature sequence is deleted by the ultiles of θj, the sequence is subtracted by We can arrange the ites in a table of rows (Table ) will delete the th row, and θj delete the (2n)od th row Thus there are ( 2) rows left in which each ite Z k od 0,(2n)od Table Set Z ( ) (s ) s s a 2 2 ( ) 2 (s ) 2 s 2 s a 2 b b ( ) b (s ) b s b s a b 2 s s will But every ites (0 r Z k od ) in any row of the first s coluns consist in a colete syste of residues odulo, because C {,,2,,( ) } and C r {C r} are both colete syste of residues odulo, where r is any (row or colun) constant There are ( 2) such rows or s ( 2) ites left These ites are effective to a nature sequence when deleting ultiles of, s s ( 2) s( 2)( 2) Let t a b, 0 b, 0 a, t will delete at ost a(a) t s ites If we add these ites by reoving those fro the end of sequence, then the sequence is again effective to a nature sequence, which has at least, M(j) s ( 2)t a(a) s ( )ts i 4a 2 t s ( ) t(s a 2) s ( ) t (s a 2) (s t)( ) (s a 2) ( ) (s a 2)
4 SHOUYU DU AND ZHANLE DU For s 2 or a 2, we have (s a 2) (s ) ( a ) 0, M(j) ( ) For s and a, the ites of t have rows, coluns and soe b ites In each of the first b rows, there are exact ites which consist in a colete syste of residues odulo, and these ites can be considered as an effective nature sequence when deleting the ultiles of (Z k od 0,λ i ) The other ites have at ost rows and coluns where the ultiles of have at ost 2( ) As before, we can add these ites to ake the t as an effective nature sequence, therefore, M(j) s ( 2)t 2( ) ( ) (s a ) ( ) Thus for any <, the original sequence of 2 j, after deleted all the ultiles of fro 2n Z k Z k, is effective to reconstruct a new nature sequence having at least ( ) ites Exale 5 2n 46, 2n 45, Z {,2,,45}, For 5, after deleted the ite of Z k od 0,(2n)od, it becoes Z Z {2,,4,7,8,9,2,,4,7,8,9,22,2,24,27,28,29,2,,4,7,8,- 9,42,4,44} we can rearrange these ites as Z {(9),2,,4,(2)7,8,9,(22),- (29),2,,4,(27),(28),7,8,,24,2,,4,7,8,9,42,4,44} ) The first ( j 8itescanbetakenasaneffectivenaturesequence({,2,,8}) fro the original one when deleting the ites of Z k od 0,(2n)od The other sequence X {24,2,,4,7,8,9,42,4,44}, having at least zero ite after deleted the ultiles of all ries, will be neglected in further rocess 6 Proof of Theore For a given 2n, consider the ossible airs of 2n, where and are both ries Lea 6 The (double) rie airs in 2n, (6) D(2n) where v v6 i v (62) W(v) v If W(v) > 9 then D(2n) 2 i Proof Fro equation (29), (), (47), (49) and (46), W(v) 9 6,
THERE IS AT LEAST ONE PAIR OF DOUBLE PRIMES FOR ANY EVEN NUMBER 5 but D 0 () (2n) od 0 2 v i2 (2n) od 0 ( ) v v 2 i2 ( ) ( ) v v 2 v 2 i2 ( v i ) 2 ) v 6 i ( i, v i( ) v and 2n 2 v, so (6) D 0 (2n ) 4 5 4 2 v v 2 6 v i Fro equation (24), (25), 4 4 v v v v v v 2 v v i, v v i D(2n) D0(2n ) 2 v v 6 i D( 2n) D 2 v v 2 i 2 Proof of Theore Because for v 5, v, W(v) 4 8 5 7 0057 > 9 Suose that for v, W(v) v v i > 9, then for v, (64) W(v ) v v i W(v) 2 v v 2 v Because, v v 2,, v, v v i v v v v 2 v v 2 v ( v 2 ) 2 ( v 2 ) 2 v 2 v 4 v 4 2 v 6 2 v ( ) v ( v 4 6) > 0 Therefore, 2 v v >, W(v ) > W(v) > 9 Fro the rincile of atheatical induction, we can conclude that for any v 5, we have W(v) 9 and 2 v D(2n), or thereis at least one airofdouble ries, such that 2n for any 2n > 2 v 2 22
6 SHOUYU DU AND ZHANLE DU Besides, when 6 2n 20, we know that there is at least one air of ries, such that 2n In fact, fro Eq (6), D(2n) aroaches infinity as n grows without bound The roof is coleted Corollary 62 Any odd nuber not less than 9 can be exressed as the su of three odd ries Proof If n is odd nuber,, then n 6 and can be reresent as the su of two ries 2 fro Theore So n 2 Exale 6 (Actual vs Silified Forula) Figure shows the iniu actual rie airs D(2n) (solid line) of 2n in the range of 2 v,2 v and the silified forula (dashed line) fro Eq (6) against v Fro this figure, it is easily seen that (65) D(2n) v v 6 i 2 Figure The iniu nuber of actual rie airs(solid) and its silified forula (dashed) against v References J R Chen, On the Reresentation of a Large Even Integer as the Su of a Prie and the Product of at Most Two Pries, Sci Sinica 6 (97), 57-76 2 J R Chen, On the Reresentation of a Large Even Integer as the Su of a Prie and the Product of at Most Two Pries II, Sci Sinica 6 (978), 42-40 J R Chen & C D Pan, The excetional set of Goldbach nubers, Sci Sinica, 2 (980), 46-40 4 H L Montgoery & R C Vaughan, The excetional set in Goldbach s robe, Acta Arith, 27 (975), 5-70 5 Kenneth H Rosen, Eleentanry Nuber Theory and Its Alications, Fifth Edition, Pearson Education Asia Liited and China Maxhine Press, 2005 Chinese Acadey of Sciences, 99 Donggang rd ShiJiaZhuang, HeBei, 0500, China E-ail address: shouyudu@yahoococn Chinese Acadey of Sciences, A20 Datun rd Chaoyang Dst Beijing 0002, China E-ail address: zldu@baoaccn