On a few Diophantine equations, in particular, Fermat's last theorem

1、 ONAFEWDIOPHANTINEEQUATIONS,INPARTICULAR,FERMATSLASTTHEOREMC.LEVESQUEReceived 20October 2002This is a survey on Diophantine equations, with the purpose being to give theavourofsomeknownresultsonthesubjectandtodescribeafewopenproblems.We will come across Fermats last theorem and its proof by Andrew W

2、iles usingthemodularity ofelliptic curves, andwewillexhibit other Diophantine equationswhich weresolved laWiles. Wewillexhibit many families ofThueequations, forwhich Bakers linear forms inlogarithms andtheknowledge oftheunitgroups ofcertainfamilies ofnumber eldsproveusefulforndingalltheintegral sol

3、utions.Oneofthemostdicultconjecture innumbertheory,namely,theABCconjecture,will also be described. We will conclude by explaining in elementary terms thenotion ofmodularity ofanelliptic curve.2000 Mathematics Subject Classication: 11-02, 11-06.1. Introduction. On June 23, 1993, at the Isaac Newton I

4、nstitute of Cam-bridge(England), Professor Andrew Wiles(Princeton University) madeastrik-ingannouncement. HehadfoundaproofofFermats lasttheorem.Fermats last theorem. Let n be an integer greater than or equal to 3.ThentherearenononzerointegersA,B,C suchthatAn+Bn=Cn.(1.1)Themathematical community beca

5、meveryexcitedandthenewsspreadallovertheworldinamatterofdays.Formorethan350years,manymathemati-ciansandagreaternumberofamateurshadtriedwithoutsuccesstoobtainaproof ofthisconjecture. Already in1918, PaulWolfskehl poured some oilontherebypromisinga100000Germanmarksrewardtowhoeverprovidestherstproofofth

6、istheorem.Themarkdevaluationalmostannihilatedthevalueof the prize, but did not decrease the interest of specialists in this questionandsimilarproblems.In this survey article, we oer you an excursion over centuries into thisfantastic world of Diophantine equations. Though we will not deal with theequ

7、ationsconsideredbyMordellandbyShoreyandTijdemanintheirclassicalbooks 31, 39, we plan to make you more familiar with other Diophantineequations. You will encounter some equations which have integral solutions, 4474C. LEVESQUEsometimes a nite number, sometimes an innite number; you will also seesomeeq

8、uations whichhavenosolutionatall,andyouwillcomeacrosssomeequationsaboutwhichtheonlythingweknowisthatweknownothingaboutthem.WewillsayafewwordsabouttheFermatequationandinSection9,youwillgettheavourofWilesproof.WewillexhibitotherDiophantineequationswhich weresolved la Wiles, namely, byusing some modula

9、r elliptic curves.You will also see that Bakers linear forms in logarithms and the knowledgeoftheunitgroupsofsomefamilies ofnumber eldsprovedusefulinsolvingsomefamiliesoftheso-called Thueequations. Bytheway,thedelightsoftheABCconjecturemaymakeyourmouthwater:theassumptionofthisconjectureprovidesashor

10、tproofofFermatslasttheorem(forallnbutanitenumber).InSection10,wewilldaretoopenaparenthesis onthemodularity notion ofanellipticcurve,butwewillrushtocloseitinordertoavoidgettinginvolvedintechnicalities.2. Diophantus andFermat. TheGreekmathematician Diophantus (bornin325) got interested in nding soluti

11、ons of a given equation belonging to theset Q of rational numbers. However, under modern terminology, solving aDiophantine equation is looking for integral solutions, that is, for solutionsbelonging tothesetZofintegers.Ontheonehand,youmayagreewiththefactthattheequationX+Y =Z(2.1)iseasytosolve,butthe

12、rearestillopenproblemsconcerningthisequation(aswillbeseen inSection6).Ontheother hand, thesituation often getscompli-catedifpowersXn,wherenisanintegergreaterthanorequalto2,comeintoplay. Some solutions may be easily exhibited; for instance, a solution of theDiophantine equationX3+Y2=Z2(2.2)is X =2, Y

13、 =1, and Z =3, while another is X =3, Y =3, and Z =6. SomeDiophantineequationsmayhappentohavenointegralsolutionatall,liketheequation X22Y2=0.Theso-called Fermat-Pell equationX2DY2=1(2.3)hasbeenaroundformanycenturies,andthecontinued fractionexpansion ofDleadstoitssolution.ThisequationgoesbacktoArchim

14、edes(withthecattleproblem) andwasstudied bytheIndian mathematician Brahmagupta around1630andbytheEnglishmathematician WilliamBrouncker around1650. ON A FEW DIOPHANTINE EQUATIONS4475Figure 3.1Pierre deFermat (16011665) had acopy oftheLatin translation (made byBachet) of Diophantus book Arithmetica. Q

15、uite often, Fermat used to writepersonalnotesinthemarginofthisbook,andaLatinannotationofhim(oncetranslated) reads“Itisimpossibletowriteacubeasasumoftwocubes,afourthpoweras a sum of two fourth powers, and in general, a power (except asquare) asasumoftwopowers withthesameexponent. Ipossess atrulywonde

16、rfulproofofthisresult,thatthismarginistoonarrowtocontain.”This is equivalent to stating that the equations X3+Y3 =Z3, X4+Y4 =Z4,X5+Y5=Z5,andsoon,havenosolutions inpositiveintegralintegers.3. Pythagoras. Stating his theorem, Fermat assumed n3, precisely be-causeforn=2,theDiophantineequationX2+Y2=Z2ha

17、sintegralsolutions.Asamatteroffact(seeFigure3.1),allthesolutions oftheequationX2+Y2=Z2(3.1)are given by X = kU2kV2, Y = 2kUV, and Z = kU2+kV2, in which onesubstitutes any integer for k, U, and V. Indeed, we come across a very oldresult.Pythagoras theorem. Suppose that in a given rectangle triangle,

18、thelengthofthebaseisa,theheightb,andthediagonal c .Thena2+b2=c2.(3.2)TheproofappearsasPropositionXLVIIoftherstbookofEuclidsElements36,page38.Amodern proofofthistheorem istoletc=dinthestatementof a result called the parallelogram law (easy to prove with some use of thescalarproduct). 4476C. LEVESQUEa

19、dcbbaFigure 3.2c/2bmc/2aFigure 3.3Parallelogramlaw. Letaandbbethetwosidesofaparallelogram,andletc anddbethetwodiagonals (asinFigure3.2).Then2a2+2b2=c2+d2.(3.3)Theparallelogram lawimmediately leadstothemedianformula.Median formula. Let a,b,and c bethe sides ofatriangle and let m bethemedian(asinFigur

20、e3.3).Then2a2+2b2=4m2+c2.(3.4)Fortheproof,expandthistriangleintoaparallelogram (seeFigures3.3and3.4).OnendsinPropositionXLVIIIofEuclidsElements36,page39theproofofthefollowing result.ConverseofPythagoras theorem. Ifa2+b2=c2,thenthetriangleofsidesa,b,andc isrectangle withc asthehypotenuse.Thankstothem

21、edianformula,onecansupplyashortproof.Leta2+b2=c2;then2m2+(1/2)c2=c2,thatis,c=2m.Hence,thetwodiagonalsofFigure3.4areequal,andtheparallelogramisarectangle,thatis,ismadeoftworectangletriangles. ON A FEW DIOPHANTINE EQUATIONS4477ac/2mbbmc/2aFigure 3.4DCABFigure 3.5Babylonians applied theconverse ofPytha

22、goras theorem tobuildanangleof 90 degrees. Indeed, they used ropes having knots at intervals of the samelength andusedthemasinFigure3.1.Theyweresuretoobtain arightangleof90degreesbetweenthehorizontal lineandtheverticalline.Nowadays, when it comes to xing the wooden rectangle (wooden rail) onthe foun

23、dations of a building to be built (see Figure3.5), where the lengthbetweenAandDisequaltothelengthbetweenBandC,andwherethelengthbetweenAandBisequaltothelengthbetweenC andD,homebuildersmakesure that the length between A and C is equal to the length between B andD.Though theymaynotbeaware ofit,they“use

24、”theparallelogram law(seeFigure3.2)andmakesurethatc2 (=a2+b2)=d2,andthenusetheconverseofPythagoras theorem toconclude thatthetwogluedtriangles arerectangletriangles.Thanks to Pythagoras (and to the converse of his theorem), we can solveaproblem which became famous in San Francisco on July 28, 1993,

25、during apublicconference onFermat.Pizzaproblem. Theownerofarestaurant advertizes asmallpizzaat$6,amediumsizepizzaat$9,andalargepizzaat$15.Spending$15,doyougetabetterdealbybuyingalargepizza,orbybuyingasmallpizzaandamediumsizeone?Youmayuseonlyapizzaknifetomakeyourdecision. 4478C. LEVESQUEstrFigure 3.6

26、t2s2r2Figure 3.7Justcutthethreepizzasintwoequalpartsandplacethethreehalfpizzasofdierent sizesastoformatriangle.Threecasescanoccur.Case3.1. Ifyougetarectangletriangle(Figures3.6and3.7),yourchoiceisasgoodasminesince8r2+8s2= 8t2, thatis, r2+s2=t2.(3.5)Case3.2. Ifthetriangle isobtuse(Figure3.8),yourbest

27、dealistotakethelargepizzasince8r2+8s2 8t2, thatis, r2+s2t2.(3.6)WithFigures3.9and3.10,oneseesthatr2+s2=r2+u2+v2 8t2, thatis, r2+s2t2.(3.8)Figures3.12and3.13showthatr2+s2=r2+u2+v2(ru)2+v2=t2.(3.9)4. Some Diophantine equations. Diophantine equations are often myste-rious. Two very similar equations ma

28、y have very dierent solution sets, andit may happen that one is dicult to deal with, and the other one is easy tostudy.Wejustsawthatforn3,theequationXn+Yn=Znhasnonontrivialsolution.Nevertheless, foralln1,theequationXn+Yn=2Zn(4.1)possesses the positive solution X=Y =Z=1.Are there more for n3?Wewillse

29、elaterthattheanswerisno,thoughtheproofisdeep.Weexplainwhy,forn2,theDiophantine equationX2+Y2=Zn(4.2)has an innite number of (nontrivial) solutions, and that it is easy to nd allofthem.Foralln0,letAn andBn bedenedbyAn+Bni=(a+bi)n,(4.3)wherei= 1andwhereaandbarevariablesrunningthroughZ.Ontheonehand,weh

30、ave An+Bni An+Bni = An+Bni AnBni =An2+Bn2,(4.4) ON A FEW DIOPHANTINE EQUATIONS4481t2s2r2Figure 3.12t22u2(r u)2rrFigure 3.13wherezisthecomplexconjugate ofz.Ontheotherhand, An+Bni An+Bni =(a+bi)n(a+bi)n=(a+bi) n (abi)n(4.5)n= a2+b2.Hence A2n+Bn2 =(a2+b2)n. Thus the equation X2+Y2 =Zn has an innitenumb

31、er of solutions given by X = A , Y = B , and Z = a2+b2, where thenn 4482C. LEVESQUETable 4.1n012345678An1Bn0ab2aba2b2a33ab23a2bb3a46a2b2+b44a3b4ab3a510a3b2+5ab4a615a4b2+15a2b4b65a4b10a2b3+b56a5b20a3b3+6ab57a6b35a4b3+21a2b5b78a7b56a5b3+56a3b58ab7a721a5b2+35a3b47ab6a828a6b2+70a4b428a2b6+b8binomial exp

32、ansion of(a+bi)n gives(1)san2sb2s,(1)san12sb2s+1.n/2(n1)/2n2sn2s+1An=Bn=s=0s=0(4.6)Itturns outthatallthesolutions areofthatform, andthisisjustied bythefactthattheclassnumberoftheimaginaryquadraticeldQ(i)is1.WewriteafewvaluesofAn andBn inTable4.1.Itisnoweasytodeducethataninnitefamilyofsolutions ofX2Y

33、2=Zn(4.7)(4.8)(4.9)isgivenbyX=Rn,Y =Sn,andZ=a2b2,wheren/2(n1)/2n2sn2s+1Rn=an2sb2s,Sn=an12sb2s+1.s=0s=0According toEuler,theequationW3+X3+Y3=Z3hasaninnityofsolutions; itsucestoconsiderW =a3+3a2b3ab2+9b3+1,X=a3+3a2b+3ab2+9b31,Y =a4+6a2b2a+9b43b,Z=a4+6a2b2a+9b4+3b,(4.10) ON A FEW DIOPHANTINE EQUATIONS4

34、483orW =3a2+5ab5b2,X=4a24ab+6b2,Y =5a25ab3b2,Z=6a24ab+4b2.(4.11)ThankstoElkies12,weknowthattheDiophantine equationW4+X4+Y4=Z4(4.12)alsopossesses aninnite number ofsolutions (with W,X,Y,andZ dierentfrom0),withthesmallest,accordingtoR.Frye,beingW =95800,X=217519,Y =414560,andZ=422481.WerstdealwithW4+X

35、4+Y2=Z4.(4.13)Elkies 12 exhibited an innite family of solutions of the latter Diophantineequation:W =2a2+6a+20,X=a2+31,(4.14)Y=4 2a4+28a275a+80 ,Z=3 a2+11 .Using a judiciously chosen elliptic curve, he next showed that there exist aninnitenumberofintegersa(eectively computable) forwhichY isaperfects

36、quare, thus giving rise to an innity of solutions of the equation W4+X4+Y4=Z4.TheequationV5+W5+X5+Y5=Z5(4.15)possesses, for instance, the solution V =27, W =84, X =110, Y =133, andZ=144,thoughwestilldonotknowwhetheritpossessesaninnitenumberofsolutions.Alongthesamelines,sofarnobody couldexhibitaninte

37、ger n(withn6)andn1nonzerointegerssuchthatthesumofthenthpowersofthesen1integersisannthpower.Thecasewheren=6readsU6+V6+W6+X6+Y6=Z6(4.16)andnosolutionwithUVWXYZ0isknown.Weconclude thissectionwiththreeotheropenproblems. 4484C. LEVESQUEProblem 4.1. Does there exist a perfect rectangular box (Figure4.1),

38、thatis, a rectangular box such that the lengths of the three sides and of the fourdiagonals areintegers? Inother words, wedonotknow whether ornottherearepositiveintegersa,b,c,d,e,f,andgsuchthata2+b2=d2,a2+c2=f2,b2+c2=e2,b2+f2=g2.(4.17)If we do not require the interior diagonal to be integral, such a

39、 box exists:simplytakea=117,b=44,c=240,d=125,e=244,andf =267.fegdbcaFigure 4.1Problem4.2. Doesthereexistaperfectsquare(Figure4.2),thatis,asquarewithsidesoflengthA,havinganinteriorpointrespectively atdistances B,C,D, and E from the four corners such that A, B, C, D, and E are all positiveintegers?ABC

40、AAEDAFigure 4.2 ON A FEW DIOPHANTINE EQUATIONS4485Problem 4.3. Does there exist aperfect triangle (Figure4.3), that is, atri-angle such that the sides A,B,and C,the medians D,E,and F,and the areaareallpositiveintegers?CFBDEAFigure 4.35. First attempts on Xn+Yn =Zn. Between 1640 and 1850, a few math-

41、ematicians, Fermat, Euler, Lejeune Dirichlet, Legendre, Lam, and Lebesgue,successfully studiedtheequationXn+Yn=Zn(5.1)for n = 3,4,5,6,7. In 1857, Kummer settled Fermats conjecture for all theexponents n100.In 1983, a major breakthrough was made by Faltings 13 when he provedthat foraxed n4,theequatio

42、n Xn+Yn=Zn hasonly anite number ofsolutions (withnocommondivisors). Asamatteroffact,Faltingsobtained in1986theFieldsMedalforhavingprovedtheMordellconjecture:everysmoothalgebraic curve of genus g 2 over any given algebraic number eld K hasanite number ofK-rational solutions.Ifone views analgebraic cu

43、rveasaRiemann surface, the genus ofis the number of holes. Since for n4, theFermat algebraic curve Xn+Yn =Zn is of genus (n1)(n2)/2, the curvehasonlyanitenumberofpositiveintegralsolutionscoprimetooneanother.Weknowsince1993thatFermatslasttheoremistrueforn4000000;thiswas established with the help of c

44、omputers. Moreover, a result of K. Inkeriimplies that if there exist integers C B A 1 such that An+Bn = Cn,thenA400000011999996.Thislastintegerissobigthatifwewantedtowriteit at full length, it would require more than 70 million digits, which wouldmake itclose to 100 kilometers long (with 6digits per

45、 centimeter). However,Wiles wanted to prove Fermats last theorem denitely without the help of acomputer, andsohedid! 4486C. LEVESQUE6. A small detour: the ABC conjecture. As odd as this may look, thereexistsanopenproblemconcerning theequalityA+B=C,(6.1)and itiscalled the ABC conjecture. This isone o

46、fthedeepest conjectures inmathematics anditisfarfrombeingproved, though manyexperts thinkitistrue.TheABC conjecture,formulatedin1985byJ.OesterlandD.W.Masser,providesanupperboundfor|C|intermsoftheproductoftheprimedivisorsof ABC.More precisely, rst choose a real number 0 (e.g., =0.000001).Nextsupposet

47、hatA+B=C, whereAandBhavenodivisorincommon.ThentheABC conjecture assertstheexistence ofaconstant M (depending onlyon)suchthat|C|MR1+.(6.2)Assuming the ABC conjecture, one can give ashort proof of the existenceof a (noneective) constant N such that Fermats last theorem is true for allnN.Here ishow itg

48、oes. Choose and x with, for instance, 0ba0,coprimetooneanother, such thatan+bn=cn.PutA=an,B=bn,andC=cn .ThentheABC conjectureguaranteestheexistenceofaconstantM (dependingonlyon)suchthat1+anbncnMp.(6.3)(6.4)p|abcHence3+3(abc)n0.DenotebyEwhatwasC,thatis,A+B=E;use the letter C for what was R, and take

49、=1. Then the conjecture statesthat EMC2, and the relativity of this shaky conjecture will certainly scarephysicists.7. SomegeneralizedFermatequations. Foralongtime,itwasconjecturedthat 8 and 9 are the only consecutive powers. Many mathematicians contrib-uted numerous partial results till this so-cal

50、led Catalan conjecture was o-ciallyprovedbyMihailescu(see30or4),thankstoacleveruseofthearith-meticofcyclotomic elds.Theresultcanbestatedinthefollowing terms. ON A FEW DIOPHANTINE EQUATIONS4487(7.1)Theorem 7.1. TheonlypositivesolutionoftheDiophantine equationXm+1=Yn,withm,n2,is(X,Y,m,n)=(2,3,3,2).Usi

51、ng deep mathematics, namely, elliptic curves la Wiles, Darmon andMerel11solvedsomevariants oftheFermatequation. TheyprovedthattheDnesconjectureistrue:forn3,theonlypositivesolutionoftheDiophantineequationXn+Yn=2Zn,(7.2)with XYZ =0 and gcd(X,Y,Z) =1, is X =Y =Z =1. They also proved thefollowing: forn4andforq2,3,theDiophantine equationXn+Yn=Zq(7.3)hasnointegralsolutionwithgcd(X,Y,Z)=1andXYZ=0.We justify their hypothesi