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Chapter2Theunendingsequenes
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Howprimesfitintothenumberjigsaw
Howwebesurethattheprimesdonotbeerarerandrarerauallypeteroutalthtthinkthatsihereareinfinitelymanyumbersandeabebrokendorodues（somethingexplainedmorecarefullyiheremusttheelymahejob.Althoughthisistrue，itdoesnotfollowfromthepreviousobservatiinwithafiiohereishenumberofdifferentnumbersroducejustusingthosegiveors.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.Forinstance
120=12×10=（3×4）×（2×5）=（3×（2×2））×（2×5）
butrearrangingtheprimefaleasttogreateststillyieldsthesameresultasbefore:120=23×3×5.
&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.
Itiswthattheuniquenessofprimefactorizationwouldnotholdifweihenumber1amongtheprimes，asweaoafactorizatioretainsthesamevalue.Thisshowsthat1isfuallydifferentiheprimes，andshttoframethedefinitionofprimenumberinawaythatexcludes1fromthe.
Euclid'sinfinityofprimes
&urionastohothattheprimesgoohatthereisnoasttheIfsomeo101isthelargestprime，yourefutehimatoncebyshowingthat103hasnofactors（exceptfor1and103）andso1erprime.Yhtthehemadeaslipandthatheshouldhavesaidthatitwas103thatisthelargestprimeofall.Youshowhimupagaiingthat107isalsoprime，butyhtstillpersistinhiserrorbyadjustinghispositioprimenumberonview.Heretreatalittlefurtherandadmitthathedoesnotkyestprimebutioclaimthatheisthatthereisone.
&waytosettlethisquestiooshowthat，givenanyceivablefiionofprimes，roduewprime.Forexample，ifsomeoherewasalargestoddheresomewhere，youcouldrefutehimbysayingthatifnisodd，thenn+2isalargeroddhereotbealargestoddhisapproach，however，isheprimes–giveofprimes，wehavenowayofusiiontomanufactureaprimethatisdemerthanallofthePerhapsthereisabiggestprimeafterall?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）thatinnitelymanymembersofsuchasequenceareprime.（Ofcourse，thereisnohopeifaandbdohaveaonfactor，dsay，asthehelistisalsoamultipleofd，andsoisnotprime.）Whena=1ahesequenbers，byEuclid’sproof，sinnitelymanyprimenumbers.IbeshhfairlysimpleadaptationsofEuclid’sargumentthatotherspecialcasessuchasthesequenbersoftheforms3+4n，5+6n，and5+8n（asnrunsthroughthesuccessivevalues1，2，3，...），eanitelymahegeofDirichletis，however，verydifculttoprove.
2，3，57，13，23，43，83，163，317，631，1259，2503，4001.
Foreatherao4000，takethelargestprimepihatisheheherangen＜q＜2nandthistheBertrand’sPostulateholdsforallnupto4000.Forexample，forn=100，p=83，andthenq=163＜2×100.Asubtleargumentinvolvingthesizeoftheso-tralbis（introduChapter5）thenshoostulateisalserthan4000.However，wedoosimilar-soundiasyetremainunsolved.Forexample，nooneknowsifthereisalrimebetweenanytwosecutivesquares.Anotherobservationisthatthereseemstobeenoughprimestoeeveryeveerthawoofthem（Goldbajecture）.Thishasbeelyverifiedfornupto1018.Wemightthenhopefthelirand’sPostulate，thatbeyoaiegerroduparisoisknownaboutthedistributiooeherewillalwaysbeatleastoiootheequationp+q=2nforanyevehisstilleludesus，althoughthereareweakerresultsalongtheselines–forexample，ithasbeenkhateverysuffitlylargeoddhesumof atmostthreeprimehateveryevehesumof han300，000primes.ProofofthefullGoldbajecturestillseemsalongwayoff.
Asimpleresultthathassomethingoftheflavourumeoaboveisthatthereisahan4billionthatbewrittenasthesumof fourdiffereendistinctways.Itisknownthat1729=13+123=93+103isthesmallestisthesumof twotwodifferentedonotnecessarilyhavetoidentifythenumberoknowthatitmustexist.Sometimesitispossibletokhattherearesolutionstoaproblem，withoutactuallyfindinganyof thosesolutioly.
Inthiscase，webeginbynotingthatifwetakefourdifferentnumbershaegermandformthesumof theircubes，theresultislessthan4m3.However，ifm=1000，thearycalshowsthatthenumberof sumsof fourdifferentcubesismorethahenumber4m3，fromwhichitfollowsthatsomenumbern≤4m3=4，000，000，000mustbethesumof fouratleastteways.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）!theheverseisalittlehardertoprove:ifpisprimethenpisafactorof1+（p-1）!.This，however，isthesurprisiioheorem，althoughthereadereasilyverifyparticularple，theprime5isiorof1+4!=1+24=25.
Asafiion，loitfactorials，whichbydesignhavemanyfaordertoprovethatibers，thatistosayoheforma，a+b，a+2b，a+3b，…sistonlyofprimesasitispossibletoshobetweensuccessiveprimesbearbitrarilylargewhiletheoweeivemembersoftheprevioussequenceisfixedatb.Toseethis，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.
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