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

Chapter2Theunendingsequenes

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。