acyclotomicpolynomial).otherexamples,writingthefactorizationsasdierencesof squares,are (5y2)5?1 (3y2)3+1 3y2+1=(3y2+1)2?(3y)2

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1 Abstract.TheCunninghamprojectseekstofactornumbersoftheformbn1withb= 2;3;:::small.OneofthemostusefultechniquesisAurifeuillianFactorizationwherebysuch AndrewGranvilleandPeterPleasants AURIFEUILLIANFACTORIZATION gavealistofsuchidentitiesthathaveprovedusefulinthecunninghamproject;webelieve ifoneacceptsourdenitionofwhat\suchanidentity"is.wethendevelopourthemeto (2y2)2+1=(2y2?2y+1)(2y2+2y+1)wecanpartiallyfactor24k+2+1.Schinzel[Sch] similarlyfactorf(bn)foranygivenpolynomialf,usingdeepresultsoffaltingsfromalgebraic geometryandfriedfromtheclassicationofnitesimplegroups. thatschinzelidentiedallnumbersthatcanbefactoredbysuchidentitiesandweprovethis factorizationispossible.forexample,bysubstitutingy=2kintothepolynomialfactorization anumberispartiallyfactoredbyreplacingbbyapolynomialinsuchawaythatpolynomial theform2n1;3n1;etc.evidentlysuchinformationprovidesusefulexamplesfor numbershaveprovedtobefertilegroundfortheinitialdevelopmentoffactoringtechniques, whichmaysubsequentlybegeneralizabletofactoringarbitraryintegers.thebook[bls] severaltopicsinelementarynumbertheory.asthetheoryoffactoringhasdeveloped,such containsagoodhistoricalaccountuptothetimeitwaswritten;andforup-to-datedata In1925CunninghamandWoodallpublishedabookoffactorizationsofnumbersof 1.Introduction seethewebsitehttp:// eventhenumbereldsieve,thelatestgeneralfactoringtechnique,wasrstsuggestedby Pollardtoattacknumbersinthe\CunninghamProject".Theendofthesedevelopments littlefurther,justastheinventionofanewastronomicalinstrumentmaypushoalittle isnotyetinsight: theboundariesofthephysicaluniverse;buttheunknownregionsareinnite,andifwe couldcomebackathousandyearsfromnowweshouldnodoubtndworkersinthetheory ofnumbersannouncinginthejournalsnewschemesandnewprocessesfortheresolution ofagivennumberintoitsfactors. Theinventionofnew[factorization]methodsmaypushothelimitsoftheunknowna sotheymaycomewithintherangeofotherfactorizationtechniques.(similarlybn+1can 'd(b)mayfactorfurther,ofcourse,andsincetheycanbeconsiderablysmallerthanbn?1 factorization"xn?1=qdjn'd(x),where'd(x)isthedthcyclotomicpolynomial,the monic,irreduciblepolynomialwhoserootsaretheprimitivedthrootsofunity.thefactors Thenumberbn?1canbepartiallyfactoredbysubstitutingx=bintothe\algebraic D.N.Lehmer,ScienticMonthly,Sept1918. Foundation. TherstauthorisaPresidentialFacultyFellow,andissupported,inpart,bytheNationalScience 1 TypesetbyAMS-TEX

2 bepartiallyfactoredasqd'd(b)wheretheproductisoverintegersddividing2nbutnot factorsinz[y];thensubstitutingvaluesforytogetpartialfactorizationsofnumbersinthe 2Cunninghamproject.Forexample,substitutingx=y6inx2+1givesthefactorization n.)onecangeneralizethistechniquebyseekingpolynomialsg(y)2z[y],suchthatg(y)n1 ANDREWGRANVILLEANDPETERPLEASANTS y12+1=(y4+1)(y8?y4+1)andsimilarsubstitutionsx=ykallowustofactorizex2+1, providedthatkisnotapowerof2.thesefactorizationsmerelyamounttoalgebraic whereqisthelargestfactorofqprimetonand Ingeneraltheeectofthesubstitutionx=yqonthecyclotomicpolynomial'nisgiven bythecyclotomicfactorization,'n(yq)=ydjq'nq=d(y); factorizationintohigherdegreecyclotomicpolynomials,however,sogiveusnothingnew. (1) 'n(?x)=8><>:'2n(x)ifn1(mod2) allowingustofactornumbers22(2k+1)+1.thisisanaurifeuillianfactorization(which Alesstrivialexampleisthesubstitutionx=2y2intox2+1whichgives (2y2)2+1=(2y2+1)2?(2y)2=(2y2?2y+1)(2y2+2y+1); 'n=2(x)ifn2(mod4) wedenetobeafactorizationgivenbymakingsubstitutionoftheformx=ay2into 'n(x)if4dividesn: acyclotomicpolynomial).otherexamples,writingthefactorizationsasdierencesof squares,are (5y2)5?1 (3y2)3+1 3y2+1=(3y2+1)2?(3y)2 Noticethatineachcasewecantakey=pktogetapartialfactorizationofpp(2k+1)1. (7y2)7+1 5y2?1=(25y4+15y2+1)2?(5y)2(5y2+1)2 factorizationandanaurifeuillianonearebothpossibletheaurifeuillianfactorizationis usuallyfarbetter:forexample,for1025=210+1thecylotomicfactorizationis usefulrststepinndingacompletefactorization.fornumberswhereacyclotomic Aurifeuillianfactorizationsgivetwofactorsofaboutthesamesize,whichisamost 7y2+1=(7y2+1)6?(7y)2(49y4+7y2+1)2: (almosttrivial)whereastheaurifeuillianoneis (22+1)(28?26+24?22+1)=5205 (25?23+1)( )=2541:

3 Moredramatically,thefactorizationsof222+1are Inan1878paper[Lu],LucasexplainedhowAurifeuilleprovedthatthereareidentities = and = ;respectively: AURIFEUILLIANFACTORIZATION 3 likethoseaboveforeveryprimeexponentn,andschinzel[sch]foundsimilaridentities formanycompositeexponentsn.firstheprovedthatifn=n;2nor4n,withnodd, (2) un;d(x);vn;d(x)2z[x]suchthat andifdisany(positiveornegative)squarefreedivisorofnthenthereexistpolynomials '2N(x)=u2N;d(x)2+?1 'N(x)=uN;d(x)2??1 Nowbysubstitutinginx=ry2,wherer=(?1=d)d;?(?1=d)dor2d,respectively,the '4N(x)=u4N;d(x)2?2dxv4N;d(x)2: ddxvn;d(x)2; aboveidentitiesbecomeadierenceoftwosquares,andweobtainapolynomialfactorizationof'n(ry2).(later,stevenhagen[st]andbrent[br]gavenewproofsoftheseidentities ddxv2n;d(x)2; andalgorithmsforcomputingthethepolynomialsuandv.)weshallgiveamotivated descriptionofthesefactorizationsattheendofsection4. whatisdesireddoesnotseemtohavebeenpreciselydenedintheliterature.foran identitytobeasusefultothecunninghamprojectasthoseabove,onepresumablywants ndmoresuch\aurifeuillianfactorizations",orsomeanalogousconstruction.however, thefollowing DesiredProperties.Givenn,thereisag(y)2Q[y]with SincetheyhaveprovedsousefultotheCunninghamproject,ithaslongbeendesiredto (II)forsomeintegerb,thereareinnitelymanyintegersmforwhichg(m)isapower (I)'n(g(y))reducible,and (I)and(II)above.Thenwecanwriteg(t)=(a(t+c)q)NLwhere(L;n)=1andallprime Theorem.Supposethatthereisanintegernandg(t)2Q[t]withthedesiredproperties cyclotomicfactorizations,onendsallgwiththeabove\desiredproperties". divisorsofndividen,suchthataisrationalbuta1=qisirrationalifqjqandq>1. Ourmainresult(provedinSection6)isthatbycombiningSchinzel'sidentitiesand ofb. iii)ifnn`isfourtimesanoddintegerandristwiceanoddinteger. ii)ifnn`istwiceanoddintegerandr3(mod4); Then'n(g(t))=Q`jL'nN`(a(t+c)q).Thepolynomials'nN`(a(t+c)q)canbefactored dividesnn`and furtheronlyinthefollowingsituation:writea=rb2whererissquarefree.ifqiseven,r i)ifnn`isoddandr1(mod4);

4 4whichsuggestthattheremaybesomeotherwaytoextendAurifeuillianfactorizations. Forinstance,WagstapointsoutthefollowinginterestingexamplefromtheCunningham Ineachcasethislastfactorizationisgivenbytakingx=rT2,withT=b(t+c)q=2,in Amongthedatain[BLS]onnumbersfactoredbyothermethodsthereareexamples Schinzel'sidentities(2)andobtainingthedierenceoftwosquares. ANDREWGRANVILLEANDPETERPLEASANTS project: dence? wherethelasttwofactorsdierbyaboutone-fthofapercent.canthisreallybecoinci- 62+1= ; irreduciblepolynomialf(x)2z[x]wewishtofactorf(n)forallintegersn,orperhapsfor ninsomespecialsubset(suchasthepowersofsomexedinteger).firstwewillwantto determineg(t)2q[t]forwhichf(g(t))isreducible. Lemma1.Supposethatf(x)2Z[x]isirreducible.Thenf(g(t))isreducibleinQ[t]if Theideasusedabovecanbedevelopedforafarmoregeneralproblem:Foragiven 2.Generalizations andonlyifg(t)?isreducibleink[t],wherekisthesplittingeldoffoverqand f()=0. Proof.Notethatf(g(t))=Q(g(t)?).Firstsupposethatf(g(t))=A(t)B(t)for A;B2Q[t]andleta(t)=gcd(A(t);g(t)?).Thenthea(t)'sareconjugateinKand notwohavearootincommon,sotheyhavenorma.conversely,ifg(t)?factorsthen thenormofeachfactorisafactorinq[t]off(g(t)). forwhichf(g(t))isreducible.(moregenerally,ifk(t);l(t)2z[t]andh(t)isanyfactorof f(k(t))thenh(t)jf(g(t))forg(t)=k(t)+l(t)h(t).)thisleadsustothe: Question.Supposethatf(x)2Z[x]isirreducible.Canonendinnitelymanyg(t)2 Q[t],distinctunderlineartransformationst!at+b,withdegg<degf,suchthatf(g(t)) isreducibleinq[t]? Notethatf(t)jf(t+f(t)),sothatthereisnodicultyinndingg(t)withdeggdegf foragiveng(x),thesetofintegersforwhichg(x)?isreducible:triviallyt?ndivides inspecialcases.ontheotherhand,researchershavestudiedtheproblemofdetermining, suchasg(t)=p(q(t)),thenq(t)?ndividesg(t)?p(n).in1986,fried[fr]showedthatif g(t)?g(n)sog(x)?isreducibleif=g(n)forsomeintegern.alsoifgisacomposition, gisnotanon-trivialcompositionofpolynomialsandisnotamemberofacertainfamily ofdegreevepolynomials,thenthereareatmostnitelymany,notequaltog(n)for ItseemstobediculttoapplythecriterionofLemma1ingeneral,thoughitmaywork asanapplicationoftheclassicationofnitesimplegroups.similarresultscanbeproved withrestrictedtoanygiveneld.theexceptionalmonicquinticsareparametrizedby somen,forwhichg(x)?isreducible.thedeepproofinvolvesfaltings'theoremaswell

5 x5+tb2x3?(t+5)b3x2+(t2?2t?15)b4x=4+cwheret=(u2?5v2?10)=2andb;c;u;v2q. investigatingwhetherwecanndinnitelymanyg(x)2z[x],withoutrepeatedrootsbut withg(n)asquareforinnitelymanyintegersn,suchthatthereexistu(x);v(x);w(x)2 Fascinatingbutbesidethepoint. ReturningtoourattackonAurifeuillianfactorization,wenowproceedalittledierently, AURIFEUILLIANFACTORIZATION 5 Z[x]satisfying Notethatifg(n)=y2thenw(n)f(n)=(u(n)?yv(n))(u(n)+yv(n))anditislikelythat multiplythroughbyanappropriateconstant(thoughleavingfxed). gcd(f(n);u(n)yv(n))willbefactorsoff(n).notealsothatthereisnolossofgenerality inourassumptionthatthepolynomialsin(3)areinz[x]ratherthanq[x],sincewemay Equation(3)isequivalenttotheassertionthatg(x)correspondstoasquareinthe w(x)f(x)=u(x)2?g(x)v(x)2: quotienteldk=q[x]=(f(x));andthereisnolossofgeneralityintakingthatsolution withwofminimaldegree.wecanassumethatgisnotaconstantasthatwouldmerely leadtoapolynomialfactorizationofw(x)f(x).ontheotherhand,ifthereareinnitely manyintegersnforwhichg(n)isasquarethengmustbeofdegree2bysiegel's Theorem[Si].Thuswewillknowthatthereareonlynitelymanysuchgin(3)ifthe rootsoff(x).thereforeqg()2q2,andthisequalsjresultant(f;g)=fdegg K=Q[x]=(f(x)). manysquarefreepolynomialsg(x)2z[x]ofdegreedthatcorrespondtosquaresintheeld ond)suchthatiffisanirreduciblepolynomialofdegreedthenthereareonlynitely Conjecture1.Fixanintegerd1.ThereexistsanintegerD=Dd>0(depending followingconjectureholdstrueford=1andd=2: wheref0istheleadingcoecientoff. f0istheleadingcoecientoff.thusconjecture1isimpliedbythefollowing: manysquarefreepolynomialsg(x)2z[x]ofdegreedforwhichjresultant(f;g)j2jf0jdq2, ond)suchthatiffisanirreduciblepolynomialofdegreedthenthereareonlynitely Conjecture2.Fixanintegerd1.ThereexistsanintegerD=Dd>0(depending Equation(3)isalsoequivalenttotheassertionthatg()2Q()2forany,orall,ofthe 0jwhere withd1=5whend=1. above.writingg(x)=ax?b,wehaveresultant(f;g)=f(b;a).howeverf(b;a)2 [DG](whichisaconsequenceofFaltings'Theorem[Fa1]).Thusbothconjecturesaretrue f0q2foronlynitelymanypairsofcoprimeintegersa;bifdegf5,bytheorem2of aresultfromfaltings[fa2]ifweassumethatw=1,sincethenf()2q()2whenever Whend=2wehavebeenunabletoprovetheseconjectures,thoughwecandeducesuch Letf(x;y)=ydegff(x=y)bethehomogenizationoff.Wenowconsiderthecased=1 g()=0. is,wewishtofactorf(m;n),where(m;n)=1.asbefore,weinvestigatewhetherwecan ndinnitelymanyhomogeneousg1(x;y);g2(x;y)2z[x;y],withoutrepeatedfactors,for which Finallywementionspecialvaluesofbinaryhomogeneousformsf(x;y)2Z[x;y];that w(x;y)f(x;y)=g1(x;y)u(x;y)2?g2(x;y)v(x;y)2;

6 andforwhichg1(m;n)=g2(m;n)2q2forinnitelymanypairsofcoprimeintegersm;n; inotherwordsg(m;n)2q2whereg=g1g2.bytheorem2of[dg]thisimpliesthat 6correspondtoasquareinK=Q[x]=(f(x)),andsoourproblemwillagainberesolvedif degg4.dehomogenizingourequation,wendthatweareagainrequiringg(x)to Conjectures1or2aretrueforeachd4. ANDREWGRANVILLEANDPETERPLEASANTS Lemma2.Supposethatf0(t)andf1(t)aregivenpolynomials,withnocommonfactor, ofdegreesd0andd1,withd0>d1.bytheeuclideanalgorithmweobtainasequenceof polynomialsf0;f1;f2;:::;fkofdegreesd0>d1>d2>>dk=0suchthatthereexist gk(t);hk(t)with Weshallneedthefollowingtechnicalresult: 3.TheEuclideanalgorithm deggk=d1?dk?1anddeghk=d0?dk?1foreachk=1;:::;k. Proof.Findthequotientq=q(t),ofdegreedk?2?dk?1,sothatfk=fk?2?qfk?1has degree<dk?1.fork=2theresultisthustrivial(asitisfork=1).fork=3;:::;kwe havegk=gk?2?rgk?1andhk=hk?2?rhk?1andtheresultfollowsbyinduction.we terminatethesequencewhenfkisaconstant,whichisnecessarilynon-zerosinceotherwise fk(t)=gk(t)f0(t)+hk(t)f1(t); Corollary1.Supposethatf(x);g(x)2Z[x]nf0gareofdegreesD>d,respectively, fk?1wouldbeacommonfactoroff0andf1.hencedk=0. wheref(x)isirreducibleandg(x)isasquareinq[x]=(f(x)).thenthereisanon-trivial solutionu(x);v(x);w(x)2q[x]nf0gto(3)withdeg(w)d=2. Proof.Letf0=f,andf1besuchthatf1(x)2=g(x)inQ[x]=(f(x)).Applythelemma andselectthesmallestkinf1;2;:::;kgforwhichdk=d=2+d=4.(notethat sothatuvf1(modf(x)).thereforeu2v2f21gv2(modf(x)),sothatu2?gv2is divisiblebyf,andthuswhasdegreemax(2;d+2(d?))?d=d=2. d0=d>anddk=0<.)letu=fkandv=hk,whichhasdegreed?dk?1d?, HeusedthisinArticle357toestablishthat,foralloddprimesponehas pisdenedby InArticle356of[Ga],Gausshadshownthat2p=(?1)(p?1)=2pwheretheGausssum 4.GaussandAurifeuille:HistoryandMotivation 4xp?1 p:=p?1 x?1=y(x)2?(?1)(p?1)=2pz(x)2 Xa=0apapwithp=e2i=p: forsomey(x);z(x)2z[x].forexample, 4x3?1 x?1=(2x+1)2+312and4x5?1 x?1=(2x2+x+2)2?5x2:

7 Fortheproofheexpands z:=y =R?(S+T) (rp)=1?x?rp=r+sx AURIFEUILLIANFACTORIZATION 2+(S?T) (rp)=1rp+tx (np)=?1np 7 forsomer=r(x),s=s(x),t=t(x)inz[x].multiplyingthisbyaconjugateoverq thattakespto?p,andtakingy=2r?s?t,z=s?t,heobtainstheresult. willalwaysbe2xm+xm?1andthehighesttermofthefunctionzisxm?1.theremaining coecients,allofwhichwillbeintegers,willvaryaccordingtothenatureofthenumber Gausswrote:\ItiseasytoseethatthetwotermsofhighestdegreeinthefunctionY 2p wherefp(x)=pm1(mp)xmm,wheneverp3(mod4).healsogaveananalogousexpression,involvingcoshandsinh,for4'n(x)whennisanyoddsquarefreenumber>3. Aurifeuille[Lu]showedasimilarresult:Foralloddprimesponehas Y(x)=2rxp?1 x?1cospp 2fp(x)andZ(x)=2pprxp?1 x?1sinpp 2fp(x); thedelightfulformulae n,andcannotbegivenbyageneralanalyticformula."howeverin1993brent[br]gave forsomec(x);d(x)2z[x].howeverthisisratherusefultothecunninghamprojectfor (4) bytakingx=(?1)(p?1)=2py2weget (py2)p1 py21=c(py2)2?(py)2d(py2)2; xp?1 x?1=c(x)2?(?1)(p?1)=2pxd(x)2 intheintroductionaresodeducedfromtheidentities adierenceoftwosquares,andsofactorable.theexamplesofaurifeuillianfactorization Z(x)inGauss'sidentity,asfollows: Brent[Br]alsogavedirectexpressionsforC(x)andD(x),similartothoseforY(x)and x5?1 x?1=(x2+3x+1)2?5x(x+1)2andx7+1 x2+1=(x+1)2?2x12; x3+1 x+1=(x+1)2?3x12; C(x)=rxp+1 x+1cosh(ppgp(px))andd(x)=1 x+1=(x+1)6?7x(x2+x+1)2: ppxrxp+1 x+1sinh(ppgp(px));

8 wheregp(x)=pmodd1?pmxmmwheneverp3(mod4)(andasimilarexpressionwhen 8extrafactor4ontheleft):Write=(p+1)=2 pisreplacedbyanysquarefreenumber). In1992,HendrikLenstrashowedoneofusadelightfuldirectproofof(4)(withapossible ANDREWGRANVILLEANDPETERPLEASANTS wherepy2z[p][y].takingnormq(p)=qofbothsidesweobtain (?1)(p?1)=2py2?p=(py)2?2=(py?)(py+) (py2)p1 py21=a(y)b(y) p sothat2=pandthus A(y)+B(y)and(A(y)?B(y))=yarexedbythemapy!?y,sobotharefunctionsin wherea(y)=norm(py?)andb(y)=norm(py+).now,bydenition,both side. Z[y2].MoreoverAandBareproductsoftermsoftheform(py),andsoA(y)+B(y) though,slightlymodifyingtheaboveproof,itsucestonotethattheirproductisasquare. and(a(y)?b(y))=pyareinz[(py)2]=z[py2],giving(4)withafactor4onthelefthand Thisthengeneralizeseasily:Supposethatrisaninteger,n=e2i=nandthatrnisa squareinq(n),sayrn=2.thenn(rx2?n)=(x)2?2n=(x?n)(x+n),and so'n(rx2),whichisthenormofn(rx2?n)(fromq(n)toq)factorsasthenormof x?ntimesthenormofx+n.thisgivesallofthefactorizationsthatonecandeduce fromschinzel'sequations(2),providedthatweknowwhenrnisasquareinq(n).we Lenstra'sprooffollowsfromnotingthatpand(?1)(p?1)=2parebothsquaresinQ(p) iii)ifnisfourtimesanoddintegerthenr=2dforsomeodddjn. ii)ifnistwiceanoddintegerthenr=?(?1=d)dforsomeodddjn; foranym1.thenumberrnisasquareinq(n)inonlythefollowingcases: Lemma3.Supposethatrisasquarefreeinteger(positiveornegative),andm=e2i=m determineallcasesofthisinthenextresult. Proof.SincernisasquareinQ(n),weknowthatalloftheprimedivisorsofrmust ramifyinq(n)jqsothattheydividen,andthusrdividesn. (?1=p)p.Thereforeforanyoddrdividingnwehavethat(?1=r)risthesquareofQpjrp, i)ifnisoddthenr=(?1=d)dforsomedjn; soisasquareinq(n),sinceeachq(p)q(n). Aswehaveseen,foranyoddprimeptheGausssumpisinQ(p),anditssquareis soneitherare?n;2n. ThenrnisasquareinQ(n)ifandonlyifsnisasquareinQ(n).Thepossiblevalues forsare1;2;notethatthesmallestcyclotomiceldinwhich?1isasquareisq(4) since?1=(4)2,andthesmallestcyclotomiceldinwhich2(resp.,?2)isasquareis Q(8)since2=((8+8))2(resp.,2=((38+38))2). Ifnisoddthenn=((n+1)=2 Forgivenrasinthehypothesis,letdbethelargestodddivisorofr,ands=r=(?1=d)d. arenotsquaresinq(n),soneitheraren;2n. Ifnistwiceanoddnumberthen?n=((n=2+1)=2 n )2;andfromtheabove?1and2arenotsquaresinQ(n), n

9 inq(n)soneitherare?n;2n. neitheraren. -1isasquareinQ(n),so?2nisalso.Ontheotherhand2isnotsquareinQ(n),so Ifnisfourtimesanoddnumberthen2n=((1+4)=(n=4?1)=2 If8jnthen,fromtheabove?1and2aresquaresinQ(n).Howevernisnotasquare AURIFEUILLIANFACTORIZATIONn )2.Fromtheabove9 wherewedenethefeketepolynomial Were-interpretLenstra'sproofbynotingthat 5.DevelopingLenstra'sperspective 2p=p(p)2p(x)2mod(x?p) Thiscongruenceholdsforanyprimitivepthrootofunityinplaceofp,so p(x)22pmod'p(x)=p?1 p(x):=p?1 Xa=0apxa: inotherwords,p(x)2=2pinq[x]=('p(x)).applyingcorollary1withg(x)=2p= g(x)=(?1)(p?1)=2pxgivesaurifeuille'sidentityuptoaconstant,since (?1)(p?1)=2pwerecoverGauss'sidentity,uptoaconstant.Thesameproofbutwith Ya?1(x?ap); soxisasquareinq[x]=('p(x)). determiningafactorofp:=(pp?(?1)(p?1)=2)=(p?(?1)(p?1)=2),asequenceofnumbers thathavelongbeenofinterest.(forexample,(pp?1)=(p?1)isconjecturedtobetheperiod modpofthesequenceofbellnumbers;see[ld].)takingx=2p=(?1)(p?1)=2pabove,we In1994Hahn[Ha]showedhowsimilarideascouldbeusedtosimplifytheprocessof x(x(p+1)=2)2mod'p(x) (p((?1)(p?1)=2p)p(p+1)=2)=p(=+;?,say),whicharebothintegerslessthanp(when obtainp(x)2x(x(p+1)=2)2mod'p(x).notealsothatpdividesp(x)x(p+1)=2= p>3).therefore(;p)willbothbenon-trivialfactorsofp(infacttheaurifeuillian factorsofp,asiseasilydeducedbynotingthatp(x)?x(p+1)=2p(ap)?a(p+1)=2 DinAurifeuille'sidentity.AnevenfastermethodisduetoDirichlet(forGauss'sidentity, generalizedtoaurifeuille'sidentitybybrent[br]):fromgausswehave (modx?ap).) Corollary1providesafastpracticalwaytondAandBinGauss'sidentityandCand A(x)??1 pp1=2b(x)=y (ap)=1(x?ap)=p?1 2Xj=0(?1)jajxp?1 p 2?j:

10 Dirichlet(1863)usedNewton'srecurrence(1707),kak=?P0j<ksk?jajwhere 10 si=x ANDREWGRANVILLEANDPETERPLEASANTS todeterminetheakbyinduction. (ap)=1ai p=8<:12(ip)(?1 12(p?1); p)p1=2?12;ifp6ji, analogousproof(whichwethusomit): WestartbynotingthefollowingstrengtheningofLemma1,whichfollowsfroman 6.FindingallAurifeuillianfactorizations:Proofofthetheorem ifpji, f(g(t))=a1(t)a2(t):::ak(t)inq[t]whereeachai(t)isirreducibleinq[t]ifandonly Lemma1'.Supposethatf(x)2Z[x]isirreducible,andg(t)2Q[t].Wecanwrite ifg(t)?=a1;(t)a2;(t):::ak;(t)ink[t],wherekisthesplittingeldoffover Qandf()=0,andeachaj;(t)isirreducible.Infactwecantakeaj;(t)tobethe gcd(aj(t);g(t)?)ink[t],uptoascalarmultiple.similarlyaj(t)canbetakentobethe Lemma4.Letb2Qandm2N.Ifb1=misinacyclotomiceldthenb1=miseithera normofaj;(t),uptoascalarmultiple. ofthegaloisclosureofq(b1=m)mustbeabelian. rationalnumberorthesquarerootofarationalnumber. ofq(b1=m)containsthemthdihedralgroupasasubgroupwhichisnon-abelianunless m=1or2.ontheotherhandifb1=mbelongstoacyclotomiceldthenthegaloisgroup showedthatifg(x)hasmorethanonerootthentherearenosolutionstog(r)=skin integersr;s>1ifkissucientlylarge.thusifproperty(ii)holdstheng(t)isofthe ProofoftheTheorem.SchinzelandTijdeman[ST],applyingresultsofSiegelandBaker, Proof.SupposethatBisnotaperfectpower.ThentheGaloisgroupoftheGaloisclosure Thenatq=TQwhereT=btq=Q,sothat g(t)=atqforsomeintegerq>1andrationala. forma(t+c)qandbysendingt!t?cwecanassume,withoutlossofgenerality,that DeneQtobethelargestintegerdividingqforwhicha1=Qisrational,andleta=bQ. assumewithoutlossofgeneralitythata1=qisirrationalforallqjqwithq>1. Q(sothat(nN;Q=N)=1).Thuswemaybeginbyfactoringasabove,andthenwecan wherenistheproduct,overtheprimesdividingn,ofthepowerofthatprimedividing ByLemma1'thefactorizationof'n(atq)isreected,precisely,inthefactorizationof 'n(atq)='n(tq)='nn(tq=n)=y dj(q=n)'nnd(t)=y dj(q=n)'nnd(btq=q) somepropersubsetjoff0;:::;q?1g. Sinceatq?n=aQq?1 atq?ninq(n);leth(t)beanon-trivialmonicirreduciblefactorofatq?ninq(n)[t]. cyclotomiceld,sobylemma4itequalseitherarationalnumberorthesquarerootofa Nowh(0)2Q(n),andh(0)equalsa?jJj=qtimesarootofunity.ThusajJj=qisina j=0(t?nqjqa?1=q),weknowthath(t)=qj2j(t?nqjqa?1=q)for

11 rationalnumber.ifajjj=qisarationalnumber,thensoisa1=qwhereq=q=(q;jjj)divides ajjj=qisthesquarerootofarationalnumber,thena1=qisrationalwhereq=q=(q;2jjj) q:thenbyassumptionq=1andsoqdividesjjjwhichisimpossibleas1jjjq?1.if whichdividesq:byassumptionq=1andsoqdivides2jjjwhichimpliesthatqiseven isjjj=q=2since1jjjq?1. AURIFEUILLIANFACTORIZATION 11 whereeachci2q(nq).ifci6=0thenai=qa?1=2=(a?1=q)q=2?i2q(nq),whichis rootsofsuchanequationequalaxednumbertimestheq=2throotsofunity,soj= hasdegreeq=2(andsothereisexactlytwoofthem).nowh(t)=pq=2 impossiblefor1iq=2?1bylemma4.thush(t)=tq=2+c0a?1=2.nowthe f0;2;4;:::;q?2gorf1;3;5;:::;q?1g. Wehavethusprovedthatanynon-trivialmonicirreduciblefactorofatq?ninQ(n)[t] andthishappensifandonlyifanisasquareinq(n).lemma3tellsusexactlywhenthat Thereforeifatq?nisreducibleinQ(n)thenitfactorsas(a1=2tq=2?2n)(a1=2tq=2+2n); i=0ci(a?1=q)q=2?iti x=btq=2there).notethatthesegivetheonlypossiblesuchfactorizationsinceweproved happens,andthefactorizationsthatarisewerediscussedattheendofsection4(taking abovethatthereisonlyone. to(3)?(thatis,thedegreeonecaseofthequestioninsection2).sinceq[x]=(f(x))is thatu(x)2islinearmodf.wesawinsection2thatdegf4.afteralinearsubstitution whetherthereareinnitelymanymonicpolynomialsu(x)2q[x]ofdegreed?1,such aeldwhenfisirreducible,v(x)hasaninversemodf,sothisisequivalenttoasking wecanwriteanyfofdegreedasxd?axd?2?bxd?3??c. Forwhatirreduciblef(x)2Z[x]arethereinnitelymanylinearg(x)withsolutions 7.Linearg(x) u(x)2(modf(x))isr2+2s+a,whichcanbemade0,foranyr,bytakings=?(r2+a)=2. areinnitelymanysuchuandhenceinnitelymanyg. Soagainthereareinnitelymanyg. Ifdegf=2theneverypolynomialiscongruenttoalinearpolynomialmodf,sothere andt,andwendthatthecoecientofx2is0ift=?(c+a2+2sa+2rb+r2a+s2)=2rwhen isofdegreethreeinxwhenreducedmodf.thecoecientsofthisarepolynomialsinr;s r6=0.thecoecientofx3isthen1=rtimesapolynomialinrandswhichisquadratic Ifdegf=3thenwritef(x)=x3?ax?bandu(x)=x2+rx+s.Thecoecientofx2in sisasquare,andcomputationallywefoundthatthisdiscriminantisexactly4f(r).thus ins.thishasarationalrootifandonlyifitsdiscriminantasaquadraticpolynomialin thereareinnitelymanysuchgifandonlyifthereareinnitelymanyrationalpoints Ifdegf=4thenwritef(x)=x4?ax2?bx?candu(x)=x3+rx2+sx+tsothatu(x) correspondence,tosamwagstaforhisexamplementionedaboveandothersinthecomputationalnumbertheorycommunitywhoencouragedustopublishthesenotes,andto DanAbramovic,MitchRothsteinandTomTuckerforconversationsaboutapplyingFaltings'Theoremtothequadraticgcase,eventhoughweneversucceeded!ThispaperisatitydiscussedinSection4whichinspiredthispaper,toMikeFriedforhishelpful Acknowledgements.ThanksareduetoHendrikLenstraforhisproofofAurifeuille'siden- (?b=a;c=a2)onthegenusonecurvey2=f(x).

12 expandedversionoftherstauthor'slectureattheantsiiiconferenceatreedcollege, Oregonin ANDREWGRANVILLEANDPETERPLEASANTS [BLS]J.Brillhart,D.H.Lehmer,J.L.Selfridge,B.TuckermanandS.S.WagstaJr,Factorizationsof [Br]R.P.Brent,Oncomputingfactorsofcyclotomicpolynomials,Math.Comp.61(1993),131{149. [DG]H.DarmonandA.Granville,Ontheequationszm=F(x;y)andAxp+Byq=Czr,Bull.London [Fa1]G.Faltings,EndlichkeitssatzefurabelscheVarietatenuberZahlkorpern,Invent.Math.73(1983), bn1;b=2;3;5;6;7;10;11;12uptohighpowers,amer.math.soc.,providence,ri,1988. Math.Soc.27(1995),513{ {366. References [Ga]C.F.Gauss,DisquisitionesArithmeticae,YaleU.Press,NewHaven,Connecticut,1965(1801). [Fr]M.Fried,Applicationsoftheclassicationofsimplegroupstomonodromy,PartII:Davenport [Fa2]G.Faltings,Diophantineapproximationonabelianvarieties,Ann.ofMath.(2)133(1991),549{ [Ha]S.Hahn,OnAurifeuillianfactorizations,Math.Japon.39(1994),1{2. [LD]J.LevineandR.E.Dalton,Minimumperiods,modulop,ofrstorderBellexponentialintegers, andhilbert-siegelproblems(toappear) [ST]A.SchinzelandR.Tijdeman,Ontheequationym=P(x),ActaArith.31(1976),199{204. [Sch]A.Schinzel,Onprimitiveprimefactorsofan?bn,Proc.CambridgePhilos.Soc.58(1962), [Lu]E.Lucas,Theoremesd'arithmetique,Atti.Roy.Acad.Sci.Torino13(1878),271{284. [Si]C.L.Siegel,UbereinigeAnwendungenDiophantischerApproximationen,Abh.Preuss.Akad. [St]P.Stevenhagen,OnAurifeuillianfactorizations,Indag.Math.49(1987),451{ {562. Wiss.(1929). Math.Comp.16(1962),416{423. Suva,Fiji. DepartmentofMathematics,UniversityofGeorgia,Athens,GA,USA DepartmentofMathematicsandComputingScience,UniversityoftheSouthPacific,