consecutive integers II problem, and some conjectures of Erdős on Small gaps between almost primes, the parity Journal of Number Theory

(1)

Contents lists available atScienceDirect

Journal of Number Theory

www.elsevier.com/locate/jnt

General Section

Small gaps between almost primes, the parity problem, and some conjectures of Erdős on consecutive integers II

DanielA. Goldston, Sidney W. Graham,Apoorva Panidapu, Janos Pintz¹, Jordan Schettler^∗, CemY. Yıldırım

a r t i c l e i n f o a bs t r a c t

Article history:

Received25April2020 Accepted28June2020 Availableonline16July2020 CommunicatedbyS.J.Miller

Keywords:

Almostprime Smallgaps Erdos Mirsky Divisor

Exponentpattern

Weshowthatforanypositiveintegern,thereissomefixed Asuchthatd(x)=d(x+n)=Ainfinitelyoftenwhered(x) denotesthenumberofdivisorsofx.Infact,weestablishthe stronger result that both x andx+nhave thesame fixed exponent pattern for infinitely many x. Here the exponent pattern of an integer x > 1 is the multiset of nonzero exponentswhichappearintheprimefactorizationofx.

1. Introduction

Thispaperisintendedasasequelto[GGPY11] writtenbyfourofthecoauthorshere.

Inthepaper,theyprovedastrongerform oftheErdős-Mirksyconjecturementionedin [EM52] whichstatesthatthere areinﬁnitelymanypositive integersxsuchthatd(x)= d(x+ 1) whered(x) denotesthenumberofdivisorsofx.Thisconjecturewasﬁrstproven byHeath-Brownin1984[HB84],butthemethoddidnotrevealthenatureofthesetof

* Correspondingauthor.

E-mailaddress:jordan.schettler@sjsu.edu(J. Schettler).

1 ResearchsupportedbytheNationalResearchDevelopmentandInnovationOﬃce,NKFIH,K119528.

https://doi.org/10.1016/j.jnt.2020.06.002

(2)

valuesd(x) forsuchx.Inparticular,onecouldnotconcludethattherewasanyparticular valueAforwhichd(x)=d(x+ 1)=Ainﬁnitelyoften.In[GGPY11],theauthorsshowed that

d(x) =d(x+ 1) = 24 for infinitely many positive integersx. (0.1) Similarresultswereprovenfor otherrelatedarithmeticfunctionswhichcount numbers ofprimedivisors. Thegoalof thispaper isto establishresults foranarbitrary shiftn, i.e.,d(x)=d(x+n)=Ainfinitelyoftenforsomefixed A.

2. Notationandpreliminaries

Forour purposes,alinear form isan expression L(m)=am+b where a and b are integersand a> 0.We view L both as apolynomial and as a functionin m. We say L isreduced if gcd(a,b)= 1.If K(m)=cm+dis another linearform,then arelation betweenL andK is anequation oftheform |c_L·L−c_K·K|=nwhere c_L, c_K,n are allpositive integers.We call cL, cK the relationcoeﬃcients and we call nthe relation value.Wedeﬁne thedeterminant ofLandK asdet(L,K)=|ad−bc|.

Foraprimep,ak-tupleof linearforms L1,L2,. . . ,Lk iscalled p-admissible ifthere isanintegert_p suchthat

L₁(t_p)L₂(t_p)· · ·L_k(t_p)≡0 (modp)

Wesay thatak-tuple of linearforms is admissibleif itis p-admissiblefor everyprime p. Notethatak-tupleoflinearformsisadmissibleiﬀalltheformsarereducedandthe tupleisp-admissibleforeveryprimep≤k.

AnEr number isapositive integer thatisthe product ofr distinct primes.Several ofthecoauthorshereprovedthefollowing resultonE2-numbersinadmissibletriplesin [GGPY09].Later,FrankThorne[Tho08] obtainedageneralizationforEr-numberswith r≥3.

Theorem1.LetC beany constant.IfL₁,L₂,L₃ isanadmissibletriple oflinearforms, then there are two among them, say L_j and L_k such that both L_j(x) and L_k(x) are E2-numberswithboth primefactors largerthan C forinﬁnitelymanyx.

The results obtained in this paper will use Theorem 1 above in combination with Theorem2below,aspecialcaseofwhichwasproveninthepreviouspaper[GGPY11].

Weprovideaproofhereofthegeneralversionsinceitcontainsimportantideasrelevant fortherestofthepaper.

Theorem 2 (Adjoining Primes). Assume that L_i = a_im+b_i for i = 1,. . . ,k gives an admissiblek-tuplewith relations |c_i,jL_i−c_j,iL_j|=n_i,j.We can always“adjoin” prime factorstotherelationcoeﬃcientswithoutchangingtherelationvalues:foreverychoiceof

(3)

positiveintegersr1,r2,. . .,rk suchthatgcd(ri,ai)= gcd(ri,det(Li,Lj))= gcd(ri,rj)= 1 whenever i = j, there is an admissible k-tuple of linear forms K1,K2,. . . ,Kk with relations |ci,jriKi−cj,irjKj|=ni,j.

Proof. LetxbeasolutionofthecongruencesLi(x)≡ri (modr²_i) for1≤i≤k.Suchan xexistsbytheChineseRemainderTheoremsincegcd(a_i,r_i)= gcd(r_i,r_j)= 1.Thisxis uniquemodulor= (r₁r₂· · ·r_k)².Nowdeﬁneanewk-tupleviaK_i(m)=L_i(rm+x)/r_i. Byconstruction,wehave|ci,jriKi−cj,irjKj|=ni,j,soweonlyneedtocheck thatthis new k-tuple is admissible. Wewill show thatthe newk-tuple is p-admissiblefor every primep.Therearetwo cases.

Case 1:Suppose thatp|r. Sincegcd(r_i,r_j)= 1 for i=j,we havethatp|r for exactly oneindex.Now

K(0) =L(x)/r≡1 (modr)

so K(0)≡1≡0 (modp).WeclaimthatalsoKi(0)≡0 (modp) wheni=.Suppose, bywayofcontradiction,thatKi(0)≡0 (modp) forsomei=.ThenLi(x)≡0 (modp) sinceri≡0 (modp),butL(x)≡r≡0 (modp),so

det(L, Li) =|aib−abi|=|aiL(x)−aLi(x)| ≡0 (modp),

butthiscontradictstheassumptionthatgcd(r,det(L,L_i))= 1.ThusK₁(0)· · ·K_k(0)≡ 0 (modp).

Case 2: Now suppose p r. Since L1,. . . ,Lk is admissible, there is an integer tp such that L1(tp)· · ·Lk(tp) ≡ 0 (modp). Choose τp such that rτp+x ≡ tp (modp). Then L_i(rτ_p+x)≡L_i(t_p)≡0 (modp) andr_i≡0 (modp) foralli,so

K₁(τ_p)· · ·K_k(τ_p) =L₁(rτ_p+x)

r1 · · ·L_k(rτ_p+x)

rk ≡0 (modp).

Letnbeapositiveintegerandwriteitsprimefactorizationasn=p^k₁¹p^k₂²· · ·p^k_j^j where the p_i are distinct primes with k_i >0.Then the exponent pattern of n is themultiset {k₁,k₂,. . . ,k_j}where order doesnotmatter butrepetitionsare allowed.Thevalues of manyimportantarithmeticfunctionsdependonlyontheexponentpatternoftheinput;

suchfunctionsinclude:

d(x) = # of divisors ofx

Ω(x) = # of prime factors (counted with multiplicity) ofx ω(x) = # of distinct prime factors ofx

μ(x) = Möbius function = (−1)^ω(x)ifnis squarefree, zero otherwise λ(x) = Liouville function = (−1)^Ω(x)

(4)

Thus if both x and x+n have the same exponent pattern, then d(x) = d(x+n), Ω(x)= Ω(x+n), ω(x)=ω(x+n), etc. In establishingthe strong form of the Erdős- MirskyConjecture (0.1),theauthorsin[GGPY11] actually provedthefollowing result.

Theorem3.Thereareinﬁnitelymanypositiveintegersxsuchthatboth xandx+ 1have exponentpattern {2,1,1,1}.

Wewill show thatforany shift n,there are inﬁnitelymany positive integersxsuch thatboth xand x+nhaveaﬁxedsmallexponentpattern.Akeytool fordoingthisis containedinthenextremark.

Remark 4.Suppose we have an admissible triple of forms Li with relations |ci,jLi− cj,iLj| = n. For a given form Li in the triple, we call ci,j and ci,k where {i,j,k} = {1,2,3}thepairofrelationcoeﬃcientsforLiinthetriple.Supposethesepairsofrelation coeﬃcientsforeachforminthetriplehavematchingexponentpatterns,i.e.,ci,jandci,k

havethesameexponentpatternwithany choicesofi,j,ksuchthat{i,j,k}={1,2,3}. We then can choose pairwise coprime integers having any desired exponent pattern whicharerelativelyprimetoalllinearcoefficientsanddeterminants(sincedeterminants of distinct reduced forms are alwaysnonzero). In particular, we canadjoin integers to the relation coefficientsso thatthe new triple has theproperty thatall of its relation coefficientshaveanygivenexponentpattern Pwhichcontainstheexponentpatternsof everyci,j. Henceby Theorem 1, we would then getinfinitely many positive integersx suchthatbothxandx+nhaveexponentpattern_P∪ {1,1}.TheproofsofTheorems5 and7belowwillrelyheavilyonthisidea.

3. Shiftswhichareevenornotdivisibleby15

Theorem5. Letnbeapositiveintegerwith2|nor15n.Thenthereareinﬁnitelymany positiveintegers xsuchthatboth xandx+n haveexponent pattern{2,1,1,1,1}. Proof. Consider the following triple of linear forms:L₁ = 2m+n,L₂ = 3m+n, and L₃= 5m+ 2n.Wehavetherelations

3L₁−2L₂=n 5L₁−2L₃=n 3L₃−5L₂=n

Now deﬁne gi = gcd(i,n) and reduce the linearforms: takeL1 =L1/g2, L2 =L2/g3, andL3=L3/g5.Then therelationsbecome

3·g2L1−2·g3L2=n 5·g2L1−2·g5L3=n

(5)

3·g5L3−5·g3L2=n

Case 1: Suppose n is even and write n = 2n2. Then g2 = 2, so L1 = m+n2, L2 = (3/g₃)m+ 2(n₂/g₃),andL₃= (5/g₅)m+ 4(n₂/g₅).

Subcase 1a:Suppose2|n2.Then

L₁(1)L₂(1)L₃(1)≡1³≡0 (mod 2),

so the triple L1, L2, L3 is 2-admissible. Now we check this triple is also 3-admissible (and thereforeadmissible).

•If3n2,then

L1(0)L2(0)L3(0)≡n2(−n2)(n2/g5)≡0 (mod 3).

•If3|n₂,theng₃= 3,soL₁≡m≡ ±L₃ (mod 3).Nowchoosem₀∈ {1,−1}suchthat L2(m0)≡0 (mod 3).Then

L1(m0)L2(m0)L3(m0)≡m0·L2(m0)·(±m0)≡0 (mod 3).

Here therelation coeﬃcientsmatch inpairsfor agiven form inthetriple and allhave exponent patterns contained in {1,1}, so by appeal to Remark 4 we have a slightly stronger result, namely, there are inﬁnitely many positive integers xsuch thatboth x and x+nhaveexponentpattern{1,1,1,1}.

Subcase 1b:Supposenow2n2.Let

K1=L1(4m+n2)/2 = 2m+n2

K2=L2(4m+n2) = 4· 3 g3

m+ 5· n₂ g3

K₃=L₃(4m+n₂) = 4· 5 g5

m+ 9· n₂ g5

Ourrelationsthusbecome

2²·3K1−2·g3K2=n 2²·5K1−2·g5K3=n 3·g5K3−5·g3K2=n

Herethepairsofrelationcoeﬃcientsforeachforminthetriplehavematchingexponent patterns.WewillcheckthatthetripleK₁,K₂,K₃isadmissible.First,wenotethateach form isstillreduced:

K1= 2m+n2

(6)

isreducedsince2n2.

K₂= 4· 3

g₃m+ 5·n2

g₃

isreducedsincetheconstanttermisoddand notdivisibleby3 ifg₃= 1.

K₃= 4· 5

g₅m+ 9·n2

g₅

isreducedsincetheconstanttermisoddand notdivisibleby5 ifg₅= 1.

NextK1K2K3 ≡1 (mod 2), sothetriple isindeed2-admissible. Nowwecheck that thistripleis3-admissible.

•If3n₂,theng₃= 1,so

K1(−n2)K2(−n2)K3(−n2)≡(−n2)²(n2/g5)≡0 (mod 3)

•If 3| n₂, then K₁K₃ ≡ ±m² (mod 3). Choose m₀ ∈ {1,−1}such thatK₂(m₀) ≡0 (mod 3).Then

K1(m0)K2(m0)K3(m0)≡ ±(m0)²K2(m0)≡0 (mod 3).

Heretherelationcoeﬃcientsallhaveexponentpatternscontainedin{2,1,1},soadjoin- ingprimesagaingivesusthestatementofthetheorem.

Case2:Nowsupposenisodd,sog2= 1 fromnowon.OurrelationsforLibecome 3L1−2·g3L2=n

5L1−2·g5L3=n 3·g5L3−5·g3L2=n

Ifwelook atthismodulo2,wegetL₁ ≡1,L₂≡m+ 1,L₃≡m.Thusthis tripleisnot 2-admissiblehere.However,we canrestrictm (mod 2) andreduceto get2-admissible.

Todothis,wewrite

M1=L1(2m) = 4m+n M2=L2(2m) = 2· 3

g3

m+ n g3

M₃=L₃(2m)/2 = 5 g5

m+ n g5

.

ThetripleM1,M2,M3 hasreducedformsandis2-admissiblewith relations

(7)

3M1−2·g3M2=n 5M1−2²·g5M3=n 2·3·g₅M₃−5·g₃M₂=n

Note, however, that the relation coeﬃcients for M3 do not have matching exponent patterns.Wecanremedythis byrestrictingandreducingmodulo3.

Subcase 2a:Suppose3n,sog3= 1.Take

N1=M1(3m+n) = 12m+ 5n N₂=M₂(3m+n) = 18m+ 7n N3=M3(3m+n)/3 = 5

g5

m+ 2· n g5

Now wegetrelations

3N1−2N2=n 5N₁−2²·3·g₅N₃=n 2·3²·g5N3−5N2=n

Alltheseformsarereducedandthetripleisstill2-admissiblesinceN1(1)N2(1)N3(1)≡ 1³≡0 (mod 2).Infact,thetripleis3-admissible toosince

N1(0)N2(0)N3(0)≡(−n)(n)(−n/g5)≡0 (mod 3).

Heretherelationcoefficientsallhaveexponentpatternscontainedin{2,1,1},soadjoin- ing primes again gives us thestatement of the theorem. In fact, if we also have 5 n here, thentherelation coefficientsallhaveexponentpatterns containedin{2,1}sowe getinfinitelymanypositiveintegersxsuchthatxandx+nbothhaveexponentpattern {2,1,1,1}.

Subcase 2b: Suppose now 3| n,so 5n by ourassumption that 15 n. Westill must factorouta3 fromM3,butdoingso willforceus toalsofactorouta3 fromM1which thentellsustoalsofactorouta5 fromM1tomakeitspairofrelationcoeﬃcientsinthe triplehavematchingexponentpatterns.Thuswewillrestrictmodulo15:writen= 3n3

and take

J1=M1(15m−4n)/15 = 4m−n J2=M2(15m−4n)/(g9/3) = 10· 9

g9

m−23· n g9

J3=M3(15m−4n)/3 = 25m−19n3

(8)

where, as indicated above, g9 = gcd(9,n) which is either 3 or 9 in this case. Here we haverelations

3²·5J1−2·g9J2=n 3·5²J1−2²·3J3=n 2·3²J₃−5·g₉J₂=n

Alltheformsarereduced(since5n)andthetripleis2-admissiblesinceJ1(0)J2(0)J3(0)

≡1³≡0 (mod 2).

Nowwecheckthatthistripleis3-admissible.

•If3n3,theng9= 3,so

J1(−n3)J2(−n3)J3(−n3)≡(−n3)(n3)²≡0 (mod 3).

•If3|n₃,theng₉= 9 soJ₁J₃≡m² (mod 3).Choosem₀∈ {1,−1}suchthatJ₂(m₀)≡ 0 (mod 3).Then

J1(m0)J2(m0)J3(m0)≡(m0)²J2(m0)≡0 (mod 3).

Here therelation coeﬃcients allhaveexponent patterns containedin{2,1,1}(or even in{2,1}inthe casethat9|n), so adjoining primes againgives us thestatementof the theorem.

Remark6.Ifweassumethetwinprimeconjecture,thenforanypositiveintegern,there areprimespandp+ 2 suchthatneitherdivide15n.Inthiscase,wecanusethefollowing triple: L1 = 2m+n, L2 =pm+n(p−1)/2,L3 = (p+ 2)m+n(p+ 1)/2.Buildingoﬀ this triple will show—asin Subcase 2a above—that there are inﬁnitely many positive integers xsuch that xand x+n both haveexponent pattern {2,1,1,1}. We will not includethedetailsheresincewegiveanunconditionalproofofaresultfortheremaining casenotcoveredbyTheorem5.

4. Shiftswhichareoddanddivisible by15

Theorem 7. Let n be a positive integer with 2 n and 15|n. Then there are infinitely manypositiveintegersxsuchboth xandx+nhave exponentpattern {3,2,1,1,1,1,1}. Proof. Byconsideringtheadmissibletriplem,m+4,m+10,wefindthatforanyconstant C there areinfinitelymanypairsof E2 numberseachhavingprimefactorsbigger than C and which areadistance of either4,6, or 10 apart.Inparticular, there are oddE₂ numbersq₁,q₂suchthatgcd(q_i,n)= 1 fori= 1,2 andq₂=q₁+ 2j wherej ∈ {2,3,5}. Thuswemaywriteq1=p1,1p1,2 andq2=p2,1p2,2where p1,1,p1,2,p2,1,andp2,2 areall

(9)

distinct primes,noneofwhichdivide2n.There areintegersa,bwith aevenandb odd suchthat−aq2+bq1= 1.Writea= 2a2 anddeﬁnethetripleoflinearforms

L₁=q₁m+a₂n L2= 2q2m+bn L3= 4·j

gm+ (b−a)n g

where g = 1 if j = 2 and g = j otherwise. Now we check that this triple is admissible. Weonlyneedto checkfor2-admissibleand3-admissiblesinceeachform isreduced by construction. The triple is 2-admissible since L1·L2·L3 ≡ L1·1·1 (mod 2). To check thetriple is 3-admissible, choosem0 ∈ {1,−1} withL3(m0)≡0 (mod 3). Then L₁(m₀)L₂(m₀)L₃(m₀)≡(q₁m₀)(−q₂m₀)L₃(m₀)≡0 (mod 3). Moreover,thetriplesat- isﬁestherelations

q₁L₂−2q₂L₁=n

gq₁L₃−2²jL₁=n (7.1)

gq2L3−2jL2=n

However, the pairs of relation coeﬃcients for L₁, L₂ do not have matching exponent patterns inthetriple, sowewill needtoadjoin primesusing Theorem2. Wewillbreak up theproofinto casesdepending onthevalueof j,butinboth casesweneed tonote thatthepairwisedeterminantsarerelativelyprimetotheintegerswewantto adjoin:

det(L1, L2) =q1bn−2a2nq2=n det(L1, L3) =q1(b−a)n

g −4a2n· j g = n

g det(L2, L3) = 2q2(b−a)n

g −4bn· j

g = 2·n g Case 1:Supposej= 2,sog= 1.

We apply Theorem 2directlywith r₁ =p²_2,1p_2,2, r₂ =p_1,1, andr₃ = 1,so we get a newadmissibletripleofforms K_i whichsatisﬁesthefollowing relations:

|p²_1,1p_1,2K₂−2p³_2,1p²_2,2K₁|=n

|q₁K₃−2³p²_2,1p_2,2K₁|=n

|q2K3−2²p1,1K2|=n.

Here the relation coefficients of K₁ both have exponent pattern {3,2,1}, the relation coefficients of K₂ both have exponent pattern {2,1}, and the relation coefficients of K3 both have exponentpattern {1,1}. Thus by another applicationof Theorem2 via

(10)

Remark 4 we can arrange an admissible triple with common relation value n and all relation coeﬃcients having exponent pattern {3,2,1,1,1} (or even {3,2,1,1} in this case).

Case 2: Suppose j = 2, so g = j. We apply Theorem 2 directly with r1 = p2,1, and r₂=r₃= 1,so weget anew admissibletripleofforms K_i whichsatisﬁesthefollowing relations:

|q1K2−2p²_2,1p2,2K1|=n

|jq1K3−2²jp2,1K1|=n

|jq2K3−2jK2|=n.

Here the relation coefficients of K₁ both haveexponent pattern {2,1,1}, the relation coefficientsofK₂both haveexponentpattern{1,1},and therelation coefficientsof K₃ bothhaveexponentpattern{1,1,1}.ThusbyappealtoTheorem2viaRemark4wecan arrange anadmissible triplewith common relation valuen and allrelation coefficients havingexponentpattern{3,2,1,1,1}(or even{2,1,1,1}inthiscase).

Therefore, in either case, there are inﬁnitely many pairs of positive integers both havingexponentpattern{3,2,1,1,1,1,1}whichareadistance ofnapart.

References

[EM52]PaulErdős,LeonMirsky,Thedistributionofvaluesofthedivisorfunctiond(n),Proc.Lond.

Math.Soc.2 (3)(1952)257–271.

[GGPY09]D.A.Goldston,S.W.Graham,J.Pintz,C.Y.Yıldırım,Smallgapsbetweenproductsoftwo primes,Proc.Lond.Math.Soc.98 (3)(2009)741–774.

[GGPY11]D.A. Goldston,S.W.Graham,J.Pintz,C.Y.Yıldırım,Smallgapsbetweenalmostprimes, theparityproblem,andsomeconjecturesofErdősonconsecutiveintegers,Int.Math.Res.

Not.2011 (7)(2011)1439–1450.

[HB84]D.R. Heath-Brown, The divisor function at consecutive integers, Mathematika 31(1984) 141–149.

[Tho08]Frank Thorne, Boundedgaps betweenproductsof primeswithapplications toideal class groupsandellipticcurves,Int.Math.Res.Not.2008 (5)(2008)156.