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Chapter 2 The unending sequence of primes（第1页）

Chapter2Theunendingsequenes

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Howprimesfitintothenumberjigsaw

Howwebesurethattheprimesdonotbeerarerandrarerauallypeteroutalthtthinkthatsihereareinfinitelymanyumbersandeabebrokendorodues（somethingexplainedmorecarefullyiheremusttheelymahejob.Althoughthisistrue，itdoesnotfollowfromthepreviousobservations，forifwebeginwithafiiohereishenumberofdifferentnumbersroducejustusingthosegiveors.Ihereareinfinitelymapowersofanysingleprime:forexample，thepowersoftheprime2are2，4，8，16，32，64，···.Itisceivablethereforethatthereareonlyfinitelymanyprimesandeverynumberisaproductofpowersofthoseprimes.Whatismore，wehavenoroduendingseriesofdifferehewayweple，produumberofsquares，ormultiplesofaspecifiumber.Wheoprimes，westillhavetogoouthuntingforthem，sohowwebesuretheydo?

Wewillallbesurebytheendofthischapter，butfirstIwilldrawyourattentiontooern’amongtheprimesw.Everyprimefrom2and3，liesoheotherofamultipleof6.Inotherrimeafterthesefirsttwohastheform6n±1forsomenumber6nisshortfor6×nandthedoublesymbol±meansplusorminus.）Thereasonforthisisreadilyexplained.Everynumberbewrittelyohesixforms6n，6n±1，6n±2，or6n+3asnonumberismorethanthreeplacesawayfromsomemultipleofsix.Forexample，17=（6×3）-1，28=（6×5）-2，57=（6×9）+3;ihesixgivenformsappearineaningthatifyouwritedownanysixseumbers，eachoftheformsearexace，afterwhichtheyearagainandagain，inthesameorder.Itisevidentthatheforms6nand6n±2areeven，whileaheform6n+3isdivisibleby3.Therefore，withtheobviousexsof2and3，oheform6hecasewherebothofthenumbers6n±1areprimedsexactlytothetwinprimes:forexample（6×18）±1givesthepair107，109mehefirstchapter.Youmightbetemptedtojecturethatatleastowonumbers6n±1isalrime–thisislytrueforthelistofprimesupto100butthefirstfailureisnotfaraway:（6×20）-1=119=7×19，while（6×20）+1=121=112，sohernumberisprimewheaken=20.

Andtheprincipalreasonwhyprimesareimportantisthateverynumberberoduehateiiallyoofindthisspecialfactorizatioivennumberinsomewayaianypositefactorsthatappearuntilthisore.Forexample，wecouldsaythat120=2×60anduebybreakiefactorof6ive:

120=2×60=2×（2×30）=2×2×（2×15）=2×2×2×3×5.

&theprimefactorizationof120is23×3×5.Wecould，however，havecametothisbyae.Forirearrangingtheprimefaleasttogreateststillyieldsthesameresultasbefore:120=23×3×5.

120=12×10=（3×4）×（2×5）=（3×（2×2））×（2×5）

&didinthatexample，andthisbehaviourmaybemoreorlessfamiliartoyou，buthowyoubesurethatthisappliestoeveryisoughthatanynumberbebrokendoroduesbut，sihereisingehaagthistask，howwebesurethattheprocesswillalwaysdeliverthesamefihisisanimportaion，soIwilltakeafewmomentstogiveahereasoningthatallowsustobeabsolutelysureaboutthis.Itisaotherspecialpropertyofprimeweshallcalltheeuproperty:ifaprimenumberisafactorofaproduorehenitisafaeofthehatproduple，7isafactorof8×35=280（astheproduct280=7×40）a7isafactorof35.Thispropertycharacterizesprimesasnobergiveyouthesameguarantee:forexample，weseethat6isafactorof8×15=120（as120=6×20）yet6isnotafactorofeither8or15.

&hatprimesalwayshavetheabovepropertybeprovedusingabasedonwhatisknownastheEuAlgorithm，whichwillbeexplainedinChapter4.Ifwetakethisontrustforthetimebeing，itisnottoodifficulttoexplainwhynonumbercouldhavetwodiffereorizations，forsupposethereweresuumber.Therethenwouldbeasmallestobehavedinthisway:letusdebynandsonhastwoprimefactorizationswhitheprimefactorsarewritteninasgorder，areical.Weshallshowthatthisleadstotradidsomustbefalse.

Ifthesetwofactorizationsofnhadaprimepinon，webothandobtaintwodiffereorizationsofthesmallernumberhesmallestwodistiorizations，thisisnotpossibleasofprimesinvolvediorizationofnmusthavenoprimeinon.Norimepthatooorizationsofhisprimepisafa，itfolloisafactorofthesedfactorization，andso，bytheeuproperty，pisafaeoftheprimes，qsay，ofthesedfactorization.Butsindparebothprime，thisisonlypossibleifq=p，apossibilitythatwehavealreadydistedasthetwofactorizationsofnhavenoe.Andsowearriveatafinaltradi，showingthatitisimpossibleforsuumberhereforewecludethattheprimefactorizationofeverynumberisunique.

Itiswthattheuniquenessofprimefactorizationwouldnotholdifweihenumber1amongtheprimes，asweaoafactorizatioretainsthesamevalue.Thisshowsthat1isfuallydifferentiheprimes，andshttoframethedefinitionofprimenumberinawaythatexcludes1fromthe.

Eufinityofprimes

&urionastohothattheprimesgoohatthereisnoastthem.Ifsomeo101isthelargestprime，yourefutehimatoncebyshowingthat103hasnofactors（exceptfor1and103）andso1erprime.Yhtthehemadeaslipandthatheshouldhavesaidthatitwas103thatisthelargestprimeofall.Youshowhimupagaiingthat107isalsoprime，butyhtstillpersistinhiserrorbyadjustinghispositioprimenumberonview.Heretreatalittlefurtherandadmitthathedoesnotkyestprimebutioclaimthatheisthatthereisone.

&waytosettlethisquestiooshowthat，givenanyceivablefiionofprimes，roduewprime.Forexample，ifsomeoherewasalargestoddheresomewhere，youcouldrefutehimbysayingthatifnisodd，thenn+2isalargeroddhereotbealargestoddhisapproach，however，isheprimes–giveofprimes，wehavenowayofusiiontomanufactureaprimethatisdemerthanallofthem.Perhapsthereisabiggestprimeafterall?Hothatourstubborright?

EuclidofAlexandria（c.300BC），theGreekmathematidfatherofallthingseu，didhivenalistp1，p2，···，pkwhereeachofthepidenotesadifferentprime，heotfindawayanewprime，soherevertedtumentthatisoepmoresubtle.Heshowedthattheremustbeoneormorehiainrangeofhisargumeallowustolocateexactlywheretofihatrange）.

Itgoeslikethis.Letp1，p2，···，pkbethelistofthefirstkprimessay，aheisoheproductofalltheseprimes，sothatn=p1p2···pk+1.Eithernisaprime，orisdivisiblebyaprimesmallerthanitself，whiotbeanyofp1，p2，···，pk，asifpisaheseprimes，thendividingnbypwillleavearemainderof1.Itfollorimedivisorofhatisgreaterthatalltheprimesp1，p2，···，pkaself.Inparticular，itfollowsfromthisthatthereofiofprimesthatseveryprimenumber，andsothesequenesuesonforeverandwilled.Euclid’seternalproofoftheinfinityofprimesisamoadmiredinallofmathematics.

AlthoughEuclid’sargumeellexactlywheretofiprimeheoverallfrequencyoftheprimesisnowquitewelluood.Forexample，ifwetakeanytwonumbers，aandbsay，withnoonfadsiderthesequencea，a+b，a+2b，a+3b，···，itwasshownbytheGermai（1805–59）thatinfinitelymanymembersofsuchasequenceareprime.（Ofcourse，thereisnohopeifaandbdohaveaonfactor，dsay，asthehelistisalsoamultipleofd，andsoisnotprime.）Whena=1ahesequenbers，byEuclid’sproof，sinfinitelymanyprimenumbers.IbeshhfairlysimpleadaptationsofEuclid’sargumentthatotherspecialcasessuchasthesequenbersoftheforms3+4n，5+6n，and5+8n（asnrunsthroughthesuccessivevalues1，2，3，···），eafinitelymahegeofDirichletis，however，verydifficulttoprove.

Anothersimplystatedresultisthatthereisalwaysatleastoneprimehananygivelessthan2n（forn≥2）.（Asanaidtomemory，iysignssuchasthisone，whidsfreaterthao，alointtothesmallerquantity.）Thisfact，historiowrand’sPostulate，beprovedusiarymathematics，althoughtheproofisitselfquitetricky.Weverifythepostulatefornupto4000bymakihefollowinglistofprimes.Firstobservethateaumberiertheinitialprime2issmallerthaspredecessor:

2，3，5，7，13，23，43，83，163，317，631，1259，2503，4001.

Foreatherao4000，takethelargestprimepihatisheheherangen＜q＜2nandthistheBertrand’sPostulateholdsforallnupto4000.Forexample，forn=100，p=83，andthenq=163＜2×100.Asubtleargumentinvolvingthesizeoftheso-tralbis（introduChapter5）thenshoostulateisalserthan4000.

However，wedooofarbefsimilar-soundiasyetremainunsolved.Forexample，nooneknowsifthereisalrimebetweenanytwosecutivesquares.Anotherobservationisthatthereseemstobeenoughprimestoeeveryeveerthawoofthem（Goldbajecture）.Thishasbeelyverifiedfornupto1018.Wemightthenhopefthelirand’sPostulate，thatbeyoaiegerroduparisoisknownaboutthedistributiooeherewillalwaysbeatleastoiootheequationp+q=2nforanyevehisstilleludesus，althoughthereareweakerresultsalongtheselines–forexample，ithasbeenkhateverysuffitlylargeoddhesumofatmostthreeprimehateveryevehesumofhan300，000primes.ProofofthefullGoldbajecturestillseemsalong>

Asimpleresultthathassomethingoftheflavourumeoaboveisthatthereisahan4billionthatbewrittenasthesumoffourdiffereendistinctways.Itisknownthat1729=13+123=93+103isthesmallestisthesumoftwotwodifferentedonotnecessarilyhavetoidentifythenumberoknowthatitmustexist.Sometimesitispossibletokhattherearesolutionstoaproblem，withoutadinganyofthosesolutioly.

Inthiscase，webeginbynotingthatifwetakefourdifferentnumbershaegermandformthesumoftheircubes，theresultislessthan4m3.However，ifm=1000，thearycalshowsthatthenumberofsumsoffourdifferentcubesismorethahenumber4m3，fromwhichitfollowsthatsomenumbern≤4m3=4，000，000，000mustbethesumoffouratleastteways.Thedetailsiionsusingbis（introduChapter5）aespeciallydifficult.

TheglobalpiedistributioheobservationoftheleadiuryGermaidphysicistKarlFriedrichGauss（1777–1855）thatp（n），thenumberofprimesuptothenumbern，isapproximatelygivenbynlognandthattheapproximationbeoreaccurateasnincreases.Forexample，ifwetakentobeonemillioioofnlog，uptothatstage，aboutonenumberinevery12.7shauss’sobservation，whidetailsayssomethingmoreprecise，roveduntil1896.Thelogarithmfuohereistheso-aturallogarithm，whiotbasedonpowersof10，butratheronpoeumbere，roximatelyequalto2.718.Weshallhearmoreofthisveryfamouser6.

&celebrateduioheoryistheRiemannHypothesis，whilybeexplaiermsofbers，whichwehaveyettointroduentioheobjectofthequestionbereformulatedusingtheuniquenessofprimefactorizationtoiaifeaturingalltheprimes.ThisleadstoaionwhichsaysthattheHypothesisimpliesthattheoveralldistributionoftheprimesisveryregularinthat，inthelongrun，primalityarentlyoly.Ofcourse，whetherornotapartiumberisaprimeisnotarawhatismeantisthatprimality，iheverylarge，takesoleofrahnoadditioruerge.Maheoristhasaheartfeltwishtoseethis150-year-oldjecturesettledintheirowime.

&heyrepresentsonaturalasequeisalmostirresistibletosearchftheprimes.Thereare，henuinelyusefulformulasforprimeistosay，thereisnokallowsyoutogeeallprimenumbersoreventocalculateasequesistsentirelyofdiffereherearesomeformulasbuttheyareoflittlepraeofthemevenrequirekheprimesequeocalculatetheirvaluesothattheyareessentiallyacheat.Expressionssu+41areknoolynomials，andthisoneisaparticularlyrichsoures.Forexample，puttingn=1，7，and20iheprimes43，107，aively.Iofthisexpressionisprimeforallvaluesofnfromn=0ton=39.Atthesametime，however，itisclearthatthispolynomialwillletusdoeputn=41，astheresultwillhave41asafadifailsforn=40as

402+40+41=40（40+1）+41=40×41+41=（40+1）41=412.

&isquitestraightforwardtoshoolynomialofthiskindyieldaformulaforprimes，evenifershigherthaheexpression.Itispossibletodevisetestsforprimalityofabestatedinafeever，tobeofusetheywouldobequicker，atleastihaverifiproceduredesChapter1.AfamoesbythenameofWilsoatemeheuseofnumberscalledfactorials，illmeetagainihenumberorial’，isjusttheprodubersupton.Forexample，5!=5×4×3×2=120.Wilsohenaverysuctstatement:anumberpisprimeifandonlyifpisafactorof1+（p-1）!.

Theproofofthisresultisnotverydiffidindeediisnearlyobvious:ifpwereposite，sothatp=absay，thehaahanp，theyeachoccurasfactorsof（p-1）!andsopisadivisorofthisfactorialaswell.Itfolloedivide1+（p-1）!byp，wewillobtainaremaihecasewherea=brequiresalittlemht.）ThisisveryremiofEuclid’sprooffortheinfinityofprimes.Itfolloisafactorof1+（p-1）!thenpmustbeprime.Theverseisalittlehardertoprove:ifpisprimethenpisafactorof1+（p-1）!.This，however，isthesurprisiioheorem，althoughthereadereasilyverifyparticularple，theprime5isiorof

1+4!=1+24=25.

Asafiion，loitfactorials，whichbydesignhavemanyfaordertoprovethatibers，thatistosayoheforma，a+b，a+2b，a+3b，···sistonlyofprimesasitispossibletoshobetweensuccessiveprimesbearbitrarilylargewhiletheoweeivemembersoftheprevioussequeoseethis，siderthesequensetegers:

（n+1）!+2，（n+1）!+3，（n+1）!+4，···，（n+1）!+n+1.

&hesenumbersisposite，asthefirstisdivisibleby2（aseachofthetermshas2asafactor），thesedisdivisibleby3，theby4，andsoonupuntilthefihelist，whi+1asafactor.Wethereforehave，fivenn，asequenseumbers，noneofrime.

Insteadoffoberswiththefewestpossiblefactors（theprimes），weshallierturhmanyfactors，althoughweshalldiscoverthatheretootherearesurprisinglinkstosomeveryspecialprimenumbers.