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On a conjecture involving Fermat's Little Theorem

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On a conjecture involving Fermat's Little Theorem
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Number Theory
Prime Numbers
Divisibility
Congruences
Sums of powers of consecutive integers
Dissertations, Academic -- Mathematics -- Masters -- USF   ( lcsh )
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Thesis (M.A.)--University of South Florida, 2008.
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On a conjecture involving Fermat's Little Theorem
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Mode of access: World Wide Web.
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Advisor: Stephen Suen, Ph.D.
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Number Theory
Prime Numbers
Divisibility
Congruences
Sums of powers of consecutive integers
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Dissertations, Academic
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OnaConjectureInvolvingFermat'sLittleTheorembyJohnClarkAthesissubmittedinpartialfulllmentoftherequirementsforthedegreeofMasterofArtsDepartmentofMathematicsandStatisticsCollegeofArtsandSciencesUniversityofSouthFloridaMajorProfessor:StephenSuen,Ph.D.MohamedElhamdadi,Ph.D.ArthurDanielyan,Ph.D.DateofApproval:May13,2008Keywords:NumberTheory,PrimeNumbers,Divisibility,Congruences,SumsofPowersofConsecutiveIntegerscCopyright2008,JohnClark

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TomywonderfulanceeMarcia,yourloveandsupportkeptmegoing.

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ACKNOWLEDGEMENTSIwouldliketoextendaspecialthankyoutoDr.StephenSuen.Iamtrulygratefulforyourencouragement,guidance,andhardwork.

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TABLEOFCONTENTSLISTOFFIGURESiiABSTRACTiii1.INTRODUCTION12.BACKGROUND93.RESULTS193.1Overview193.2ProofofTheorem3.1203.3ProofofTheorem3.2253.4ProofofTheorem3.3273.5ProofofTheorem3.4313.6ProofofTheorem1.2333.7ProofofTheorem1.3343.8ProofofTheorem1.4354.COMPUTATIONS36BIBLIOGRAPHY48i

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LISTOFFIGURESFig.4.1:MapleCodePart142Fig.4.2:MapleCodePart243Fig.4.3:MapleCodePart344Fig.4.4:OverviewFlowchart45Fig.4.5:Top-DownFlowchart46Fig.4.6:Bottom-UpFlowchart47ii

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ONACONJECTUREINVOLVINGFERMAT'SLITTLETHEOREMJOHNCLARKABSTRACTUsingFermat'sLittleTheorem,itcanbeshownthatPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 23.359 0 Td[(1modmifmisprime.Ithasbeenconjecturedthattheconverseistrueaswell.Namely,thatPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 22.54 0 Td[(1modmonlyifmisprime.Weshallpresentsomenecessaryandsucientconditionsfortheconjecturetohold,andwewilldemonstratethatnocoun-terexampleexistsform1012.iii

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1.INTRODUCTIONMathematicsisthequeenofthesciencesandnumbertheoryisthequeenofmathe-matics."SosaysCarlFriedrichGauss,anineteenthcenturynumbertheorist.Numbertheory,thesubjectofmuchofhisstudy,isthebranchofpuremathematicsconcernedwiththepropertiesandrelationshipsofnumbers,specicallyintegers.Wewishtoconsideraproblemdealingwithtwoconceptsfromelementarynum-bertheory:primalityandcongruence.Fermat'sLittleTheoremstatesthatifpisaprimeintegerandaisapositiveintegerwithp-a,thenap)]TJ/F17 7.97 Tf 6.586 0 Td[(11modp.AknownconsequenceofFermat'sLittleTheoremisthatifpisprime,thenPpi=1ip)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 23.094 0 Td[(1modp.InKennethH.Rosen'sbookElementaryNumberTheoryandItsApplica-tionsthEdition[8],readersaretaskedtoprovethisasanexerciseonpage221.There,itisconjecturedthattheconverseoftheproblemistrueaswell.Namely,thatPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modmonlyifmisprime.Conjectureslikethesearecommoninmathematics.Forinstance,Wilson'sTheo-remstatesthatifpisaprimeinteger,thenp)]TJ/F15 11.955 Tf 12.427 0 Td[(1!)]TJ/F15 11.955 Tf 23.096 0 Td[(1modp.Itturnedoutthattheconversewastrueaswell.Namely,thatifn>1isapositiveinteger,thenn)]TJ/F15 11.955 Tf 11.968 0 Td[(1!)]TJ/F15 11.955 Tf 21.95 0 Td[(1modnonlyifnisprime.Ourgoalisverysimilar.Weknowthatifpisprime,thenPpi=1ip)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 22.342 0 Td[(1modp,andwewouldliketoprovethatifmisapositiveinteger,thenPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 21.917 0 Td[(1modmonlyifmisprime.Whileunabletoprovetheconjecture,wecanprovethatmmustsatisfysomestrongconditionsinorderforthecongruencetohold.Westatethesenecessaryandsucientconditionsasatheorem.Theorem1.1.Letm3beaninteger.Then,Pmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 23.458 0 Td[(1modmifandonlyifmisaproductofdistinctoddprimessuchthatmpmodp2p)]TJ/F15 11.955 Tf 12.487 0 Td[(1foreachprimedivisorpofm.ThenextthreethoeremsareconsequencesofTheorem1.1.Theorem1.2.Letm3beanintegerandletp1,p2beanytwoprimedivisorsofm.IfPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modm,thenp261modp1.Theorem1.3.Letm3beaninteger.IfPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 24.586 0 Td[(1modm,thenmcannotbeaproductoftwoprimes.Theorem1.4.Letm3beaninteger.IfPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 24.586 0 Td[(1modm,thenmcannotbeaproductofthreeprimes.1

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Wedelaytheseproofssothatwemaypresentsomebackgrounddenitionsandtheo-remsinChapter2.InChapter3wewillproveTheorem1.1usingaseriesofsmallertheorems,andthenwewillproveTheorems1.2,1.3,and1.4.Letusnowconsidersomeapplicationsofthesetheorems.Considerapositiveintegerm3satisfyingPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 23.607 0 Td[(1modm.If3jm,thenbyTheorem1.2nootherprimedivisorpofmmaybecongruentto1modulo3.SincemmustbeaproductofdistinctoddprimesbyTheorem1.1,p60mod3,meaningp2)]TJ/F15 11.955 Tf 22.609 0 Td[(1mod3.Thismeansthatallotherprimedivisorsofmmustbeoftheform3k)]TJ/F15 11.955 Tf 11.431 0 Td[(1,wherekissomepositiveinteger.Sincemmustbeodd,weseethatallprimedivisorsofmotherthan3mustactuallybeoftheform6k)]TJ/F15 11.955 Tf 11.257 0 Td[(1.Hence,if3jm,then7-m,13-m,19-m,31-m,etc.Similarly,if5jm,thenbyTheorem1.2nootherprimedivisorofmmaybeoftheform5k+1,wherekissomepositiveinteger.Sincemmustbeodd,weseethatnoprimedivisorofmmayactuallybeoftheform10k+1.Thus,if5jm,then11-m,31-m,41-m,etc.NotethatTheorems1.3and1.4implythatanycompositemwouldhavetobetheproductofatleastfourdistinctoddprimes,sothesmallestpossiblequadrupleofprimedivisorsofmis3;5;17;23.Infact,morecanbesaid.Considerthefollowingexample.Example1.1.Ifm3isanintegersatisfyingPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 24.024 0 Td[(1modm,thenm6=3517p,foranyprimep.Proof.Assumeforthepurposeofcontradictionthatsuchanmexists.Fromabove,m=3517pimpliesp23.ByTheorem1.1,mpmodp2p)]TJ/F15 11.955 Tf 12.515 0 Td[(1,sowemaywritem=p2p)]TJ/F15 11.955 Tf 12.226 0 Td[(1X+pforsomepositiveintegerX.Thatis,3517p=p2p)]TJ/F15 11.955 Tf 11.654 0 Td[(1X+p.Dividingthroughbypleaves3517=pp)]TJ/F15 11.955 Tf 11.655 0 Td[(1X+1.Thisimpliesthatpp)]TJ/F15 11.955 Tf 12.292 0 Td[(1X=3517)]TJ/F15 11.955 Tf 12.292 0 Td[(1=254.Butsincep23andX1,wemusthavepp)]TJ/F15 11.955 Tf 11.985 0 Td[(1X233)]TJ/F15 11.955 Tf 11.985 0 Td[(11=506.Thisisacontradiction,since2546506.Hence,nosuchmcanexist. Beforeconsideringmoreexamples,wewishtogeneralizethetechniqueweusedintheprecedingproof.Lemma1.5.Letm3andQ7beintegers,andletp1
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ItfollowsthatQp1=p2p2)]TJ/F15 11.955 Tf 11.955 0 Td[(1X+1Qp1p2p2)]TJ/F15 11.955 Tf 11.956 0 Td[(1+1sinceX1Qp1p1+2p1+2)]TJ/F15 11.955 Tf 11.955 0 Td[(1+1sincep2p1+2Qp1p1+2p1+1+1Qp1p12+3p1+3p12+3p1)]TJ/F19 11.955 Tf 11.955 0 Td[(Qp1+30p12)]TJ/F15 11.955 Tf 11.955 0 Td[(Q)]TJ/F15 11.955 Tf 11.955 0 Td[(3p1+30:Nowbythequadraticformula,thisgivesQ)]TJ/F15 11.955 Tf 11.955 0 Td[(3)]TJ/F24 11.955 Tf 11.955 10.222 Td[(p Q)]TJ/F15 11.955 Tf 11.955 0 Td[(32)]TJ/F15 11.955 Tf 11.955 0 Td[(12 2p1Q)]TJ/F15 11.955 Tf 11.955 0 Td[(3+p Q)]TJ/F15 11.955 Tf 11.955 0 Td[(32)]TJ/F15 11.955 Tf 11.956 0 Td[(12 2:Hence,p1Q)]TJ/F15 11.955 Tf 11.955 0 Td[(3+p Q)]TJ/F15 11.955 Tf 11.956 0 Td[(32)]TJ/F15 11.955 Tf 11.955 0 Td[(12 2p1
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WewilluseLemma1.5inthenexttwoexamples.Example1.2.Ifm3isanintegersatisfyingPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 24.024 0 Td[(1modm,thenm6=35p1p2,foranytwoprimesp1andp2.Proof.Assumeforthepurposeofcontradictionthatsuchanmexists.ByTheorem1.1,mmustbeaproductofdistinctoddprimes,sowemayassumewithoutlossofgeneralitythat511.Thiscontradictionshowsthatnosuchmcanexist. Example1.3.Ifm3isanintegersatisfyingPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 24.024 0 Td[(1modm,thenm6=57p1p2,foranytwoprimesp1andp2.Proof.Assumeforthepurposeofcontradictionthatsuchanmexists.ByTheorem1.1,mmustbeaproductofdistinctoddprimes,andbyTheorem1.2,p1;p26=3since71mod3,sowemayassumewithoutlossofgeneralitythat731ifp16=13;17;19;23.Letusconsiderthesefoursituations.Ifp1=13,then5713=p2p2)]TJ/F15 11.955 Tf 11 0 Td[(1X+1forsomepositiveintegerX.Ifp2=17;19;or23,weseethattheequalityfailstohold,sop237,contradictingthefactthatp233.Hence,p16=13.Similarly,ifp1=17,then5717=p2p2)]TJ/F15 11.955 Tf 12.94 0 Td[(1X+1forsomepositiveintegerX.Ifp2=19or23,weseethattheequalityfailstohold,sop237,contradictingthefactthatp233.Hence,p16=17.Ifp1=19,then5719=p2p2)]TJ/F15 11.955 Tf 11.007 0 Td[(1X+1forsomepositiveintegerX.Ifp2=23,weagainseethattheequalityfailstohold,meaningp237.Thiscontradictionshowsthatp16=19.Finally,ifp1=23,thenp237,contradictingthefactthatp233.Hence,p16=23.Sincep16=13;17;19;23,weseethatp1>31,contradictingthefactthatp131.Thiscontradictionshowsthatnosuchmcanexist. Usingthissametechnique,withonlytwocasestocheck,wecanshowthatm6=311p1p2.Withsignicantlymorecasestocheck,wecanshowthatm6=711p1p2.Usingaslightlydierentstrategy,wecanshowthatm6=3517p1p2.Thisisournextexample.4

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Example1.4.Ifm3isanintegersatisfyingPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 24.024 0 Td[(1modm,thenm6=3517p1p2,foranytwoprimesp1andp2.Proof.Assumeforthepurposeofcontradictionthatsuchanmexists.ByTheo-rem1.1,mmustbeaproductofdistinctoddprimes,andbyTheorem1.2,p1;p26=7;11;13,sowemayassumewithoutlossofgeneralitythat17<>:m3mod32)]TJ/F15 11.955 Tf 11.955 0 Td[(1m5mod52)]TJ/F15 11.955 Tf 11.955 0 Td[(1m17mod1727)]TJ/F15 11.955 Tf 11.955 0 Td[(1:Thatis,8><>:m3mod18m5mod100m17mod4624:Wewishtosolvethissystemofcongruences.Thethirdlinetellsusthatm=4624q+17forsomepositiveintegerq.Substitutingthisintothesecondlinegives4624q+175mod100,whichreducesto24q+175mod100since462424mod100.Now,24q+175mod10024q)]TJ/F15 11.955 Tf 21.918 0 Td[(12mod100122q12)]TJ/F15 11.955 Tf 9.299 0 Td[(1mod1002q)]TJ/F15 11.955 Tf 21.918 0 Td[(1mod25byTheorem2.212q24mod25since)]TJ/F15 11.955 Tf 9.299 0 Td[(124mod25q12mod25byCorollary2.22q=25`+12;forsomenonnegativeinteger`.Sincem=4624q+17,weseem=4624`+12+17=115600`+55505.Substi-tutingthisintotherstlineofoursystemofcongruencesgives115600`+555053mod18,whichreducesto4`+113mod18.Now,5

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4`+113mod184`)]TJ/F15 11.955 Tf 21.917 0 Td[(8mod18`)]TJ/F15 11.955 Tf 21.917 0 Td[(2mod9byTheorem2.21`7mod9`=9k+7;forsomenonnegativeintegerk.Sincem=115600`+55505,weseem=115600k+7+55505=1040400k+864705=3517391+4080k.Moreover,m=3517p1p2,sothisimpliesthatp1p2=3391+4080k,forsomenonnegativeintegerk.ByLemma1.5,p1251andp2253,sop1p263503.Thismeansthatk14.Thus,inordertoderiveacontradiction,weneedonlycheckthefteencases,k=0;1;2;:::;14.First,ifk=0,wewouldhavep1p2=3391,aprimenumber.Thiscontradictionshowsthatk6=0.Ifk=1,wewouldhavep1p2=3391+4080=7471=31241.But311mod5,contradictingTheorem1.2.Hence,k6=1.Ifk=2,wewouldhavep1p2=3391+24080=11551,aprimenumber.Thiscontradictionshowsthatk6=2.Ifk=3,wewouldhavep1p2=3391+34080=15631=721129,aprod-uctoffourprimes.Thiscontradictionshowsthatk6=3.Ifk=4,wewouldhavep1p2=3391+44080=19711=23857,contradict-ingthefactthatp2253.Hence,k6=4.Ifk=5,wewouldhavep1p2=3391+54080=23791=37643,contradict-ingthefactthatp2253.Hence,k6=5.Ifk=6,wewouldhavep1p2=3391+64080=27871=47593,contradict-ingthefactthatp2253.Hence,k6=6.Ifk=7,wewouldhavep1p2=3391+74080=31951=89359,contradict-ingthefactthatp2253.Hence,k6=7.Ifk=8,wewouldhavep1p2=3391+84080=36031=137264,contradict-ingthefactthatp2253.Hence,k6=8.6

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Ifk=9,wewouldhavep1p2=3391+94080=40111,aprimenumber.Thiscontradictionshowsthatk6=9.Ifk=10,wewouldhavep1p2=3391+104080=44191=759107,aproductofthreeprimes.Thiscontradictionshowsthatk6=10.Ifk=11,wewouldhavep1p2=3391+114080=48271,aprimenumber.Thiscontradictionshowsthatk6=11.Ifk=12,wewouldhavep1p2=3391+124080=52351=134027,contra-dictingthefactthatp2253.Hence,k6=12.Ifk=13,wewouldhavep1p2=3391+134080=56431,aprimenumber.Thiscontradictionshowsthatk6=13.Finally,ifk=14,wewouldhavep1p2=3391+144080=60511=115501,contradictingthefactthatp2253.Hence,k6=14.Thus,nopositiveintegerkexistssothatm=3517391+4080k,meaningm6=3517p1p2. Ingeneral,ifmisapositiveintegersatisfyingPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 26.011 0 Td[(1modm,andp1p2pk)]TJ/F17 7.97 Tf 6.586 0 Td[(2aredistinctoddprimes,thenbycheckingnitelymanycaseswecanshowthatm6=p1p2pk)]TJ/F17 7.97 Tf 6.586 0 Td[(2pk)]TJ/F17 7.97 Tf 6.587 0 Td[(1pkforanytwoprimespk)]TJ/F17 7.97 Tf 6.586 0 Td[(1;pk.ThisfollowsdirectlyfromLemma1.5,whichplacesupperboundsonpk)]TJ/F17 7.97 Tf 6.586 0 Td[(1;pk.Wementionthatthedicultyinproducingmoreexampleslikethelastthreeisthatastheknownprimedivisorsofmincreaseinmagnitude,Theorem1.2becomeslessusefulatrulingoutotherpotentialprimedivisors,leavingmorecasesforustocheck.Sowhileweneedonlychecknitelymanycases,thismaystilltakequitesometime.Notethatifweallowmtohavemorethantwounknownprimedivisors,thedicultyincreasessignicantly.Luckily,computerscandothesetypesofcalculationsquitequickly,soinChapter4wewillcombineourresultswithsomecomputerprogrammingtodemonstratethatnocounterexampletotheconjectureexistsformlessthanorequaltoonetrillion.Inournalexample,wewishtoexploitthefactweusedearlierthatanyintegermustbecongruenttoeither)]TJ/F15 11.955 Tf 9.298 0 Td[(1;0;or1;modulo3.Thespecialness"ofthesmallmagnitudeof3willallowustopartiallyovercomethedicultyofconsideringcaseswhenmhasmorethantwounknownprimedivisors.Example1.5.Ifm3isanintegersatisfyingPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 24.024 0 Td[(1modm,thenm6=3p1p2p3pr,foranyoddnumberofprimesp1;p2;p3;:::pr.7

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Proof.Letm=3p1p2p3prbeanintegersatisfyingPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modm.Wewillshowthatrmustbeeven.ByTheorem1.1,misaproductofdistinctoddprimesandm3mod32)]TJ/F15 11.955 Tf 11.955 0 Td[(1.Thus,m3mod18im 31mod6;byTheorem2.21,since3jm.Nowletp>3beaprimedivisorofm.Recall,byTheorem1.2,p=6k)]TJ/F15 11.955 Tf 11.955 0 Td[(1forsomepositiveintegerk.So,m 3=k1)]TJ/F15 11.955 Tf 11.955 0 Td[(1k2)]TJ/F15 11.955 Tf 11.955 0 Td[(1kr)]TJ/F15 11.955 Tf 11.955 0 Td[(1;forsomepositiveintegersk1;k2;:::;kr.Nowobservethatk1)]TJ/F15 11.955 Tf 11.955 0 Td[(1k2)]TJ/F15 11.955 Tf 11.955 0 Td[(1kr)]TJ/F15 11.955 Tf 11.955 0 Td[(1)]TJ/F15 11.955 Tf 9.299 0 Td[(1rmod6:Thus,m 31mod6k1)]TJ/F15 11.955 Tf 11.955 0 Td[(1k2)]TJ/F15 11.955 Tf 11.955 0 Td[(1kr)]TJ/F15 11.955 Tf 11.955 0 Td[(11mod6)]TJ/F15 11.955 Tf 9.298 0 Td[(1r1mod6;whichimpliesthatrmustbeeven. Specically,thisshowsthatifmisanintegersatisfyingPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 22.665 0 Td[(1modm,thenm6=3p1p2p3,foranythreeprimesp1p2p3.WeconcludethischapterbymentioningthattheconjecturethatPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 23.446 0 Td[(1modmonlyifmisprimeissimilarinasensetotheconjecturethatapositiveintegerisperfectonlyifitisevenanintegerisperfectifitisequaltothesumofitsproperpositivedivisors.Whilemathematicianshavebeenunabletoprovethatnooddperfectnumbersexist,theyhaveprovenmanyconditionsthatwouldhavetobemetinorderforanoddintegertobeperfect,andtheyhaveshownthatnooddperfectnumberlessthan10300existsformoreinformationonthesearchforanoddperfectnumber,see[7].Inourcase,whilewecannotprovetheconjecturethatPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 23.142 0 Td[(1modmonlyifmisprime,wecanprovesomestrongconditionsonmthatmustbesatisedformtobecomposite,andwewillshowthatnocounterexampleexistsform1012.8

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2.BACKGROUNDInthischapter,wepresentthedenitionsandtheoremsthatserveasthefoundationforourwork.Restatedhereinourownwords,theseknownresultscanbefoundinsuchnumbertheorytextsasElementaryNumberTheorybyGarethJones[4],ElementaryNumberTheorywithApplicationsbyThomasKoshy[5],ElementaryIn-troductiontoNumberTheorybyCalvinLong[6],andElementaryNumberTheoryandItsApplicationsbyKennethRosen[8].Forbrevity,weshallonlyincludesomeofthemoreinterestingproofs.Webeginwiththeconceptofdivisibility.Denition.Ifaandbareintegers,thenwesaythatadividesbifthereisanintegercsuchthatb=ac.Ifadividesb,wealsosaythataisadivisororfactorofbandthatbisamultipleofa.Ifadividesbwewriteajb,andifadoesnotdividebwewritea-b.Thefollowingarethreewell-knownpropertiesofdivisibility.Theorem2.1.Ifa;b;andcareintegerswithajbandbjc,thenajc.Theorem2.2.Ifa;b;m;andnareintegers,andifcjaandcjb,thencjma+nb.Theorem2.3.TheDivisionAlgorithm.Ifaandbareintegerssuchthatb>0,thenthereareuniqueintergersqandrsuchthata=bq+rwith0r
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Theorem2.4.Everypositiveintegergreaterthan1hasaprimedivisor.Theorem2.5.Thereareinnitelymanyprimes.Theorem2.6.Ifnisacompositeinteger,thennhasaprimefactornotexceedingp n.Denition.Ifxisarealnumber,thentheprimecountingfunctionxdenotesthenumberofprimesnotexceedingx.Theorem2.7.ThePrimeNumberTheorem.Ifxisarealnumber,thenxisasymptotictox=lnx.ThePrimeNumberTheoremhasaninterestingcorollary.Considerthesetofallpositiveintegerslessthanorequaltosomelargepositiveintegerx.Bythetheorem,wewouldexpectaboutx=lnxprimestobeinthiscollectionofxintegers.So,ifwerandomlyselectanintegerfromthisset,theprobabilitythatwewillselectaprimenumberisapproximatelyx=lnx x=1=lnx:Wenowconsidergreatestcommondivisors.Denition.Thegreatestcommondivisoroftwointegersaandb,whicharenotboth0,isthelargestintegerthatdividesbothaandb.Thegreatestcommondivisorofaandbisdenotedgcda;b.Wedenegcd;0=0.Theorem2.8.Letaandpbeintegerswithpprime.Ifp-a,thengcda;p=1.Proof.Sincepisprime,theonlypositivedivisorsofpare1andp.Soeithergcda;p=1,orgcda;p=p.Ifgcda;p=p,thenpja,contrarytoouras-sumption.Hence,wemusthavegcda;p=1. Denition.Twointegersaandbarerelativelyprimeifgcda;b=1.Theorem2.9.Ifaandbareintegerswithgcda;b=dandd1,thengcda=d;b=d=1.Theorem2.10.Ifa;b;andcareintegers,thengcda+cb;b=gcda;b.Denition.Ifaandbareintegers,thenalinearcombinationofaandbisasumoftheformma+nb,wherebothmandnareintegers.Theorem2.11.Thegreatestcommondivisoroftheintegersaandb,notboth0,istheleastpositiveintegerthatisalinearcombinationofaandb.Theorem2.12.Ifaandbareintegers,notboth0,thenapositiveintegerdisthegreatestcommondivisorofaandbifandonlyif:idjaanddjb.10

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iiifcisanintegerwithcjaandcjb,thencjd.Theorem2.13.Ifa,b,andcareintegerswithgcda;c=1,thengcda;b=gcda;bc.Proof.Letd=gcda;b.Clearlydgcda;bc.Wewanttoshowthatdgcda;bc.ByTheorem2.11,sincegcda;b=dandgcda;c=1,wemaywrited=as+btand1=au+cv,forsomeintegerss;t;u;v.Now,d=as+bt=as+bt1=as+btau+cv=aQ1+bcQ2forsomeintegersQ1;Q2.Thisisalinearcombinationofaandbc,andweknowfromTheorem2.11thatgcda;bcisthesmallestlinearcombinationofaandbc,sodgcda;bc.Sincedgcda;bcanddgcda;bc,weconcludethatd=gcda;bc. Thenexttwodenitionsallowustoconsiderthegreatestcommondivisorofmorethantwointegers.Denition.Leta1;a2;:::;anbeintegers,notallzero.Thegreatestcommondivisoroftheseintegersisthelargestintegerthatisadivisorofalloftheintegersintheset.Thegreatestcommondivisorofa1;a2;:::;anisdenotedgcda1;a2;:::;an.Wedenegcd;0;:::;0=0.Denition.Wesaythattheintegersa1;a2;:::;anaremutuallyrelativelyprimeifgcda1;a2;:::;an=1.Theseintegersarecalledpairwiserelativelyprimeifforeachpairofintegersaiandajfromthesetwithi6=j,wehavegcdai;aj=1.ThefollowingtheoremmakesuseofthegreatestcommondivisorandwillbeusefultousinChapter3.Theorem2.14.Letaandbbeintegerswithgcda;b=d.Then,theequationax+by=chasnointegersolutionsforxandyifd-c.Ifdjc,thenthereareinnitelymanyintegersolutions.Thenextthreetheoremsconcernthefunadmentaltheoremofarithmetic.Theorem2.15.Ifa;b;andcarepositiveintegerssuchthatgcda;b=1andajbc,thenajc.Theorem2.16.Ifpdividesa1a2anwherepisaprimeanda1;a2;:::;anarepositiveintegers,thenthereisanintegeriwith1insuchthatpdividesai.Theorem2.17.TheFundamentalTheoremofArithmetic.Everypositiveintegergreaterthan1canbewrittenuniquelyasaproductofprimes,withtheprimefactorsintheproductwritteninnondecreasingorder.11

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Proof.Ourproofisintwoparts.First,weshallproveexistence.Assumeforthepurposeofcontradictionthatthereisatleastonepositiveintegergreaterthan1thatcannotbewrittenastheproductofprimes.Letnbethesmallestsuchcounterex-amplethepositiveintegersarewell-ordered,sotheleastelementofanonemptysetofpositiveintegersexists.Ifnisprime,thenwearedone,soassumethatniscomposite.Writen=ab,with10.Ifmja)]TJ/F19 11.955 Tf 12.18 0 Td[(bwesaythataiscongruenttobmodulom,denotedabmodm.Ifm-a)]TJ/F19 11.955 Tf 11.98 0 Td[(b,wesaythataandbareincongruentmodulom,denoteda6bmodm.12

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Thenexttwotheoremscoversomebasicpropertiesofcongruence.Theorem2.19.Ifaandbareintegers,thenabmodmifandonlyifthereisanintegerksuchthata=b+km.Theorem2.20.Letmbeapositiveinteger.Congruencesmodulomsatisfythefollowingproperties:iReexiveproperty.Ifaisaninteger,thenaamodm.iiSymmetricproperty.Ifaandbareintegerssuchthatabmodm,thenbamodm.iiiTransitiveproperty.Ifa;bandcareintegerswithabmodmandbcmodm,thenacmodm.Thenextthreeitemsconcernmodulararithmetic.Theorem2.21.Leta;b;c;andmbeintegerswithm>0;andletd=gcdc;m.Then,acbcmodmiabmodm=d:Corollary2.22.Leta;b;c;andmbeintegerswithm>0;andletgcdc;m=1.Then,acbcmodmiabmodm:Theorem2.23.Ifa;b;c;d;andmareintegerssuchthatm>0;abmodm,andcdmodm,thenia+cb+dmodm,iia)]TJ/F19 11.955 Tf 11.956 0 Td[(cb)]TJ/F19 11.955 Tf 11.956 0 Td[(dmodm,iiiacbdmodm.Wenowconsidersystemsofresidues.Denition.Acompletesystemofresiduesmodulomisasetofintegerssuchthatforeachx2Z,xiscongruentmodulomtoexactlyoneintegeroftheset.Lemma2.24.Asetofmincongruentintegersmodulomformsacompletesetofresiduesmodulom.Theorem2.25.Ifr1;r2;:::;rmisacompletesetofresiduesmodulom,andifaisapositiveintegerwithgcda;m=1,thenar1+b;ar2+b;:::;arm+bisacompletesystemofresiduesmodulomforanyintegerb.Thenextthreeitemsconcernadditionalwaystomanipulatecongruences.13

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Theorem2.26.Ifa;b;kandmareintegerssuchthatk>0;m>0;andabmodm,thenakbkmodm.Theorem2.27.Leta;b;m1;m2;:::;mkbeintegerswithm1;m2;:::;mkpositive.Then,abmodm1;abmodm2;:::;abmodmkifandonlyifabmodlcmm1;m2;:::;mk.Corollary2.28.Leta;b;m1;m2;:::;mkbeintegerswithm1;m2;:::;mkpositiveandpairwiserelativelyprime.Then,abmodm1;abmodm2;:::;abmodmkifandonlyifabmodm1m2mk.Wenowmoveontotheconceptoflinearcongruences.Denition.Acongruenceoftheformaxbmodm,wherexisanunknowninteger,iscalledalinearcongruenceinonevariable.Thefollowingtheoremstateshowmanysolutionsagivenlinearcongruencehas,ifany.Theorem2.29.Leta;b;x;andmbeintegerssuchthatm>0,andletgcda;m=d.Ifd-b,thenaxbmodmhasnosolutions.Ifdjb,thenaxbmodmhasexactlydincongruentsolutionsmodulom.Thistheoremcanbegeneralizedtoapplytoasystemoflinearcongruences.Theorem2.30.Letx;a1;a2;:::;arbeintegersandm1;m2;:::;mrbepositiveinte-gers.Then,thesystemofcongruencesxa1modm1xa2modm2...xarmodmrhasasolutionifandonlyifgcdmi;mjjai)]TJ/F19 11.955 Tf 13.094 0 Td[(ajforallpairsofintegersi;j,where1i
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Proof.Considerthep)]TJ/F15 11.955 Tf 11.847 0 Td[(1integersa;2a;:::;p)]TJ/F15 11.955 Tf 11.846 0 Td[(1a.Sincep-a,noneoftheseinte-gersisdivisiblebyp,andnotwooftheseintegersarecongruentmodulop.Becausetheintegersa;2a;:::;p)]TJ/F15 11.955 Tf 12.464 0 Td[(1aareasetofp)]TJ/F15 11.955 Tf 12.463 0 Td[(1-manyintegersallincon-gruentto0,andnotwoarecongruentmodulop,byLemma2.24weknowthattheleastpositiveresiduesofa;2a;:::;p)]TJ/F15 11.955 Tf 10.926 0 Td[(1a,takeninsomeorder,mustbetheintegers1;2;:::;p)]TJ/F15 11.955 Tf 12.584 0 Td[(1.Asaconsequence,theproductoftheintegersa;2a;:::;p)]TJ/F15 11.955 Tf 12.584 0 Td[(1aiscongruentmoduloptotheproductoftherstp)]TJ/F15 11.955 Tf 11.955 0 Td[(1positiveintegers.Hence,a2ap)]TJ/F15 11.955 Tf 11.956 0 Td[(1a12p)]TJ/F15 11.955 Tf 11.955 0 Td[(1modp:Thatis,ap)]TJ/F17 7.97 Tf 6.587 0 Td[(1p)]TJ/F15 11.955 Tf 11.955 0 Td[(1!p)]TJ/F15 11.955 Tf 11.956 0 Td[(1!modp:NowusingCorollary2.22,wemaycancelthep)]TJ/F15 11.955 Tf 11.955 0 Td[(1!frombothsidesleavingap)]TJ/F17 7.97 Tf 6.586 0 Td[(11modp: WenowintroduceEuler'sphi-functionandthreerelatedtheorems.Denition.Letnbeapositiveinteger.TheEulerphi-functionnisdenedtobethenumberofpositiveintegersnotexceedingnthatarerelativelyprimeton.Theorem2.34.Ifpisprime,thenp=p)]TJ/F15 11.955 Tf 11.956 0 Td[(1.Theorem2.35.Ifpisprimeandnisapositiveinteger,thenpn=pn)]TJ/F19 11.955 Tf 11.955 0 Td[(pn)]TJ/F17 7.97 Tf 6.587 0 Td[(1.Theorem2.36.Euler'sTheorem.Ifmisapositiveintegerandaisanintegerwithgcda;m=1,thenam1modm.UsingEuler'sphi-function,wecandeneareducedresiduesystem.Denition.Areducedresiduesystemmodulonisasetofnintegerssuchthateachelementofthesetisrelativelyprimeton,andnotwodierentelementsofthesetarecongruentmodulon.Letusnowconsiderordersofintegers.Denition.Letaandnberelativelyprimepositiveintegers.Then,theleastpositiveintegerxsuchthatax1modniscalledtheorderofamodulon.Theorem2.37.Leta;n;i;andjbeintegerssuchthataandnarerelativelyprimeandn;i;andjarenonnegative.Ifxistheorderofamodulon,thenaiajmodnifandonlyifijmodx.Wenowintroduceprimitiveroots.Denition.Letrandnberelativelyprimeintegerswithn>0.Iftheorderofrmodulonisequalton,thenriscalledaprimitiverootmodulon.15

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Theorem2.38.Letrandnberelativelyprimepositiveintegerswithn>0.Ifrisaprimitiverootmodulon,thentheintegersr1;r2;:::;rnformareducedresiduesetmodulon.Now,letusconsidertherootsofpolynomialcongruences.Denition.Letfxbeapolynomialwithintegercoecients.Anintegercisarootoffxmodulomiffc0modm.Theorem2.39.Lagrange'sTheorem.Letpbeprimeandletfx=anxn+an)]TJ/F17 7.97 Tf 6.587 0 Td[(1xn)]TJ/F17 7.97 Tf 6.587 0 Td[(1++a1x1+a0beapolynomialofdegreen1withintegercoecients.Ifp-an,thenfxhasatmostnincongruentrootsmodulop.Proof.Weuseproofbyinduction.Whenn=1,wehavefx=a1x+a0withp-a1.Arootoffxmodulopisasolutionofthelinearcongruencea1x)]TJ/F19 11.955 Tf 22.603 0 Td[(a0modp.ByTheorem2.29,sincegcda1;p=1,weknowthislinearcongruencehasexactlyonesolution.Thisimpliesthatthereisexactlyonerootmodulopoffx.Hence,Lagrange'sTheoremistruewhenn=1.Now,supposethatthetheoremistrueforpolynomialsofdegreen)]TJ/F15 11.955 Tf 13.211 0 Td[(1,andletfxbeapolynomialofdegreenwithleadingcoecientnotdivisiblebyp.Assumeforthepurposeofcontradictionthatfxhasn+1-manyincongruentrootsmodulop,sayc0;c1;:::;cn,sothatfck0modpfork=0;1;:::;n.Wehavefx)]TJ/F19 11.955 Tf 11.955 0 Td[(fc0=anxn)]TJ/F19 11.955 Tf 11.955 0 Td[(cn0+an)]TJ/F17 7.97 Tf 6.586 0 Td[(1xn)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F19 11.955 Tf 11.955 0 Td[(cn)]TJ/F17 7.97 Tf 6.587 0 Td[(10++a1x)]TJ/F19 11.955 Tf 11.955 0 Td[(c0=anx)]TJ/F19 11.955 Tf 11.955 0 Td[(c0xn)]TJ/F17 7.97 Tf 6.587 0 Td[(1+xn)]TJ/F15 11.955 Tf 11.955 0 Td[(2c0++xcn)]TJ/F17 7.97 Tf 6.586 0 Td[(20+cn)]TJ/F17 7.97 Tf 6.586 0 Td[(10+an)]TJ/F17 7.97 Tf 6.587 0 Td[(1x)]TJ/F19 11.955 Tf 11.955 0 Td[(c0xn)]TJ/F17 7.97 Tf 6.587 0 Td[(2+xn)]TJ/F15 11.955 Tf 11.955 0 Td[(3c0++xcn)]TJ/F17 7.97 Tf 6.586 0 Td[(30+cn)]TJ/F17 7.97 Tf 6.586 0 Td[(20++a1x)]TJ/F19 11.955 Tf 11.955 0 Td[(c0=x)]TJ/F19 11.955 Tf 11.956 0 Td[(c0gx;wheregxisapolynomialofdegreen)]TJ/F15 11.955 Tf 11.707 0 Td[(1withleadingcoecientan.Wenowshowthatc1;c2;:::;cnareallrootsofgxmodulop.Letkbeanintegerwith1kn.Becausefckfc00modp,wehavefck)]TJ/F19 11.955 Tf 11.955 0 Td[(fc0=ck)]TJ/F19 11.955 Tf 11.955 0 Td[(c0gck0modp:Itfollowsthatgck0modp,becauseck)]TJ/F19 11.955 Tf 10.74 0 Td[(c060modp.Hence,ckisarootofgxmodulop.Thisshowsthatthepolynomialgx,whichisofdegreen)]TJ/F15 11.955 Tf 10.68 0 Td[(1andhasleadingcoecientnotdivisiblebyp,hasn-manyincongruentrootsmodulop.Thiscontradictsourinductionhypothesis.Hence,fxmusthavenomorethann-manyincongruentrootsmodulop.Thus,Lagrange'sTheoremistrueforalln1. TheremainingitemsareallconsequencesofLagrange'sTheorem.NotethatTheorem2.40isourownworkbecausewedonotnditinstandardtexts.16

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Theorem2.40.Letpbeprimeandletfxbeapolynomialofdegreen1withintegercoecients.Letan60modpbetheleadingcoecientoffxandassumethatx1;x2;:::;xnarenincongruentrootsoffxmodulop.Now,letRx=fx)]TJ/F19 11.955 Tf -422.701 -13.948 Td[(anx)]TJ/F19 11.955 Tf 11.994 0 Td[(x1x)]TJ/F19 11.955 Tf 11.994 0 Td[(x2x)]TJ/F19 11.955 Tf 11.994 0 Td[(xn.Then,thecoecientsofRxareallcongruentto0modulop.Proof.LetmbethedegreeofRx,andwriteRx=amxm+am)]TJ/F17 7.97 Tf 6.586 0 Td[(1xm)]TJ/F17 7.97 Tf 6.587 0 Td[(1++a1x+a0,wheremm,soitcannotbethecasethatp-am.Thismeansthatwemusthaveam0modp.ThisimpliesthatRxam)]TJ/F17 7.97 Tf 6.587 0 Td[(1xm)]TJ/F17 7.97 Tf 6.586 0 Td[(1+am)]TJ/F17 7.97 Tf 6.586 0 Td[(2xm)]TJ/F17 7.97 Tf 6.587 0 Td[(2++a1x+a0modp.Butx1;x2;:::;xnarestillnincongruentrootsofam)]TJ/F17 7.97 Tf 6.587 0 Td[(1xm)]TJ/F17 7.97 Tf 6.586 0 Td[(1+am)]TJ/F17 7.97 Tf 6.587 0 Td[(2xm)]TJ/F17 7.97 Tf 6.586 0 Td[(2++a1x+a0modulop,soagainbyLagrangewemusthaveam)]TJ/F17 7.97 Tf 6.587 0 Td[(10modp.Continuinginthismanner,wehaveai0modpfori=1;2;:::;m.Thatis,Rxa0modp.ButaswedemonstratedinthecasewhenthedegreeofRxwasassumedtobe0,thismustmeanthata00modp.Thus,allthecoecientsofRxmustbecongruentto0modulop. Theorem2.41.Letpbeprimeandletdbeadivisorofp)]TJ/F15 11.955 Tf 11.34 0 Td[(1.Then,thepolynomialfx=xp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 d)]TJ/F15 11.955 Tf 11.955 0 Td[(1hasexactlyp)]TJ/F17 7.97 Tf 6.587 0 Td[(1 dincongruentrootsmodulop.Proof.Sinceddividesp)]TJ/F15 11.955 Tf 12.333 0 Td[(1,wemaywritep)]TJ/F15 11.955 Tf 12.333 0 Td[(1=dq,forsomepositiveintegerq.Then,xp)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 11.956 0 Td[(1=xq)]TJ/F15 11.955 Tf 11.955 0 Td[(1xqd)]TJ/F17 7.97 Tf 6.587 0 Td[(1+xqd)]TJ/F17 7.97 Tf 6.587 0 Td[(2++xq+1=xq)]TJ/F15 11.955 Tf 11.955 0 Td[(1gx:ByFermat'slittletheorem,xp)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 9.584 0 Td[(1hasp)]TJ/F15 11.955 Tf 9.584 0 Td[(1incongruentrootsmodulop.Fromabove,anyrootofxp)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 9.848 0 Td[(1modulopmusteitherbearootofxq)]TJ/F15 11.955 Tf 9.848 0 Td[(1modulop,orarootofgxmodulop.NowbyLagrange'stheorem,gxhasatmostqd)]TJ/F15 11.955 Tf 10.708 0 Td[(1=dq)]TJ/F19 11.955 Tf 10.708 0 Td[(q=p)]TJ/F15 11.955 Tf 10.708 0 Td[(1)]TJ/F19 11.955 Tf 10.709 0 Td[(qincongruentrootsmodulop.Sincexp)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 11.685 0 Td[(1hasp)]TJ/F15 11.955 Tf 11.685 0 Td[(1incongruentrootsmodulop,andgxhasatmostp)]TJ/F15 11.955 Tf 11.685 0 Td[(1)]TJ/F19 11.955 Tf 11.685 0 Td[(qrootsmodulop,thismeansthatxq)]TJ/F15 11.955 Tf 10.853 0 Td[(1hasatleastp)]TJ/F15 11.955 Tf 10.853 0 Td[(1)]TJ/F15 11.955 Tf 10.853 0 Td[(p)]TJ/F15 11.955 Tf 10.853 0 Td[(1)]TJ/F19 11.955 Tf 10.853 0 Td[(q=qincongruentrootsmodulop.ButbyLagrange'stheorem,xq)]TJ/F15 11.955 Tf 11.965 0 Td[(1hasatmostqincongruentrootsmodulop.Hence,xq)]TJ/F15 11.955 Tf 11.955 0 Td[(1hasexactlyqincongruentrootsmodulop.17

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Nowp)]TJ/F15 11.955 Tf 12.919 0 Td[(1=dqimpliesq=p)]TJ/F17 7.97 Tf 6.586 0 Td[(1 d,soxp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 d)]TJ/F15 11.955 Tf 12.919 0 Td[(1hasexactlyp)]TJ/F17 7.97 Tf 6.587 0 Td[(1 dincongruentrootsmodulop,asdesired. Lemma2.42.Letpbeprimeandletdbeapositivedivisorofp)]TJ/F15 11.955 Tf 12.738 0 Td[(1.Then,thenumberofpositiveintegerslessthanpoforderdmodulopdoesnotexceedd.Theorem2.43.Letpbeprimeandletdbeadivisorofp)]TJ/F15 11.955 Tf 11.74 0 Td[(1.Then,thenumberofincongruentintegersoforderdmodulopisequaltod.Corollary2.44.Everyprimehasaprimitiveroot.18

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3.RESULTS3.1OverviewThetoolswehavepresentedinthepreviouschapterwillallowustoproveTheorems1.1,1.2,1.3,and1.4.Recall,Theorem1.1states:Letm3beaninteger.Then,Pmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 23.96 0 Td[(1modmifandonlyifmisaproductofdistinctoddprimessuchthatmpmodp2p)]TJ/F15 11.955 Tf 11.477 0 Td[(1foreachprimedivi-sorpofm.Ourproofofthistheoremreliesonfoursmallerresults.Welistthemhere,andprovetheminthesubsequentsectionsofthischapter.Theorem3.1.Letm3beaninteger.IfPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modm,thenmmustbeaproductofdistinctprimes.Theorem3.2.Letm3beaninteger.IfPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modm,thenmmustbeodd.Theorem3.3.Letm3beaninteger.IfPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 22.416 0 Td[(1modm,thenmpmodp2p)]TJ/F15 11.955 Tf 11.956 0 Td[(1foreachprimedivisorpofm.Theorem3.4.Letm3beaninteger.Ifmisaproductofdistinctoddprimessuchthatmpmodp2p)]TJ/F15 11.955 Tf 10.616 0 Td[(1foreachprimedivisorpofm,thenPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modm.ClearlythesefourtheoremsimplyTheorem1.1.Succeedingtheproofsofthesefourtheorems,wewillprovethethreeconsequencesofTheorem1.1:Theorem1.2.Letm3beanintegerandletp1,p2beanytwoprimedivi-sorsofm.IfPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 21.917 0 Td[(1modm,thenp261modp1.Theorem1.3.Letm3beaninteger.IfPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 24.083 0 Td[(1modm,thenmcannotbeaproductoftwoprimes.Theorem1.4.Letm3beaninteger.IfPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 24.083 0 Td[(1modm,thenmcannotbeaproductofthreeprimes.19

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3.2ProofofTheorem3.1Inthissection,wewishtoprovethatifm3isanintegerwithPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 23.072 0 Td[(1modm,thenmmustbeaproductofdistinctprimes.Tothisend,wewillusethefollowinglemmawhichwillprovehelpfulthroughouttherestofthischapter.Lemma3.5.Ifmanddarepositiveintegerssuchthatdjm,thenPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1m dPdi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1modd:Proof.Becausedjm,wemaywritemXi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1=m d)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=0dXi=1jd+im)]TJ/F17 7.97 Tf 6.586 0 Td[(1m d)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xj=0dXi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1modd;sincejd+iimodd.And,sincejassumesm=ddierentvalues,wehavem d)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xj=0dXi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1=m ddXi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1: WearenowreadytoproveTheorem3.1,whichisstatedabove.Proof.Withmasinthetheorem,writem=pe11pe22pekkitsprime-powerfactoriza-tionandletpe2fpe11;pe22;:::;pekkg.Assumeforthepurposeofcontradictionthate2.SincePmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modmandpjm,byLemma3.5wehavem pPpi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modp.Recall,pejm,sowemaywritem=peqforsomeq2Z.Nowbecausee2itfollowsthatm ppXi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1=peq ppXi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1=pe)]TJ/F17 7.97 Tf 6.586 0 Td[(1qpXi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(10modp:Sincepjm,wehavemXi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 21.917 0 Td[(1modmmXi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modp;20

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andsince06)]TJ/F15 11.955 Tf 22.318 0 Td[(1modp,wehaveacontradiction.Hencee62,meaningmmustbeaproductofdistinctprimes. Intheprecedingproof,weshowedthatPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(10modp.Infact,whenm=pn,wherepisanoddprimeandn2isaninteger,wecanactuallyprovesomethingstronger.Theorem3.6.Ifpisanoddprimeandn2isaninteger,thenPpni=1ipn)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F19 11.955 Tf 21.918 0 Td[(pn)]TJ/F17 7.97 Tf 6.587 0 Td[(1modpn.Proof.First,note:pnXi=1ipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1=pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=0pXi=1jp+ipn)]TJ/F17 7.97 Tf 6.587 0 Td[(1=pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=0p)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1jp+ipn)]TJ/F17 7.97 Tf 6.587 0 Td[(1+pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xj=0jp+ppn)]TJ/F17 7.97 Tf 6.586 0 Td[(1pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=0p)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1jp+ipn)]TJ/F17 7.97 Tf 6.587 0 Td[(1+0modpn:Thiscongruenceholdsbecausepn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=0jp+ppn)]TJ/F17 7.97 Tf 6.587 0 Td[(1=pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=0j+1ppn)]TJ/F17 7.97 Tf 6.587 0 Td[(1=pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1Xj=1jppn)]TJ/F17 7.97 Tf 6.587 0 Td[(1=ppn)]TJ/F17 7.97 Tf 6.587 0 Td[(1pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1Xj=1jpn)]TJ/F17 7.97 Tf 6.586 0 Td[(1;whichiscongruentto0modulopnsinceitcanbeshownbyinductionthatpn)]TJ/F15 11.955 Tf 10.469 0 Td[(1>n.Thus,pnXi=1ipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=0p)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1jp+ipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1modpn:Nowusingbinomialexpansion,wegetpnXi=1ipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=0p)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1jp+ipn)]TJ/F17 7.97 Tf 6.587 0 Td[(1modpn=pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=0p)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1pn)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xk=0pn)]TJ/F15 11.955 Tf 11.955 0 Td[(1kjpkipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F20 7.97 Tf 6.587 0 Td[(k=pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=0p)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1pn)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xk=0pn)]TJ/F15 11.955 Tf 11.955 0 Td[(1kpkjkipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F20 7.97 Tf 6.587 0 Td[(k:21

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Notethatwhenkn,thissumiscongruentto0modulopnbecauseofthepkterm.Hence,pnXi=1ipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=0p)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1n)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xk=0pn)]TJ/F15 11.955 Tf 11.955 0 Td[(1kpkjkipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F20 7.97 Tf 6.586 0 Td[(kmodpn:Wewillnowshowthatwhenk>0,thissumiscongruentto0modulopn.Consider,pn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xj=0p)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1n)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xk=1pn)]TJ/F15 11.955 Tf 11.955 0 Td[(1kpkjkipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F20 7.97 Tf 6.587 0 Td[(k=p)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1n)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xk=1pn)]TJ/F15 11.955 Tf 11.955 0 Td[(1kipn)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F20 7.97 Tf 6.587 0 Td[(kpkpn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xj=0jk;andsincePt)]TJ/F17 7.97 Tf 6.586 0 Td[(1j=0jk=1 k+1Pkj=0)]TJ/F20 7.97 Tf 5.48 -4.379 Td[(k+1jBjtk)]TJ/F20 7.97 Tf 6.586 0 Td[(j+1,whereBjisthejthBernoullinumberseeforinstancepage283ofGraham,Knuth,andPatashnik'sConcreteMathematics[3],or[9],wehavep)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1n)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xk=1pn)]TJ/F15 11.955 Tf 11.955 0 Td[(1kipn)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F20 7.97 Tf 6.587 0 Td[(kpkpn)]TJ/F18 5.978 Tf 5.756 0 Td[(1)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xj=0jk=p)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1n)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xk=1pn)]TJ/F15 11.955 Tf 11.955 0 Td[(1kipn)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F20 7.97 Tf 6.587 0 Td[(kpk k+1kXj=0k+1jBjpn)]TJ/F17 7.97 Tf 6.586 0 Td[(1k)]TJ/F20 7.97 Tf 6.586 0 Td[(j+1=p)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1n)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xk=1pn)]TJ/F15 11.955 Tf 11.955 0 Td[(1kipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F20 7.97 Tf 6.587 0 Td[(k1 k+1kXj=0k+1jBjpn)]TJ/F17 7.97 Tf 6.586 0 Td[(1k)]TJ/F20 7.97 Tf 6.586 0 Td[(j+1pk=p)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1n)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xk=1pn)]TJ/F15 11.955 Tf 11.955 0 Td[(1kipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F20 7.97 Tf 6.587 0 Td[(k1 k+1kXj=0k+1jBjpn)]TJ/F17 7.97 Tf 6.586 0 Td[(1k)]TJ/F20 7.97 Tf 6.586 0 Td[(j+1+k=p)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1n)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xk=1pn)]TJ/F15 11.955 Tf 11.955 0 Td[(1kipn)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F20 7.97 Tf 6.587 0 Td[(k1 k+1kXj=0k+1jBjpn)]TJ/F17 7.97 Tf 6.586 0 Td[(1k)]TJ/F20 7.97 Tf 6.586 0 Td[(j+n)]TJ/F17 7.97 Tf 6.587 0 Td[(1+k:Nowinordertoseethatthissumiscongruentto0modulopn,considerthelargestintegerrsuchthatprdividesk+1,thedenominatorofthefraction.Wewanttoshowpn)]TJ/F17 7.97 Tf 6.586 0 Td[(1k)]TJ/F20 7.97 Tf 6.586 0 Td[(j+n)]TJ/F17 7.97 Tf 6.587 0 Td[(1+k)]TJ/F20 7.97 Tf 6.586 0 Td[(r0modpnfork=1;2;:::;n)]TJ/F15 11.955 Tf 11.426 0 Td[(1andj=0;1;:::;k.So,weneedtoshown)]TJ/F15 11.955 Tf 11.956 0 Td[(1k)]TJ/F19 11.955 Tf 11.956 0 Td[(j+n)]TJ/F15 11.955 Tf 11.955 0 Td[(1+k)]TJ/F19 11.955 Tf 11.955 0 Td[(rn.22

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Aftersubtractingnfrombothsidesofthisinequality,weseethatweonlyneedtoshown)]TJ/F15 11.955 Tf 11.955 0 Td[(1k)]TJ/F19 11.955 Tf 11.955 0 Td[(j+k)]TJ/F15 11.955 Tf 11.955 0 Td[(1)]TJ/F19 11.955 Tf 11.955 0 Td[(r0.Recall,itcanbeshownbyinductionthatforanyoddprimepandanypositiveintegerk,pk>k+1.Nowsinceprjk+1impliesprk+1,wemusthaver
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soPp)]TJ/F17 7.97 Tf 6.587 0 Td[(1i=1ipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1=Qp)]TJ/F15 11.955 Tf 11.955 0 Td[(1,forsomepositiveintegerQ.Thus,pnXi=1ipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1pn)]TJ/F17 7.97 Tf 6.586 0 Td[(1p)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1ipn)]TJ/F17 7.97 Tf 6.587 0 Td[(1modpn=pn)]TJ/F17 7.97 Tf 6.586 0 Td[(1Qp)]TJ/F15 11.955 Tf 11.955 0 Td[(1=Qpn)]TJ/F19 11.955 Tf 11.955 0 Td[(pn)]TJ/F17 7.97 Tf 6.587 0 Td[(10)]TJ/F19 11.955 Tf 11.955 0 Td[(pn)]TJ/F17 7.97 Tf 6.587 0 Td[(1modpn=)]TJ/F19 11.955 Tf 9.299 0 Td[(pn)]TJ/F17 7.97 Tf 6.587 0 Td[(1:Ourproofiscomplete. WementionthatTheorem3.6canalternativelybeprovedusingStirlingnumbersofthesecondkind,insteadofBernoullinumbersformoreinformationonStirlingnumbers,seepage50ofAignerandMartin'sDiscreteMathematics[1].NotethatTheorem3.6canbeviewedasageneralizationofanearlierresult.Recall,foranyprimepitfollowsthatPpi=1ip)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 23.441 0 Td[(1modp.Sincep=p)]TJ/F15 11.955 Tf 12.564 0 Td[(1,thisimpliesthatPpi=1ip)]TJ/F17 7.97 Tf 6.587 0 Td[(1pmodp.Moregenerally,sincepn=pn)]TJ/F19 11.955 Tf 12.766 0 Td[(pn)]TJ/F17 7.97 Tf 6.587 0 Td[(1,Theorem3.6impliesthatforanypositiveintegern,wehavePpni=1ipn)]TJ/F17 7.97 Tf 6.586 0 Td[(1pnmodpn.24

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3.3ProofofTheorem3.2Nowwewillshowthatifm3isanintegerwithPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 22.621 0 Td[(1modm,thenmmustbeodd.Proof.First,mmustbeaproductofdistinctprimesbyTheorem3.1.Nowassumeforthepurposeofcontradictionthatmiseven.Writem=2k,wherek3isanoddpositiveinteger.Then,m)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1=m 2)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1+m 2m)]TJ/F17 7.97 Tf 6.586 0 Td[(1+m)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=m 2+1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1=m 2)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1+m 2m)]TJ/F17 7.97 Tf 6.586 0 Td[(1+m 2)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1m)]TJ/F19 11.955 Tf 11.955 0 Td[(im)]TJ/F17 7.97 Tf 6.586 0 Td[(1m 2)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1+m 2m)]TJ/F17 7.97 Tf 6.586 0 Td[(1+m 2)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1)]TJ/F19 11.955 Tf 9.299 0 Td[(im)]TJ/F17 7.97 Tf 6.587 0 Td[(1modm=m 2)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1+m 2m)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F21 5.978 Tf 13.151 20.266 Td[(m 2)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1=m 2m)]TJ/F17 7.97 Tf 6.586 0 Td[(1:Recall,m=2k,som 2m)]TJ/F17 7.97 Tf 6.587 0 Td[(1=2k 2m)]TJ/F17 7.97 Tf 6.587 0 Td[(1=km)]TJ/F17 7.97 Tf 6.587 0 Td[(1:Nowontheonehand,km)]TJ/F17 7.97 Tf 6.586 0 Td[(1kmodkbecausekjkm)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F19 11.955 Tf 11.955 0 Td[(k.Ontheotherhand,km)]TJ/F17 7.97 Tf 6.586 0 Td[(1kmod2because2jkm)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F19 11.955 Tf 12.021 0 Td[(ksincethedierenceoftwooddintegersiseven.Hence,km)]TJ/F17 7.97 Tf 6.587 0 Td[(1kmod2kbyCorollary2.28.Since2k=m,wehavemXi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1km)]TJ/F17 7.97 Tf 6.586 0 Td[(1kmodm:25

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ButthisisacontradictionsincewesaidPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 24.1 0 Td[(1modm,andk6)]TJ/F15 11.955 Tf 24.1 0 Td[(1modmform3.Hence,mmustbeodd. 26

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3.4ProofofTheorem3.3Inthissection,wewanttoproveTheorem3.3,thatifm3isanintegerwithPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 22.627 0 Td[(1modm,thenmpmodp2p)]TJ/F15 11.955 Tf 12.239 0 Td[(1foreachprimedivisorpofm.Wepresenttheproofastwolemmas.Lemma3.7.Letm3beaninteger.IfPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 23.165 0 Td[(1modm,thenm1modp)]TJ/F15 11.955 Tf 11.955 0 Td[(1foreachprimedivisorpofm.Proof.ByTheorems3.1and3.2,weknowthatmmustbeaproductofdistinctoddprimes.ByLemma3.5,Pmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1m pPpi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1modp,andsincep0modp,wemaywritem ppXi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1m pp)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1modp:Thus,mXi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1m pp)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modp:NowbyCorollary2.44,thereexistsaprimitiverootmodulop.Letrbesuchaprimitiveroot.Recall,byTheorem2.34p=p)]TJ/F15 11.955 Tf 13.119 0 Td[(1,sobyTheorem2.38theintegersr1;r2;:::;rp)]TJ/F17 7.97 Tf 6.586 0 Td[(1formareducedresiduesetmodulop.Thus,wemaywritem pp)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1m pp)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=1rjm)]TJ/F17 7.97 Tf 6.587 0 Td[(1=m pp)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xj=1rjm)]TJ/F17 7.97 Tf 6.586 0 Td[(1modp:Sincerisaprimitiveroot,theorderofrmodulopisp)]TJ/F15 11.955 Tf 11.512 0 Td[(1.Hence,byTheorem2.37wehaverj1m)]TJ/F17 7.97 Tf 6.587 0 Td[(1rj2m)]TJ/F17 7.97 Tf 6.586 0 Td[(1modpij1m)]TJ/F15 11.955 Tf 11.956 0 Td[(1j2m)]TJ/F15 11.955 Tf 11.955 0 Td[(1modp)]TJ/F15 11.955 Tf 11.955 0 Td[(1:Weshallshowm1modp)]TJ/F15 11.955 Tf 11.813 0 Td[(1byshowingthatgcdp)]TJ/F15 11.955 Tf 11.814 0 Td[(1;m)]TJ/F15 11.955 Tf 11.814 0 Td[(1=p)]TJ/F15 11.955 Tf 11.814 0 Td[(1,whichimpliesp)]TJ/F15 11.955 Tf 12.134 0 Td[(1jm)]TJ/F15 11.955 Tf 12.134 0 Td[(1.So,letd=gcdp)]TJ/F15 11.955 Tf 12.134 0 Td[(1;m)]TJ/F15 11.955 Tf 12.134 0 Td[(1andassumeforthepurposeofcontradictionthatd
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soitfollowsthatrj1m)]TJ/F17 7.97 Tf 6.586 0 Td[(1rj2m)]TJ/F17 7.97 Tf 6.586 0 Td[(1modpij1j2modp)]TJ/F15 11.955 Tf 11.955 0 Td[(1 d:Now,recall:mXi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1m pp)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1m pp)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xj=1rjm)]TJ/F17 7.97 Tf 6.587 0 Td[(1=m pp)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xj=1rjm)]TJ/F17 7.97 Tf 6.586 0 Td[(1modp:Sincedjp)]TJ/F15 11.955 Tf 11.955 0 Td[(1,wemaywritem pp)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xj=1rjm)]TJ/F17 7.97 Tf 6.586 0 Td[(1=m pd)]TJ/F17 7.97 Tf 6.586 0 Td[(1X`=0p)]TJ/F18 5.978 Tf 5.756 0 Td[(1 dXj=1r`p)]TJ/F18 5.978 Tf 5.756 0 Td[(1 d+jm)]TJ/F17 7.97 Tf 6.587 0 Td[(1:Notethatfor`=0;1;2;:::;d)]TJ/F15 11.955 Tf 12.132 0 Td[(1,wehave`p)]TJ/F17 7.97 Tf 6.587 0 Td[(1 d+jjmodp)]TJ/F17 7.97 Tf 6.587 0 Td[(1 d.So,referringtotheistatementabove,andlettingj1=`p)]TJ/F17 7.97 Tf 6.586 0 Td[(1 d+jandj2=j,weseethatr`p)]TJ/F18 5.978 Tf 5.756 0 Td[(1 d+jm)]TJ/F17 7.97 Tf 6.587 0 Td[(1rjm)]TJ/F17 7.97 Tf 6.586 0 Td[(1modp:Thus,mXi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1m pd)]TJ/F17 7.97 Tf 6.587 0 Td[(1X`=0p)]TJ/F18 5.978 Tf 5.756 0 Td[(1 dXj=1r`p)]TJ/F18 5.978 Tf 5.756 0 Td[(1 d+jm)]TJ/F17 7.97 Tf 6.586 0 Td[(1modpm pd)]TJ/F17 7.97 Tf 6.587 0 Td[(1X`=0p)]TJ/F18 5.978 Tf 5.756 0 Td[(1 dXj=1rjm)]TJ/F17 7.97 Tf 6.586 0 Td[(1modp=dm pp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 dXj=1rjm)]TJ/F17 7.97 Tf 6.587 0 Td[(1:Inordertoderiveacontradiction,wewillshowthatPp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 dj=1rjm)]TJ/F17 7.97 Tf 6.586 0 Td[(10modp:Toseethis,rstconsiderthepolynomialcongruencefx=xp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 d)]TJ/F15 11.955 Tf 12.177 0 Td[(10modp.Sincedjm)]TJ/F15 11.955 Tf 11.955 0 Td[(1andj=1;2;:::;p)]TJ/F17 7.97 Tf 6.587 0 Td[(1 d,itfollowsthatrjm)]TJ/F17 7.97 Tf 6.586 0 Td[(1p)]TJ/F18 5.978 Tf 5.756 0 Td[(1 d=rp)]TJ/F17 7.97 Tf 6.586 0 Td[(1jm)]TJ/F18 5.978 Tf 5.756 0 Td[(1 d1jm)]TJ/F18 5.978 Tf 5.756 0 Td[(1 dmodp;byFermat'sLittleTheorem.28

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So,wemaythinkofx=rm)]TJ/F17 7.97 Tf 6.587 0 Td[(1;r2m)]TJ/F17 7.97 Tf 6.586 0 Td[(1;:::;rp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 dm)]TJ/F17 7.97 Tf 6.586 0 Td[(1asp)]TJ/F17 7.97 Tf 6.586 0 Td[(1 dsolutionstothepolyno-mialcongruencefx=xp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 d)]TJ/F15 11.955 Tf 11.955 0 Td[(10modp.Nowsinceanypairofdistinctelementsfromthesetf1;2;:::;p)]TJ/F17 7.97 Tf 6.587 0 Td[(1 dgareincongruenttoeachothermodulop)]TJ/F17 7.97 Tf 6.587 0 Td[(1 d,itfollowsfromtheistatementabovethatrj1m)]TJ/F17 7.97 Tf 6.586 0 Td[(16rj2m)]TJ/F17 7.97 Tf 6.586 0 Td[(1modpforj=1;2;:::;p)]TJ/F17 7.97 Tf 6.587 0 Td[(1 d.Hence,x=rm)]TJ/F17 7.97 Tf 6.587 0 Td[(1;r2m)]TJ/F17 7.97 Tf 6.586 0 Td[(1;:::;rp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 dm)]TJ/F17 7.97 Tf 6.586 0 Td[(1arep)]TJ/F17 7.97 Tf 6.586 0 Td[(1 dincongruentsolutionstothepoly-nomialcongruencefx=xp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 d)]TJ/F15 11.955 Tf 12.152 0 Td[(10modp.Sincethispolynomialhasexactlyp)]TJ/F17 7.97 Tf 6.587 0 Td[(1 dincongruentsolutionsbyCorollary2.41,weknowwehavethemall.NowwemayuseTheorem2.40,whichstatesthatthecoecientsofRx=fx)]TJ/F15 11.955 Tf 11.955 0 Td[(x)]TJ/F19 11.955 Tf 11.955 0 Td[(rm)]TJ/F17 7.97 Tf 6.587 0 Td[(1x)]TJ/F19 11.955 Tf 11.955 0 Td[(r2m)]TJ/F17 7.97 Tf 6.587 0 Td[(1x)]TJ/F19 11.955 Tf 11.956 0 Td[(rp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 dm)]TJ/F17 7.97 Tf 6.586 0 Td[(1areallcongruentto0modulop.SincePp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 dj=1rjm)]TJ/F17 7.97 Tf 6.586 0 Td[(1isthecoecientofthexp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 d)]TJ/F17 7.97 Tf 6.587 0 Td[(1terminRx,itfollowsfromThe-orem2.40thatp)]TJ/F18 5.978 Tf 5.756 0 Td[(1 dXj=1rjm)]TJ/F17 7.97 Tf 6.587 0 Td[(10modp:Thus,wehavemXi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1dm pp)]TJ/F18 5.978 Tf 5.757 0 Td[(1 dXj=1rjm)]TJ/F17 7.97 Tf 6.586 0 Td[(10modp:Thisisacontradictionsince06)]TJ/F15 11.955 Tf 21.917 0 Td[(1modp,andwesaidPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modp.Thusd6
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Thatis,Pmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F19 11.955 Tf 21.918 0 Td[(m=pmodp.Now,Pmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 22.44 0 Td[(1modmimpliesPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 22.44 0 Td[(1modp,sowemusthavem=p1modp.Thisimpliesm=p=pk+1,forsomek2Z+.Multiplyingthroughbypgivesm=p2k+p,whichmeansmpmodp2. ByLemma3.7m1modp)]TJ/F15 11.955 Tf 12.468 0 Td[(1,andsince1pmodp)]TJ/F15 11.955 Tf 12.469 0 Td[(1,wehavempmodp)]TJ/F15 11.955 Tf 11.666 0 Td[(1.ByLemma3.8mpmodp2,sobyCorollary2.28weconcludethatmpmodp2p)]TJ/F15 11.955 Tf 11.955 0 Td[(1,andourproofofTheorem3.3iscomplete.30

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3.5ProofofTheorem3.4OurproofofTheorem3.4willcompletetheremainingdirectionofTheorem1.1.Weneedtoprovethatifm3isanintegersuchthatmisaproductofdistinctoddprimeswithmpmodp2p)]TJ/F15 11.955 Tf 11.653 0 Td[(1foreachprimedivisorpofm,thenPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 9.298 0 Td[(1modm.Proof.Recall,Pmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1m pPpi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1m pPp)]TJ/F17 7.97 Tf 6.586 0 Td[(1i=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1modp.Nowsincempmodp2p)]TJ/F15 11.955 Tf 11.709 0 Td[(1byhypothesis,m=p2p)]TJ/F15 11.955 Tf 11.71 0 Td[(1X+pforsomenonnegativeintegerX.Thus,m pp)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1=p2p)]TJ/F15 11.955 Tf 11.955 0 Td[(1X+p pp)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1ip2p)]TJ/F17 7.97 Tf 6.587 0 Td[(1X+p)]TJ/F17 7.97 Tf 6.587 0 Td[(1=pp)]TJ/F15 11.955 Tf 11.956 0 Td[(1X+1p)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1ip)]TJ/F17 7.97 Tf 6.587 0 Td[(1p2X+p)]TJ/F17 7.97 Tf 6.586 0 Td[(1p)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1ip)]TJ/F17 7.97 Tf 6.587 0 Td[(1p2X+p)]TJ/F17 7.97 Tf 6.586 0 Td[(1modp=p)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1ip)]TJ/F17 7.97 Tf 6.587 0 Td[(1p2X+1=p)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=1ip)]TJ/F17 7.97 Tf 6.586 0 Td[(1p2X+1:NowbyFermat'sLittleTheorem,wehavep)]TJ/F17 7.97 Tf 6.586 0 Td[(1Xi=1ip)]TJ/F17 7.97 Tf 6.587 0 Td[(1p2X+1p)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=11p2X+1modp=p)]TJ/F17 7.97 Tf 6.587 0 Td[(1Xi=11=p)]TJ/F15 11.955 Tf 11.955 0 Td[(1)]TJ/F15 11.955 Tf 28.559 0 Td[(1modp:31

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Sincethisholdsforallprimedivisorspofm,andmisaproductofdistinctprimesbyhypothesis,Corollary2.28impliesthatmXi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modm: OurproofofTheorem1.1isnowcomplete.32

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3.6ProofofTheorem1.2WearenowreadytoprovesomeconsequencesofTheorem1.1.WebeginwithTheorem1.2.Wewanttoshowthatthatifm3isanintegersuchthatPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 9.298 0 Td[(1modm,thenanytwoprimedivisorsofm,sayp1andp2,mustsatisfyp261modp1.Proof.Thisistrivialwhenp1=p2,soassumethatmiscompositeandp16=p2.Clearlyp261modp1ifp2
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3.7ProofofTheorem1.3WewillnowproveTheorem1.3,thatifm3isanintegerwithPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 23.118 0 Td[(1modm,thenmcannotbeaproductoftwoprimes.Proof.FromTheorem1.1,weknowthatmmustbeaproductofdistinctoddprimes.Assumeforthepurposeofcontradictionthatmisaproductoftwosuchprimes,sayp1andp2withp1p2,whichistosayp1>p2.Thisisacontradictionbecauseweassumedp1
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3.8ProofofTheorem1.4Wecanalsoshowthatifm3isanintegerwithPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 22.702 0 Td[(1modm,thenmcannotbeaproductofthreeprimes.HereisourproofofTheorem1.4.Proof.FromTheorem1.1,weknowthatmmustbeaproductofdistinctoddprimes.Assumeforthepurposeofcontradictionthatmisaproductofthreesuchprimes,sayp1,p2,andp3withp1
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4.COMPUTATIONSWhilewehaveprovensomestrongconditionsonmforPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modm,wehavenotbeenabletoverifytheconjecturethatmmustbeprimeifPmi=1im)]TJ/F17 7.97 Tf 6.586 0 Td[(1)]TJ/F15 11.955 Tf 21.918 0 Td[(1modm.Nevertheless,usingMapleandthetheoremsweprovedinChapter3,wecandemon-stratethatthereisnocounterexampletotheconjecturelessthanorequaltoonetrillion.Figures4.1,4.2,and4.3onpages42,43,and44containtheactualMaplecodefortheprogramwedevelopedtoshowthis.Weprovideasummaryofhowtheprogramworks.TheuserinputsthelargestintegertobeveriedcalledN,andtheprogramoutputsanycounterexamplestotheconjecturelessthanorequaltoNifany,aswellasthenumberoffactoriza-tionsperformed.Wecountthenumberoffactorizationsbecausefactorizationisatime-consumingstep.Theprogramissplitintoatop-downportionandabottom-upportionseeFigure4.4onpage45inordertominimizethetotalruntimeoftheprogram.Theowchartdetailingthetop-downpartislistedasFigure4.5onpage46.WearesearchingforahypotheticalcompositeintegermNthatsatisesPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 21.953 0 Td[(1modm.Sincemiscomposite,Theorem1.1impliesthatm=p2p)]TJ/F15 11.955 Tf 12.139 0 Td[(1Xp+pforallprimedivisorspofm,whereXpissomepositiveinteger.Letpmaxbethelargestprimedivisorofm.IntheowchartsandtheMaplecode,wearecallingpmaxBP;"whichstandsforbiggestprime,"becausethisnotationiseasiertoenterintoMaple.Weclaimthatpmaxcannotexceedb3p Nc+1.Toseethis,wewillshowthatpmax<3p N+1,whichgivespmaxb3p Nc+1.Assumeforthepurposeofcontradictionthatmhasalargestprimedivisorpmax3p N+1.Then,36

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m=pmax2pmax)]TJ/F15 11.955 Tf 11.956 0 Td[(1Xpmax+pmaxbyTheorem1.1mpmax2pmax)]TJ/F15 11.955 Tf 11.955 0 Td[(1+pmaxsinceXpmax1m3p N+123p N+1)]TJ/F15 11.955 Tf 11.955 0 Td[(1+3p N+1byassumptionm[3p N2+23p N+1]3p N+3p N+1m3p N3+23p N2+3p N+3p N+1m3p N3+23p N2+23p N+1mN+23p N2+23p N+1m>N:ThisisacontradictionbecausewechosemsuchthatmwaslessthanorequaltoN.Thus,pmax63p N+1pmax<3p N+1pmaxb3p N+1cpmaxb3p Nc+1:Withourclaimproved,wecontinuetodiscussthetop-downportionoftheprogram.Theorem1.1anditsconsequencestellusthatsincemiscomposite,mmustbeaproductoffourormoredistinctoddprimes.So,ourrstgoalistosetpmaxtobeaslargeaspossiblegivenN,andthenshowthatnomcanexistwhoselargestprimedivisorispmax.Then,wewillsetpmaxtobethelargestprimelessthanpmax'scurrentvalue,andrepeattheprocess.Wewillcontinueaslongaspmax>75.Oncepmax<75,wewillexitthetop-downportionandbeginthebottom-upportionoftheprogram.Aftermuchexperimentation,wechose75asthecut-ointegerinordertominimizethetotalruntimeoftheprogram.Ifthecut-onumberistoolargeortoosmall,eithertherstpartorthesecondpartoftheprogramwilltakemuchlongertorun.Onalate-2006notebookcomputer2.16GHzIntelCore2Duoprocessorwith2.00GBofRAMrunningWindowsXP,theprogramtakes43minutestorunwhenN=10122minutesforthetop-downportionand11minutesforthebottom-upportion.Inthetop-downportion,foreachmN,wehavem=pmax2pmax)]TJ/F15 11.955 Tf 9.811 0 Td[(1Xpmax+pmax.IntheowchartsandMaplecode,wesimplywriteXpmaxasX".NowletmaxXpmaxrepresentthelargestpossiblevaluethatanyXpmaxcanassumeforagivenm.IntheowchartsandMaplecode,wecallmaxXpmaxmaxX:"ItfollowsthatmaxXpmax=N)]TJ/F19 11.955 Tf 11.956 0 Td[(pmax pmax2pmax)]TJ/F15 11.955 Tf 11.955 0 Td[(1:37

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Now,foreachXpmaxfrom1tomaxXpmaxweletm=pmax2pmax)]TJ/F15 11.955 Tf 11.453 0 Td[(1Xpmax+pmax,andthenwefactorm=pmaxtocheckthatmisaproductofdistinctoddprimesthisisfasterthanjustfactoringm.Ifmisaproductofdistinctoddprimes,thenforeachprimedivisorpofmwecheckthatm=p2p)]TJ/F15 11.955 Tf 11.152 0 Td[(1Xp+pforsomepositiveintegerXp.IntheMaplecode,wesimplycallXpx".IfXpisapositiveinteger,weaddmtoourlistofcounterexamples.OncewehaverunthroughallXpmaxforeachpmax>75,weexitthetop-downportionoftheprogram.Thepurposeofthebottom-upportionoftheprogramistoshowthatmcannotbeaproductofdistinctoddprimeslessthan75.ItsowchartislistedasFigure4.6onpage47.Forthispart,weletSP1standforthesmallestprimedivisorofm.WestartwithSP1=3sincemmustbeodd.LetSP2representthesecondsmallestprimedivisorofm.ByTheorem1.2,SP261modSP1.NowletLISTbethesetofallprimespwithSP267westopherebecauseifSP2>67,thenmwouldhavetohaveaprimedivisorlargerthan75.Oncethisoccurs,weadvanceSP1tothenextprimelargerthanSP1'scurrentvalue,resetSP2,andrepeatthesteps.Then,weadvanceSP1again,andagain,untilSP1>61ifSP1>61,thenmwouldhavetohaveaprimedivisorlargerthan75.Oncethishappens,wehavenishedthebottom-upportionoftheprogram,andtogetherwiththetop-downportion,wehaveexhaustedoursearchforacom-positeintegermNsatisfyingPmi=1im)]TJ/F17 7.97 Tf 6.587 0 Td[(1)]TJ/F15 11.955 Tf 22.377 0 Td[(1modm.Theprogramconcludesbyprintingthenumberoffactorizationsperformedandlistinganycounterexamplesdiscovered.Wecanshowheuristicallythatthenumberoffactorizationsperformedisapproxi-matelyequaltoNR1a1=x3lnxdx,wherearepresentsthecut-ointegerseparatingthetop-downandbottom-uppartsoftheprograme.g.a=75above.ThisshowsthatthenumberoffactorizationsperformedincreaseslinearlyasNincreaseswhenaisindependentofN.NotethattherearemaxXpmaxfactorizationsforeachvalueofpmax,andthatpmaxassumesavalueforeachprimebetweenaandb3p Nc+1.Furthermore,maxXpmax=N)]TJ/F19 11.955 Tf 11.955 0 Td[(pmax pmax2pmax)]TJ/F15 11.955 Tf 11.955 0 Td[(1N pmax3N pmax3:38

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So,ifweletM=b3p Nc+1andletFNequalthetotalnumberoffactorizationsperformedforagivenN,thenFNMXx=a[xN=x3]=NMXx=ax x3;wherex=1;ifxisprime0;otherwise.Atthispoint,weturntoacommonheuristicargumentinordertocontinueapproxi-matingthissum.Recall,forarealnumbern,thefunctionndenotesthenumberofprimesnotexceedingn,andbytheprimenumbertheorem,nisasymptoticallyequalton=lnn.Nowconsideralargepositiveintegerxandasmallrealnumber.Bytheprimenumbertheorem,wehavethefollowingthreefacts,x=x lnx+o;)]TJ/F19 11.955 Tf 11.955 0 Td[(x=)]TJ/F19 11.955 Tf 11.955 0 Td[(x ln)]TJ/F19 11.955 Tf 11.955 0 Td[(x+o=)]TJ/F19 11.955 Tf 11.956 0 Td[(x lnx+ln)]TJ/F19 11.955 Tf 11.955 0 Td[(+o;+x=+x ln+x+o=+x lnx+ln++o;whereogxmeansogx gx!0asx!1.Inparticular,odenotesafunctionofxthatgoesto0asx!1.Eachappearanceofocanrepresentadierentfunction.Nowconsidertheinterval[)]TJ/F19 11.955 Tf 12.076 0 Td[(x;+x].Thenumberofprimesinthisintervalisgivenbyx+)]TJ/F19 11.955 Tf 11.955 0 Td[(x)]TJ/F19 11.955 Tf 11.955 0 Td[(.Fromabove,thisisequalto+x lnx+ln++o)]TJ/F15 11.955 Tf 31.111 8.088 Td[()]TJ/F19 11.955 Tf 11.955 0 Td[(x lnx+ln)]TJ/F19 11.955 Tf 11.955 0 Td[(+o;whichisequalto2x lnx+o:Sincethereare2xintegersintheinterval[)]TJ/F19 11.955 Tf 12.229 0 Td[(x;+x],wemaysaythattheprobabilitythatarandomlyselectedintegerfromthisintervalisprimeisabout2x lnx+o 2x=1 lnx+o:39

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Inotherwords,givenalargeenoughxandarelativelysmallenoughintervalaroundx,theprobabilitythatarandomlyselectedintegerfromthatintervalisprimeisaymptoticto1=lnx.Usingthisheuristic,wehaveFNNMXx=ax x3NMXx=a1=lnx x3=NMXx=a1 x3lnx:Wewishtoapproximatethissumbyanintegralsothatwemaystudyitmore.ConsidereachterminthesumasanN=x3lnxby1rectangle.Ifweorienttheserectanglesverticallyandcenterthemontheintegersonthex-axisofacoordinateplane,thenNPMx=a1=x3lnxisequaltotheareaoccupiedbyalloftheserectangles.NowforaxM,thefunctionfx=N=x3lnxpassesthroughthemidpointofthetopofeachrectangle,sotheareaunderthiscurveisaboutequaltotheareatherectanglesoccupy.Hence,FNNMXx=a1 x3lnxFNNZMa1 x3lnxdxFN NZMa1 x3lnxdxFN NZ1a1 x3lnxdx)]TJ/F24 11.955 Tf 11.955 16.273 Td[(Z1M1 x3lnxdx:However,0Z1M1 x3lnxdxZ1M1 x3dx=1 2M2;and1=M2!0asM!1.NowsinceM=b3p Nc+13p N,wemaysaythat1=M2!0asN!1.Thus,asN!1,FN NZ1a1 x3lnxdx:40

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NowletusevaluateR1a1=x3lnxdxusingsubstituationofvariables.Lett=2lnx.Then,lnx=t=2andx2=et.Also,dx=xdt=2.Thus,Z1a1 x3lnxdx=Z1a1 xx2lnxdx=Z12lna1 xett=2xdt 2=Z12lna1 ettdt=Z12lnae)]TJ/F20 7.97 Tf 6.586 0 Td[(t tdt=E1lna;whereE1lnaistheexponentialintegralE1xevaluatedat2lnaformorein-formationonthisfunction,see[10],[11],and/orpage2ofBleisteinandHandlesman'sAsymptoticExpansionsofIntegrals[2].Thus,FN NZ1a1 x3lnxdx=E1lna;whichimpliesFNNE1lna:WithN=1012anda=75,Maplegives1012E1ln518;621;914.NotethattheactualnumberoffactorizationsperformedwithN=1012anda=75was16;580;031.41

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Fig.4.1:MapleCodePart142

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Fig.4.2:MapleCodePart243

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Fig.4.3:MapleCodePart344

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Fig.4.4:OverviewFlowchart45

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Fig.4.5:Top-DownFlowchart46

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Fig.4.6:Bottom-UpFlowchart47

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BIBLIOGRAPHY[1]Aigner,Martin.07.DiscreteMathematics.AmericanMathematicalSociety.[2]Bleistein,N.andHandlesman,R.75.AsymptoticExpansionsofIntegrals.Dover.[3]Graham,Knuth,andPatashnik.994.ConcreteMathematics:AFoundationforComputerSciencendEdition.AddisonWesley.[4]Jones,GarethA.998.ElementaryNumberTheory.Springer.[5]Koshy,Thomas.2007.ElementaryNumberTheorywithApplications,SecondEdition.AcademicPress.[6]Long,CalvinT.1995.ElementaryIntroductiontoNumberTheory.WavelandPress.[7]OddPerfectNumberSearch.008.:[8]Rosen,KennethH.2005.ElementaryNumberTheoryandItsApplicationsthEdition.AddisonWesley.[9]Weisstein,EricW.008.BernoulliNumber."MathWorld{AWolframWebResource..[10]Weisstein,EricW.008.En-Function."MathWorld{AWolframWebResource..[11]Weisstein,EricW.08.ExponentialIntegral."MathWorld{AWolframWebResource..48