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2Idonealnumbersahistoricaloverview21EulerAswasmentionedintheintroducti


2InArticle231heintroducestheimportantnewnotionofthegenusofabinaryquadraticformqthissetconsistsofthoseformsq0whosevalueshavethesamequadraticcharactersandthesamesignatureasqAnicediscussionofthisconceptc

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1 2Idonealnumbers:ahistoricaloverview2.1Eu
2Idonealnumbers:ahistoricaloverview2.1EulerAswasmentionedintheintroduction,idonealnumberswerediscoveredaround1778byEuler,whousedthemasatoolfortestinglargenumbersforprimality.Onewaytode nethemisasfollows(cf.Cox[11],p.65):De nition.Anintegern1iscalledidonealifithasthefollowingproperty:ifmisanoddnumberprimetonwhichisproperlyrepresentedbyq(x;y)=x2+ny2,andiftheequationq(x;y)=mhasonlyonesolutionwithx;y0,thenmisaprimenumber.Whilethisde nitionisnotidenticaltotheonethatEulergave,itissimilar(andequivalent)toit.Indeed,Euler'soriginalde nitionissomewhatconfusingandisoftenmisstated.TheproblemswithEuler'sde nitionandhowtocorrectlyinterpretitarediscussedindetailinSteinig[52].(SeealsoWeil[61],pp.222-226.)Eulerdiscoveredthatthereareprecisely65idonealnumbersn10;000,thelargestbeingn=1848.Moreprecisely,heproved(bydirectcomputation):Theorem1If1n10;000,thennisidonealifandonlyifnisoneofthefollowing65numbers:1;2;3;4;5;6;7;8;9;10;12;13;15;16;18;21;22;24;25;28;30;33;37;40;42;45;48;57;58;60;70;72;78;85;88;93;102;105

2 ;112;120;130;133;165;168;177;190;210;232
;112;120;130;133;165;168;177;190;210;232;240;253;273;280;312;330;345;357;385;408;462;520;760;840;1320;1365;1848:Thefactthathecouldnot ndanyotheridonealnumberscameasacompletesurprisetohim:\J'aidoncetebiensurprisdemevoirarr^eteaudernier1848,...",ashewrotetoBeguelinin[15].Laterhereferredtothisphenomenonasa\paradox"in[17].However,hewasconvincedthattherearenoothers,forheassertedattheendof[16]:\Quiaautemusqueaddeciesmillenullialliasemihiobtulerunt,multomagisverisimillimumvidetur,posthuncterminumnullospraeteraexistere...".Inotherwords,wehave:Conjecture2(Euler)Thelargestidonealnumberisn=1848.Inordertoconvincehimselfofthevalidityofthisconjecture,Euler[16]statedandprovedtentheoremsaboutidonealnumbers.Manyoftheseshowthatifweimposeextraconditionsonn,thenthereareonly nitelymanysuchidonealnumbers.Thesetentheoremsarethefollowing:2 2)InArticle231heintroducestheimportant(new)notionofthegenusofabinaryquadraticformq:thissetconsistsofthoseformsq0whosevalueshavethesamequadraticcharacters(andthesamesignature)asq.(Anicediscussio

3 nofthisconceptcanbefoundinCox[11],pp.58-
nofthisconceptcanbefoundinCox[11],pp.58-59;seealsoBuell[8],ch.4.)Thekeypointhereisthattheprimesp-whicharerepresentedbysomeformina xedgenuscanbecharacterizedbycongruenceconditionsmodulo.Itisimmediatefromthede nition(andfromthe nitenessofh())thatthegenusofqconsistsof nitelymanyequivalenceclassesofforms.Wedenotethissetofequivalenceclassesbygen(q)andwritec(q)=#(gen(q)).3)InArticles234{249,GaussintroducesacompositionofquadraticformsandshowsthatthismakesthesetCl()ofequivalenceclassesofprimitivequadraticformsofdiscriminantinto(whatwenowcall)anabeliangroup.Itsidentityis(theclassof)theprincipalform1whichisde nedby1(x;y):=x2� 4y2,ifiseven,andby1(x;y)=x2+xy+1� 4y2,ifisodd.Fromthisitfollowseasilythattheprincipalgenusgen(1)isasubgroupofCl(),andthateachgenusgen(q)=gen(1)qisacosetinCl().Thus,eachgenushasthesamenumberofelements,soc(q)=c(1)=:c(),forallq2Cl();cf.Article252.4)InArticle256Gaussnotesthatthereisanexplicitrelationbetweentheclassnumbersh(&#

4 1;)andh(t2).Inthecasethatt=pisaprime
1;)andh(t2).Inthecasethatt=pisaprime(and0),wehaveh(p2)=1 up� ph();(1)whereu=1if�4(andu�4=2,u�3=3)and pdenotestheKronecker-Legendresymbol.Notethatthisformulaisalsocorrectifisanodddiscriminantorifp=2;cf.Cox[11],p.148.5)InArticles266{285hedevelopsatheoryofternaryforms,whichhethenusesinArticle286toprovethefundamentalresultthatabinaryquadraticformqliesintheprincipalgenusifandonlyifqisasquareinCl(),i.e.gen(1)=Cl()2:(2)Fromthisitfollowsthatthenumberg()ofgeneraofprimitiveformsofdiscriminantisg()=[Cl():Cl()2]=jCl()[2]j,whereCl()[2]denotesthesubgroupofambiguousclasses,i.e.Cl()[2]=fq2Cl():q21g.Theorderofthissubgroupiseasilycomputed(cf.Article257),andsoGaussobtainstheformulag()=2!()�1+"();(3)where!()denotesthenumberofdistinctprimedivisorsofand"()=1if0(mod32),"()=�1if4(mod16)and"()=0otherwise.(Again,thisformulaisalsotrueforodddiscriminants;cf.[11],p.52.)4

5 2.3GrubeInhisbeautifulbutoftenoverlooke
2.3GrubeInhisbeautifulbutoftenoverlookedpaper,Grube[23]analyzesEuler'sresultsonidonealnumbersbyusingGauss'stheoryofbinaryquadraticforms.Inparticular,itfollowsfromhisworkthatEuler'sConjecture2andGauss'sConjecture5are,infact,equivalent.Moreprecisely,heestablishesthefollowingfundamentalconnectionbetweenidonealnumbersanddiscriminantswithone-classgenera:Theorem6(Grube)Anintegern1isanidonealnumberifandonlyifc(�4n)=1,i.e.ifandonlyifeachgenusofformsofdiscriminant�4nconsistsofasingleclass.Proof.Cox[11],pp.61{62.Remark7(a)Inhispaper,Frei[18]attributesthistheoremtoGaussandcallsit\Gauss'scriterion".Heassertsthat\Gausscertainlyhadaproofofthistheorem",butdoesnotexplainwhyhebelievesthis.PerhapsheisfollowingR.Fueter,whoexpressesasimilarsentimentintheintroductiontoEuler'sOperaOmnia,I,4.Whatisclearfrom[21]isthatGausshadaproofoftheeasyimplication\c(�4n)=1)nisidoneal".Indeed,thisisthebasisofhisprimalitytestinArticle334,andthiscertainlysucesforhispurposessinceheknewthatthelistsagreeforn10;000.However,theotherimplicationisless

6 obvious.GrubehimselfusesSchering'sextens
obvious.GrubehimselfusesSchering'sextensionofDirichlet'sTheoremthataprimitivebinaryquadraticformrepresentsin nitelymanyprimes(whichDirichlethadprovedin1840onlyforquadraticformswithprimedeterminant).Grube'sproofwascriticizedbyKeller[38]becauseScheringdidnotprovehisassertedextension,andsuggeststhatthisresultstillisunproved.ItiscuriousthatKellerwasnotawarethatthisgaphadalreadybeen lledbyWeber[58]in1882,orthatBriggs[6]gavean\elementary"proofin1954,orthatitfollowsnowadaysfromtheChebotarevdensitytheorem;cf.[11],p.188.NotethattheDirichlet/Schering/WeberTheoremisalsousedbyCox[11]intheabove-mentionedproofofGrube'sTheorem.(b)Thereareseveralequivalentwaysofstatingtheconditionc()=1;thesearelistedinTheorem3.22of[11],p.59.Forexample,byGauss'sresult(2)weseethatc()=1,Cl()2=f1g,Cl()isanelementaryabelian2-group,everyclassinCl()isambiguous.Grubepointsoutinx7ofhispaper[23]thatitiseasytoseethatifc(�4n)=1andn3(mod4),thenn=3;7,or15,andusesthisfactintheproofofhistheorem.Wethusobtain:Corollary8Ifn3(mod4)isidoneal,then

7 n=3;7or15.Inhispaper,Grube[23]usestheabo
n=3;7or15.Inhispaper,Grube[23]usestheabovetheoremtogivecorrectandshorterproofsofthetenassertionsofEuleraboutidonealnumbers.Inaddition,hegivesextensionsofmanyofEuler'sresults.Thebasicprincipleinalltheseistocomparec()with6 Theaboveresultsleadtothefollowingclassi cationoflargeidonealnumbers:Corollary14Ifn�288isanidonealnumber,thenn63(mod4).Moreover,ifn1;2(mod4),thennissquarefree,whereasifn0(mod4),thenm:=n 4isasquarefreeidonealnumberwithm2(mod4).Proof.The rstassertionfollowsfromCorollary8.Ifn1;2(mod4),thennissquarefreebyTheorem12(a),(b).Nowsuppose4jn,andputm=n 4.ByTheorem12(c),(d)wehavethat8jjn,i.e.,thatm2(mod4).ThusmisidonealbyCorollary10(b).Moreover,sincem�288 4=72,itfollowsfromTheorem12(a),(b)thatmissquarefree.GrubealsodiscussesEuler'scriterionfortestingwhetherornotagivennum-bernisidoneal,andproposestwovariants.Inallthesecriteria,oneconsidersthefactorizationofthenumbersn+k2,fork=1;:::;[p n 3].Euler'scriterionisthefollowing:Criterion15(Euler)Anintegern1isidonealifandonlyi

8 fforeveryk=1;:::;[p n 3]with(k;n)=1wehav
fforeveryk=1;:::;[p n 3]with(k;n)=1wehavethatn+k2=p,p2,2por2s,forsomeoddprimepandsomeintegers1.Remark16Eulerstatesthiscriterionintheaboveforminhisletter[15],butnotinhispaper[16]whereheleavesoutthehypothesis(k;n)=1.Grube[23]pointsoutthatthishypothesiscannotbedroppedbyprovidingcounterexamples;cf.[23],p.503.HealsocriticizesEuler'sproofofhiscriterion,andgivesapartialcorrection.Frei[18],p.57,pointsoutthatGrubeonlyprovedonedirectionofthiscriterionandsays,\whether[thiscriterion]isalsosucientisstillanopenproblem".InordertocorrecttheproblemswithEuler'scriterion,Grubeo ersthefollowingtwovariants:Proposition17(a)Anintegern1isidonealifandonlyifforeveryb=1;:::;[p n 3],andeveryfactorizationn+b2=acwithca2b(5)wehavethateithera=cora=2b.(b)Supposen1issquarefreeandn6=3;7;15.Thennisidonealifandonlyifforeveryk=1;:::;[p n 3]wehavethatn+k2=tp,tp2,or2tp,wheretjnandpisanoddprime.8 (iii)Thenumberq:='(x;y)isprimewheneverqisoddand1q'(p;1)=p(p+2),andsimilarly,q0= (x;y)isprimewheneverq0isoddandq0 (p;1)=p(2p+1).Iftheseconditionsho

9 ld,thenwehavethat2x2+pisprime,forallinte
ld,thenwehavethat2x2+pisprime,forallintegersjxjp.Remark20(a)Notethat(i)canbeviewedasasharpeningofGrube'ssecondcriterion(Proposition17(b))foridonealnumbers,soinparticularanynumbernsatisfying(i)isidoneal.Similarly,itiseasytoseethatanynsatisfying(ii)mustbeidoneal.AsFrobeniusremarks,itfollowsfromEuler'slist(andGauss'sclassnumbercom-putations)thatthispropositionappliestotheprimesp=3;5;11;29andnootherprimesp10;000.Infact,wenowknowthattherecanbenoothersbecausethe(fundamental)discriminants0withh()2areallknown;cf.Stark[50],Watkins[57]andthereferencestherein.(NotethatBaker'sresult[4]isnotquitestrongenoughtobeabletomakethisassertion.)(b)Theimplication(iii))(i)isnotexplicitlyformulated(norproved)in[20],butitiseasytoverifyitasfollows.Letkp n 3.Ifk1(mod2),thenk2+n='(k;1)isprimebythecriterionin(iii).Ifk=2x,then (x;1)isprimebythecriterionin(iii)andsok2+n=2 (x;1)istwiceaprime.Thus(i)holds.(c)FrobeniusalsohasastatementsimilartoProposition19inthecasethatn=pisprimeandp1(mod4).Inthiscase (x;y)=2x2+2xy+py2andtheboundsin(iii)aresligh

10 tlydi erent.Ashepointsout,thiscaseap
tlydi erent.Ashepointsout,thiscaseappliestop=5;13;37.(d)Frobeniusalsoconsidersthepolynomialx2+x+p,wherep3(4)isprimeandshowsthatf(x)isprimeforjxjpifh(�p)=1.Thisstatementisoftencited;cf.Ribenboim[46],p.137,foradiscussionofitshistory.Inaddition,motivatedbysome(unpublished)workofRemak,Frobeniusalsostud-iestheinde niteformx2+xy�py2andtheassociatedprime-producingpolynomialx2+x�p.2.5DicksonandHallInhisHistoryofNumberTheory,Dickson[13]discussesthehistoryofidonealnumbers(upto1919)onpp.361{365.Steinig[52],p.86,callsthis\aratherconfusedaccount"ofidonealnumbersandexplainsthemistakeinDickson'sde nitionofanidonealnumber.Inaddition,IfeelthatDickson'sdiscussionofGrube'spaperonp.363of[13]issubstandard.NotonlyareGrube'smostimportantresults(Theorems6and12)notmentioned,buthealsomakesthefollowingverypuzzlingstatementinconnectionwithEuler'scriterion:\Grubecouldonlyprovethefollowingmodi cation:Let bethesetofnumbersD+n24DinwhichnisprimetoD.Accordingasallornotallnumbersof areoftheformq,2q,q2or2(qaprime),Disorisnotan10

11 2.6ChowlaandWeinbergerAlreadyin1918Hecke
2.6ChowlaandWeinbergerAlreadyin1918HeckeandLandauobservedthattheGeneralizedRiemannHypothe-sis(GRH)forquadraticcharactersgivese ectivelowerboundsfortheclassnumberh()andthatthisleadstoa(conditional)proofofGauss'sConjecture5(a).ThisobservationwaswrittenupbyLandau[40].HereHeckeandLandaurestricttheirattentiontofundamentaldiscriminants0,i.e.,tothosediscriminantsthatareat-tachedtotheringsofintegersofimaginaryquadratic elds.(Thus,isfundamentalifandonlyifeitherisoddandsquarefree,orif 461(mod4)andissquarefree.)Moreprecisely,Hecke'sresult(cf.[29])isthefollowing:thereexistsa(computable)constantc�0suchthatforeveryfundamentaldiscriminant0wehaveL(s;):=1Xn=1 nn�s6=0fors�1�1 logjj)h()�cp jj logjj:(6)Notethatitisstillunknownwhetherthehypothesis(onthelefthandside)of(6)holdsforevery.ThiswouldfollowfromGRH(forquadraticcharacters)whichconjecturesthatL(s;)6=0forRe(s)�1 2.In1934Heilbronn[28],inspiredbyworkofDeuring(andothers),showedthat

12 ifGRHisfalse,thenonecanstillconcludethat
ifGRHisfalse,thenonecanstillconcludethath()!1,if�!1(where0;1(mod4)),andsoGauss'sConjecture5(a)(anditsextensiontoodddiscriminants)isunconditionallytrue.Thisproofisnon-e ective,andane ectiveproofwasfoundonlyin1986.(Formoreinformation,seeStark'sessay[51].)Inthesameyear,Chowla[9]modi edHeilbronn'sargumenttoshowthatc()!1,if�!1;infact,hisarticleappearsrightafterHeilbronn'sinthesamejournal.Thus,thereareonly nitelymanyidonealnumbers.AlthoughChowladoesnotmentionidonealnumbersexplictly,hedoesrefertothisasaconjectureofGauss.LaterthatyearHeilbronnandLinfoot[29]re nedtheargumentof[28]toprovethatbesidesthe9knownfundamentaldiscriminants0withh()=1,therecanbeatmostonemore.Itwasnotuntil1952thatHeegnershowedthatnosuchtenthdiscriminantexists.At rstitwasthoughtthatHeegner'sproofcontainedagap,andhencecreditforthisresultwasgiventoStark[49],butlateritwasfoundthatHeegner'sproofwascorrect;cf.Stark[51]and/orCox[11],p.271,formoreinformationandreferences,andCox[11],x12E,foranicepresentationofHeegner'

13 sproof.Inthisconnectiononeshouldalsoment
sproof.InthisconnectiononeshouldalsomentionBaker[3]who,inthewordsofSerre[48],p.296,\provedanimportantgeneraltheoremwhichreducestheproblemofitsexistencetoa niteamountofcomputation."In1954ChowlaandBriggs[10]provedasimilarresultforfundamentaldiscrim-inants0withc()=1:thereisatmostonemoresuchdiscriminantwithjj&#x]TJ/;༕ ;.9;Ւ ;&#xTf 1;.94; 0 ;&#xTd[0;1060.(ThereisanannoyingtypointhestatementoftheirLemma4:k1=27shouldbereplacedbyk�1=27.)TheyalsoshowthatGRH(orasomewhatweakerhypothesis)impliesthatnosuchexampleexistsforjj&#x]TJ/;༕ ;.9;Ւ ;&#xTf 1;.94; 0 ;&#xTd[0;1014.12 Proof.Supposethatn�1848isidonealandsquarefree.Sincen63(mod4)byCorollary8,itfollowsthat�4nisfundamentaldiscriminant,andsobyTheorem22(a)thereisatmostonesuchn.Thisprovesthe rstassertion.Nowsupposethatcase(i)doesnothold,sothereexistsanidonealnumbern�1848.Ifallsuchnumbersnaresquarefree,thenn=nisuniquebywhatwasjustsaid.ItthusfollowsfromCorollary14thatn1(mod4)becauseot

14 herwisen2(mod4)andthen4n�nis
herwisen2(mod4)andthen4n�nisanidonealnumberwhichisnotsquarefree,contradiction.Wethusseethatinthiscaseweareincase(ii).Nowsupposethatthereisatleastoneidonealn�1848whichisnotsquarefree.Thenn0(mod4)byCorollary14.SincebyTheorem1therearenoidonealnumbersintherange1848n7392,wemanyassumethatn&#x-362;7392.ThenbyCorollary14wehavethatm=n 4�1848isasquarefreeidonealnumber,andsom=nandhencen=4n.Thusnistheuniquenon-squarefreeidonealnumberwithn�1848.Sincem2(mod4)byCorollary14,wethusseethatwearenowincase(iii).Thus,theabovecases(i){(iii)aretheonlypossibilities,andhencethereareatmost67=65+2idonealnumbers.Finally,ifGRHholds,thenTheorem22(b)showsthatcases(ii)and(iii)areimpossible,soweareleftwithcase(i).Thus(cf.Theorem1)wehave65idonealnumbers.Remark24(a)Frei[18]claimsthatWeinberger'sresultimpliesthatthereisatmostoneidonealnumbern�1848,andthisassertionisfrequentlyrepeatedintheliterature;cf.e.g.[47],p.339,340.However,aswasmentionedabove,thiscannotbetrueunlesswecouldshowthattherearenoevenidonea

15 lnumbersforn�1848.ItislikelythatF
lnumbersforn�1848.ItislikelythatFrei'serrorisduetohisearlierwrongassertionmentionedinRemark13.(b)Weinberger'sproofofTheorem22usesforbothpartstheassertionthatc()�1for�5400���d11,whered11=23531=200;560;490;130istheproductofthe rst11primes.(Thus,21011d112:11011.)Hementionsin[62]thatthisassertionwasveri edby(undocumented)machinecomputationsofLehmer.Forpart(b),however,onecanavoidtheseextensivecomputationsbyusingthesharperestimatesobtainedbyLouboutin[41]whoshowsthatunderGRHwehavethatc()&#x]TJ/;༕ ;.9;Ւ ;&#xTf 1;.64; 0;&#x Td[;1if0isafundamentaldiscriminantwithjDj2107;cf.[41],Theorem2.(Itiseasytoverifywiththehelpofacomputer(cf.x2.7)thattherearenoidonealdiscriminants0intherange105jj108.)(c)AswasmentionedintheproofofTheorem22,Weinberger'sstatementofhistheorem[62]usesinplaceofthehypothesis\c()=1"theequivalenthypothesis\E(Cl())2",whereE(A)denotestheexponentofanabeliangroupA.Thisallowshimtogeneralizehismethod

16 toprovethatthereareonly nitelymanyfu
toprovethatthereareonly nitelymanyfun-damentaldiscriminants0forwhichE(Cl())3.14 (cf.Remark7(b)and/orCox[11],p.59).Hereisanothertestwhichisveryusefulforcomputations:Proposition25Ifnisanidonealnumber,thentheLegendresymbolsatis esthecondition�n p6=1;foralloddprimespp n:(7)Proof.Ifnot,thenthereexistsanoddprimepp nwith�n p=1,andhencethecongruenceb2�n(modp)hasasolutionbwith0bp 2.Writeb2+n=cpforsomec�0.Sincep2n,weseethatc=b2+n p�b2+p2 p�p�2b.Sincebp 2p n 4p n 3,wethusseethatb2+n=pcisafactorizationwhichviolatesGrube'scriterionofProposition17(a),andhencencannotbeidoneal.Remark26(a)ThispropositioncanbeviewedasacorrectionofthetestusedbySwift[53]inhiscomputations.Inhistestheassertsthatifnisidoneal,then(7)holdsforp2n+1 4(p�1)2.This,however,isfalseifnhastheformn=p2�1,wherepisanoddprime,forthenclearlyp2=n+1n+1 4(p�1)2and(�n p)=(1 p)=1.Thus,theidonealnumbersn=8;24;48;120;168;840and1848(withp=3;5;7;11;13;29and43,respectively)showthattheboundpp nisq

17 uitesharpandthatSwift'sassertionisfalseu
uitesharpandthatSwift'sassertionisfalseunlesswemakefurtherassumptionsonn.Indeed,aswillbeshownbelow(cf.Corollary30),ifweassumethatn2;4,or6(mod8),thenSwift'scriterionistrue,evenforp24n+1.Notethatthiserrordoesnota ectthevalidityofSwift'scalculationsbecauseifnisnotidoneal,theninpracticeonecan ndalreadyaverysmallprimepwhichviolates(7).Swiftremarksthatforjj107thelargestprimeneededwasp=79,andinmycomputationsforn109(andn2(mod4))thelargestprimewasp=103.(b)Thistestcanbeextendedtoodddiscriminantsaswell;i.e.wehavethatifc()=1,(0),then( p)6=1forallprimesp1 2p jj.Indeed,suchastatement(forfundamentaldiscriminants)constitutesonepartoftheproofofWeinberger'sTheorem1;cf.[62],p.119.AnaturalquestioniswhethertheconverseofProposition25holds.Strictlyspeak-ing,theanswerisno:thenumbersn=19andn=43satisfy(7),yetarenotidonealbyCorollary8.Thus,extraconditionshavetobeimposed.Webeginwiththefol-lowingpartialconversetoProposition25.Sincethereisnoextraworkinvolved,wealsoincludethecaseofodddiscriminants:Proposition27Let0beafunda

18 mentaldiscriminant.If( p)6=1forallpr
mentaldiscriminant.If( p)6=1forallprimespwithp21 3jj,thenc()=1.16 Beforegivingtheproof,weobservethattheaboveresultimpliesthatSwift'stest(cf.Remark26)iscorrectwhenn2;4;6(mod8).ItalsoimpliesthattheconverseofProposition27holds,providedthat=�4n,wheren2(mod4).Corollary30(a)Ifn2;4;6(mod8)isanidonealnumber,then(�n p)6=1,foralloddprimespp 4n+1.(b)Ifn2(mod4)issquarefree,thenthefollowingconditionsareequivalent:(i)nisidoneal;(ii)(�n p)6=1,foralloddprimespq 4 3n;(iii)(�n p)6=1,foralloddprimespp 4n+1.Proof.(a)Supposethatpisanoddprimewith(�n p)=1.ThenbyProposition29thereexistt;k1suchthatnt2=k(p�k).Thenclearlykp,andbyreplacingkbyp�k,ifnecessary,wecanassumethatkp�1 2.Sincethefunctionk(p�k)isincreasingintherange1kp�1 2,weobtainthatnnt2=k(p�k)(p�1 2)(p+1 2)=1 4(p2�1),orp24n+1,andso(a)holds.(b)Theimplication(i))(iii)followsfrompart(a),thatof(iii))(ii)istrivial,and(ii))(i)isProposition27;cf.Remark28(a).ProofofProposition29.RecallfromRemark7(

19 b)thatnisidonealifandonlyifeveryprimitiv
b)thatnisidonealifandonlyifeveryprimitiveformofdiscriminant�4nisambiguous.Sincenisevenbutnotdivisibleby8,everyambiguousformisequivalenttoax2+by2(cf.[21],Art.257-8),soweseethatinthissituationnisidonealifandonlyifwehave:q2Cl(�4n))q(x;y)ax2+cy2;whereac=nand(a;c)=1.(11)Thus,itisenoughtoshowthat(11)isequivalentto(10).Suppose rstthat(11)holds,andletpbeanoddprimewith(�n p)=1.Then9b60(modp)suchthatb2�n(modp),sob2+n=pm,forsomem.Considerq(x;y)=px2+2bxy+my2.Then(q)=(2b)2�4pm=�4n,soq2Cl(�4n).(Notethatqisprimitivebecause(2b;p)=1.)Thus,bycondition(11)wehavethatqax2+cy2,forsomea;cwithac=nand(a;c)=1.Sinceqproperlyrepresentsp(forp=q(1;0)),itfollowsthatthisisalsotrueforq0:=ax2+cy2,sothereexistt1;t22Zwith(t1;t2)=1suchthatp=at21+ct2.Clearlyt16=0(forotherwisep=ct22=cjn,contradiction)andsimilarlyt26=0.Putk=at21.Thenp�k=ct22andsok(p�k)=at21ct22=nt2,wheret=jt1t2j�0.Thusproperty(10)holds.Nowsupposeconverselythat(10)holds,andletqbeaprimitivequadraticformofdiscriminant�4n.ByDirichlet'sTheorem([11

20 ],p.188),qrepresentsin nitelymanypri
],p.188),qrepresentsin nitelymanyprimesp,sothereisaprimep-4nrepresentedbyq.Foranysuchprimewehave(�n p)=1;cf.[11],p.30.18 Remark33Inhispaper,Weber[59]doesnotcomputej(p �n)directly(whennisidoneal),butinstead ndsexpressionsforpowersoff(p �n)(oroff1(p �n),f2(p �n)),wheref,f1andf2denotetheWeberfunctions(cf.[11],x12.B).Fromthisonegetsanexpressionforj(p �n)becausej()3= 2()=(f()24�16)=f()8;cf.[11],p.249,257.AsCox[11]remarksonp.263,Weber'smethodisbasedontheKroneckerLimitFormula.Coxworksoutthedetailsofthecomputationinthecasethatn=14toobtainj(p �14)3= 2(p �14)=2(323+228p 2+(231+161p 2)p 2p 2�1);cf.[11],x12.D.NotethatinTableVIattheendofhistextbook[60],Weberliststhevaluesof(powersof)theWeberfunctionsfor=p �nfor1n52andforalltheidonealnumbersn1848(togethersomeothervaluessuchasn=55;63;67;:::;163;193).CloselyconnectedwiththistheoremisthefollowingcharacterizationofidonealnumberswhichwasmentionedinFrei[18]withoutprooforreference:Corollary34Letn1.Thentheprimesnumbersof

21 theformp=x2+ny2withx;y2Zcanbecharacteriz
theformp=x2+ny2withx;y2Zcanbecharacterizedbycongruencerelationsmodulomforsomemifandonlyifnisanidonealnumber.Proof.Bygenustheory(andquadraticreciprocity)weknowthattheprimenumberspwhicharerepresentedbysomeformintheprincipalgenusgen(1)ofdiscriminant=�4ncanbecharacterizedbycongruenceconditionsmodulojj.Thus,ifnisidoneal,theassertionholdswithm=4n.Toprovetheconverse,weshallusethefollowingfact(provedinCox[11],p.189):theprimespwhicharerepresentedby1=x2+ny2are(upto nitelymanyexceptions)theprimespwhichsplitcompletelyintheringclass eldLnofZ[p �n].(Thisisthesame eldasthatoftheproofofTheorem32by[11],Theorem11.1).Supposenowthattheprimesprepresentedby1�4ncanbedescribedbycongruenceconditionsmodulom.ThentheprimeswhichsplitcompletelyinLnaredescribedbythesameconditionsmodm,andsobyclass eldtheoryoverQ,the eldLn=Qisabelian.ByTheorem32,thismeansthatnisidoneal.Remark35(a)Therealsoexistcharacterizationsofidonealnumbersand/oridonealdiscriminantsintermsofpropertiesofthebinarythetaseries#q(z)=Xx;y2Ze2iq(

22 x;y)zattachedtoabinaryquadraticformqofdi
x;y)zattachedtoabinaryquadraticformqofdiscriminant.Itiswell-known(cf.e.g.[37]andthereferencestherein)that#q2M1(jj;( ))isamodularformofweight1,leveljj,andNebentypus( ),where( )denotestheLegendre-Kroneckercharacter.By[37],Remark17(b),wehavethefollowingcharacterizationsofidonealdiscrim-inants0:20 Thefollowingresult(whichisarestatementofTheorem5of[35])givesthecon-nectionbetweencertainidonealnumbersandthesolutionoftheabovequestioninthenon-CMcase:Theorem37If(E1;E2)isapairofellipticcurvesoftypen,thenthefollowingconditionsareequivalent:(i)thereisnogenus2curveonE1E2;(ii)n=1ornisanevenidonealnumberwhichisnotdivisibleby8;(iii)n63(mod4)and4nisanidonealnumber;(iv)n2f1;4;12;18;28;60gorn2(mod4)isasquarefreeidonealnumber.Remark38Fromcondition(iv)weseethatthesetofidonealnumberssatisfyingtheequivalentconditionsofTheorem37equalstheunionofthesetf12;28;60gandofthesetofidonealnumbersappearinginTheorem1ofLouboutin/Newman[42];cf.Remark35(b).Proof(Sketch)ofTheorem37.Note rstthatconditions(ii)and(iii)a

23 reequivalentbyCorollary10,andthatconditi
reequivalentbyCorollary10,andthatconditions(ii)and(iv)areequivalentbyTheorem12.Inordertoprovetheequivalenceofconditions(i)and(iii),weshalloutlinethemainideasoftheproofandreferto[35]forthedetails.LetPirr(A)denotethesetofirreduciblegenus2curvesConA:=E1E2.Bytheadjunctionformula,theself-intersectionnumberofCisC2=2,soPirr(A)P(A):=f2Div(A):0;2=2g;whereDiv(A)denotesthegroupofdivisorsonA.HerethenotationPirr(A)re ectsthefact(duetoWeil)thatif2P(A),then2P(A)irrifandonlyifisanirreduciblecurve.Todecidewhetherornot2P(A)isirreducible,weshallusethere nedHumbertinvariantqde nedin[35](whichiscloselyrelatedtotheclassicalHumbertinvariantde nedbyHumbert).Thisisthequadraticformde nedbytheformulaq(D)=(D:)2�2D2;D2Div(A);(14)where(:)denoteintersectionnumbers.ItisclearthatqcanbeviewedasaquadraticformontheNeron-SeverigroupofA,i.e.onthequotientgroupNS(A)=Div(A)=,wheretheequivalencerelationD1D2(numericalequivalence)meansthat(D1:D)=(D2:D),8D2Div(A).Moreover,asho

24 rtcomputation(usingthefactthat2=2)s
rtcomputation(usingthefactthat2=2)showsthatqisactuallyaquadraticformonNS(A;):=NS(A)=Z,andtheHodgeIndexTheoremshowsthatqispositivede nite(onNS(A;)).22 Proof.SinceHom(E1;E2)=Zh,wheredeg(h)=n,theonlycyclicisogeniesfromE1toE2areh.Thus,by[19],Proposition6.5,thereexistintegersz;msuchthatz2n�1(modm)andsuchthat(inthenotationthere)wehaveC'C ,where =zhjE1[m].Thus,the rstassertionfollowsdirectlyfromCorollary6.6of[19].Toprovetheotherassertions,note rstthatthequadraticformq(x;y):=nrx2�2nzxy+my2hasdiscriminant(q)=4n2z2�4nrm=�4n,andthat(nr;2nz;m)=(r;m;2).Thus,qisprimitiveifandonlyif(r;m;2)=1.Thisisalwaystrueifn63(mod4),forif(r;m;2)6=1,thenz2n+10(mod4),son3(mod4).Ifqisprimitive,thenbyDirichlet'stheoremqrepresentsin nitelymanyprimesp,i.e.p=q(x;y).Ifp-n,thenalso(nx;y)=1,andthend=pisoneofthevaluesof(17).Notethatsince(q)=�4n,wehave(�n p)=1(ifp6=2);cf.[11],p.30.Now xpandwritep=q(x;y).Thentheproofof[19],Corollary6.6,constructsintegersr0andz0suchthatr0p=(z0)

25 2n+1andsuchthatC'C 0,where 0=(z0) jE1[p]
2n+1andsuchthatC'C 0,where 0=(z0) jE1[p].Supposetheequationt2n=k(p�k)hadasolutionk;t�0.Then(kz0)2�k2t2n(modp),andsotkz0(modp)(replacingkbyp�k,ifnecessary).Thusk 0=h0jE1[p],whereh0=th.Notethatdeg(h0)=t2deg(h)=t2n=k(p�k).Bythereducibilitytheoremof[34],Theorem3,thismeansC isareduciblecurve,contradiction.Thus,nosuchsolutiont;kcanexist,andso(18)holds.Remark40NotethattheaboveProposition39combinedwithTheorem37givesaquickproofofthefactthat(10)impliesthatnisidoneal,whichisthemoredicultdirectionofProposition29.Indeed,ifn2;4;6(mod8)werenotidoneal,thenbyTheorem37therewouldexistagenus2curveConE1E2(forsomepair(E1;E2)oftypen)andthenbyProposition39wewouldhavethat(18)holdsforin nitelymanyprimesp,whichviolates(10).Thusnisidoneal.Ofcourse,thisproofismuchmorecomplicatedthantheonegiveninsection3,butitpointedthewaytosuggestthecriterion(10).5GeneralizationsofidonealnumbersA rstgeneralizationofidonealnumberswasalreadytacitlyusedabove.Indeed,sinceidonealnumbersarethosenumbersn1forwhichc(�4n)

26 =1(Theorem6),theproblemofclassifyingalld
=1(Theorem6),theproblemofclassifyingalldiscriminants0whichsatisfyc()=1constitutesa rstgeneralizationofidonealnumbers.(DiscriminantswiththispropertyarecalledidonealdiscriminantsinBuell[8],p.193.)AswasnotedbyGrube,Chowlaandothers,theproblemofclassifyingidonealdiscriminantsreducestothatofclassifyingthosethatarefundamental.Thus,aswasmentionedabove,anequivalentproblemistoclassifythoseimaginaryquadratic eldsKwhoseclassgroupCl(K):=Cl(OK)isanelementary2-group;thiswastheapproachtakenbyWeinberger[62].24 where,asbefore,1denotestheprincipalformofdiscriminant.This,therefore,suggeststhefollowing(partial)generalizationofWatson'sproblem.Problem44Findalltheequivalenceclassesofpositivede niteprimitiveformsqinrvariableswhichsatisfy:q0!1;forallq02gen(q):(20)Indeed,inviewof(19),itisclearthatforr=2condition(20)holdsforq=1ifandonlyifc(1)=1,i.e.ifandonlyifisanidonealdiscriminant.Moreover,foranyrweseethatifq!1andc(q)=1,thenqsatis es(20).Thus,everysolutionqofWatson'sProblem41withq!1isasolutionofProblem44.Itisnotatal

27 lobviousthattheconverseholds,butTheorem4
lobviousthattheconverseholds,butTheorem46belowsuggeststhatthisistrueatleastinsomespecialcases.BeforestatingtheresultofTheorem46,letuslookatyetanothergeneralizationofidonealnumbersand/orforms.Forthis,we rstintroducethefollowingterminology.De nition.Letq(x1;:::;xr)beapositivede nitequadraticforminrvariables.Wesaythatqisanidoneal-valuedformifitsonly(proper)valuesj(q)jareidonealnumbers,i.e.ifq!nj(q)j)nisanidonealnumber;(21)here,thesymbolq!nmeansthatqproperlyrepresentsn,i.e.thatthereexistx1;:::;xr2Zwithgcd(x1;:::;xr)=1suchthatq(x1;:::;xr)=n.Inaddition,wesaythatqisaspecialidoneal-valuedformifq!nj(q)j)4nisanidonealnumber;(22)andifinadditionwehavethatq6!n,foranyn3(mod4)when(q)61(mod4),andthatq6!n,foranyn3(mod4)withnj(q)jwhen(q)1(mod4).Clearly,the1-variableformq(x)=nx2isidoneal-valuedifandonlyifnisanidonealnumberbecause(nx2)=n(cf.[55],p.2)andnistheonlyvaluewhichisprimitivelyrepresentedbyq.Moreover,nx2isaspecialidoneal-valuedformifandonlyifnsatis escondition(iii)ofT

28 heorem37.Thus,thisconceptgivesanothergen
heorem37.Thus,thisconceptgivesanothergeneralizationofidonealnumbers,andhencethefollowingproblemgeneralizesthesearchforidonealnumbers.Problem45Classifyalltheequivalenceclassesofpositivede nitequadraticformswhichareidoneal-valued.At rstsightitmightseemunlikelythatidoneal-valuedformsinr2variablesexistatall.However,suchformsdoexist,asfollowingresultshows.Italsoshowsthatthereisa(perhapssurprising)connectionbetweenProblems41,44and45.26 Theorem47LetE1E2betwoisogenousellipticcurvesoverK,andassumethatE1isnotsupersingular.Thenthefollowingconditionsareequivalent:(i)thereisnogenus2curveonE1E2;(ii)qE1;E2isaspecialidoneal-valuedform.Ifthisisthecase,theneitherqE1;E2(x)=nx2and4nisidoneal(non-CMcase)orqE1;E2isoneofthe15binaryformslistedinTheorem46(CMcase).Remark48NotethatifE1isnotisogenoustoE2,thenitiseasytoseethatthereisnogenus2curveonE1E2,sothiscasecanbeexcludedfromourconsiderations.Furthermore,thecasethatE1issupersingularwastreatedin[31].The rstpartoftheproofofthistheoremisverysimilartothatofTheorem37.LetqAbetheintegr

29 alquadraticformde nedbytheintersecti
alquadraticformde nedbytheintersectionpairingontheNeron-SeverigroupNS(A),i,e.qA(D)=1 2(D:D);ifD2NS(A):InthecasethatA=E1E2,wecanchooseabasisofNS(A)suchthatqE1E2xy�qE1;E2;(23)cf.[35],Proposition22.Forany2NS(A)suchthatqA()=1wehavetheassociatedre nedHumbertinvariantqA;de nedby(14).Thenbythesamereasoningasinthe( rstpartofthe)proofofTheorem37weobtainthefollowingresult.Proposition49IfA=E1E2,thencondition(i)ofTheorem47isequivalenttothecondition(i0)qA;!1,forall2NS(A)withqA()=1.Notethatthisreducesthegeometricproblemof ndingcurvesonthesurfaceAtoapurelyarithmeticprobleminvolvingthequadraticformqAxy�qE1;E2.Beforestatingthesolutiontothisproblem,letustreatatrivialcase:Lemma50If(qE1;E2)61(mod4)andthereisanintegern3(mod4)whichisrepresentedbyqE1;E2,thenthereisa2NS(A)suchthatqA;isnotprimitive.Inparticular,qA;6!1.Proof.(Sketch)Inthenon-CMcase,thisfollowsimmediatelyfrom(16).IntheCM-case,asimilarargument(using(16))yieldstheassertion.Inviewof

30 thislemmaandTheorem37,Theorem47followsfr
thislemmaandTheorem37,Theorem47followsfromthefollowingresultwhichsharpensTheorem46:28 Toverifythisobservation,note rstthatbyreplacingqbeanequivalentform,wemayassumewithoutlossofgeneralitythatqisreduced,i.e.thatjbjac.Sinceq!a;candB:=a+jbj+c,anda;c;Bj(q)j,weseethata;c;B2S.Nowifthereisad2SnS,thenbyEulerandGrube(Corollaries10and14)weknowthatdissquare-free,andsofromWeinberger'sTheorem22(a)itfollowsthatS=S[fdg,whered�108(bynumericalcomputations).Thus,ifc2SnS=fdg,thenB=a+jbj+c�c=d,soB=2S,contradiction.Thus,a;c462,soj(q)j=4ac�b244622106d,andhence(24)holds.Wethereforehavethata;b;ccantakeononly nitelymanyexplicitvalues.Byastraightforwardbutextremelytediouscheck(butsuitableforacomputer)onecanshowfromthisthatthe(reduced)specialidoneal-valuedq'sarepreciselytheq2L.In[36]amoreintrinsicargumentisgiven,butitstillrequirestheconsiderationofmanycases.(iv))(i):HereweshallapplythemassformulaofEisenstein/Smith/Brandt(cf.Brandt[5])totheternaryformQq.Thishas

31 theformXf2gen(Qq)1 jAut(f)j=�kd 6
theformXf2gen(Qq)1 jAut(f)j=�kd 62Ypj1�1 p2Ypjkd1+d p1 p1+�4kr p1 p;(25)wherek=gcd(a;b;c),d=(q) k2,=gcd(4k2;d)andrisanyintegerrepresentedbyq0:=1 kqwith(r;d)=1.Inaddition,2=4g(�16k)g(d),where,asinx2.2,g()denotesthenumberofgenera.Calculatingtheexpressionontherighthandsideof(25)we ndthatforeachq2Lthisexpressionequals1 jAut(q)j=1 jAut(Qq)j,andsoitfollowsthatc(Qq)=1,forallq2L.Thus(i)holds.Acknowledgments.IwouldliketothankClaudeLevesqueforhisinterestinthispaperandtothankhimandtherefereesfortheirmanyhelpfulcommentsandsuggestionswhichimprovedthepaper.Inaddition,IwouldliketogratefullyacknowledgereceiptoffundingfromtheNaturalSciencesandEngineeringResearchCouncilofCanada(NSERC).References[1]E.Bach,Explicitboundsforprimalitytestingandrelatedproblems.Math.Comp.55(1990),355-380.[2]P.Bachmann,DieanalytischeZahlentheorie.Teubner,Leipzig,1894.[3]A.Baker,Linearformsinthelogarithmsofalgebraicnumbers.Mathematika13,204{216.30 [18]G.Frei,LeonhardEuler'sco

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