TheDiophantineequation

1、 thediophantineequationax2+2bxy4ay2=±1lionel bapounguÉreceived 30april 2002we discuss, with the aid of arithmetical properties of the ring of the gaussianintegers, the solvability of the diophantine equation ax2+2bxy 4ay2 = ±1,where aandbarenonnegative integers. thediscussion isrelati

2、ve tothesolutionofpells equation v2(4a2+b2)w2=4.2000 mathematics subject classication: 11d09.1. introduction. the objective of this paper is the expansion and also theextensionof1,section2.moreprecisely,itdealswiththecompletetreatmentofthesolvability ofthediophantine equationax2+2bxy4ay2=±1,(1.

3、1)where a and b are positive integers. from 2, proposition 1, (1.1) is alwayssolvable if a = 1. hence, we may assume that a > 1. moreover, we restrictoneselftogcd(a,2b)=1.intheopposite case,(1.1)isinsolvable.we denote by =4a2+b2 the discriminant of the quadratic form ax2+2bxy4ay2.if a>1, b 0,

4、and gcd(a,2b) =1, 2, theorem 1 shows that (1.1) is in-solvable ifisasquare in z.hence, we will assume also that =4a2+b2 isa nonsquare in z, which requires b to be odd. then veries 5(mod8).thus,(cf.4)thealgebraicintegersofq( )arethenumbers(1/2)(v+w)withv,wzofthesameparity.consequently, if(1/2)(v+w )i

5、saunitofq( ),wemusthavev2w2=±4.(1.2)conversely, if(v,w) isaninteger solution of(1.2),then(1/2)(v+w)isaninteger ofq( )(itstrace isv anditsnorm, by(1.2), is±1)and,hence, a00unitofq( ).writing (1/2)(v +w )forthefundamental unitofq( ),weseethatthesolutionsinpairsofnaturalnumbers (v,w) of(1.2)c

6、omprisethevaluesofthesequence (vn,wn)(n1)denedbysettingn 12v0+w0vn+wn=.(1.3)2 2242lionel bapounguÉhence,weremarkeasily(cf.2)thatif(x,y) isasolutionof(1.1),thenv=bx2+8axy+4by2,w=x2+4y2(1.4)verifywith=4a2+b2 (a>1, b1, and gcd(a,2b)=1)thepellsequationv2w2=4 withv,w odd.(1.5)hence, our study wil

7、l be based on (1.5). thus, assuming its solvability, wegive in section2 a necessary and sucient condition for (1.1) to be solvable(theorem 2.3) by methods using the arithmetic of the order z2i of index2, included in the principal ring z+zi. in the remainder of section2, weestablish thatif(1.1)issolv

8、able, then±aisthenormofanelement ofz (proposition 2.6). next, we prove insection3that when isgiven, among allthepairsofpositivecoprimeoddintegers(a,b)satisfying=4a2+b2 ,thereisexactly onepair forwhich (1.1) issolvable (theorem 3.1). that unique pairwill be constructed (theorem 4.1) in section4

9、with the aid of the followingresultprovedin5.theorem1.1(thue). ifandareintegerssatisfying>1,gcd(,)=1,andmtheleastintegergreaterthan,thereexistxandyin0,msuchthaty±x(mod).when solutions exist, we show using any of them in section5 that (1.1)possessesaninnityofsolutions(theorem 5.1);afterwards,

10、wedescribeusingitafamily (proposition 5.2). wegive theconclusion ofourpaper insection6withsomenumerical examples.2. thecasea>1,oddnonsquare, and(1.5)solvable withv,w odd2.1. preliminaries. letv,wbeodd integersgreaterthanorequalto1suchthatv2w2=4.itisclearthatgcd(v,w)=1(2.1)becauseifv andw haveacom

11、monprimefactord,thenddividesv2w2=4and,therefore, ddividesalso2.write(1.5)intheformw2=(v+2i)(v2i).(2.2)thetwofactorsoftheright-handsideof(2.2)arerelativelyprimeinzisinceany common divisor would divide 4i, but w is odd, hence, gcd(w,4i) = 1.hence,inzi,wehavegcd(v+2i,v2i)=1.(2.3)moreover, since (1.5) i

12、s written in the form (2.2), we will manipulate the ele-ments of the nonmaximal order z2i of index 2, for which we have shown the diophantine equation ax2+2bxy4ay2=±12243(2.4)in3thatthehalfgroupfdenedbyf= v+2iz2i:gcd n(v+2i),2 =1isfactorial,wheren() denotesthenormof.thus,theremarkof2,propo-siti

13、on4appliedtofenablesustostatethefollowing denition.denition 2.1. anoddsolution(v,w)z2 of(1.5)issaidtobe(i) violainif,inf,b+2aidividesv+2iorv2i;(ii) monicif,inf,b+2ai=gcd(v+2i,)orgcd(v2i,).proposition 2.2. anyoddviolainsolution(v,w)z2 of(1.5)ismonic.2.2. onecriterionofsolvabilityfor(1.1). weprovethef

14、ollowingtheorem.theorem 2.3. ifa3andb1areoddintegers with gcd(a,2b)=1and=4a2+b2 nonsquare inz,thefollowing statements areequivalent:(i) (1.1)hasasolution(x,y)z2;(ii) (1.5)hasanoddviolainsolution(v,w)z2;(iii) theoddminimalsolution(v ,w )z2(v >0,w0>0)of(1.5)ismonic.000proof.(i)(ii).let(x,y)z2 be

15、asolutionof(1.1).weset=sgn ax2+2bxy4ay2 ,(2.5)v= bx28axy4by2 ,w=x2+4y2.asbandx areodd,v andw arealsoodd.thenwehavev+2i= bx28axy4by2 +2 ax2+2bxy4ay2 i(2.6)(2.7)sothatv+2i=(b+2ai)(x+2iy)2,whereweseethat(v,w)z2isanoddviolainsolutionof(1.5).further,takingthenormofthetwosidesof(2.7),weobtainx2+4y22=w2v2+

16、4= b2+4a2(2.8)sothat(v,w) isanoddintegersolutionof(1.5).(ii)(iii). let (v,w) z2 be an odd integer violain solution of (1.5). thenfromequality(2.7),wehavegcd(v+2i,)=(b+2ai)gcd (x+2iy)2,b2ai .(2.9)now,weshowthatgcd (x+2iy)2,b2ai =1.(2.10) 2244lionel bapounguÉifthereexistsf,isnotaunit,thatis,±

17、;1suchthat|x+2iy,|b2ai,(2.11)(2.12)thenas(2.7)impliesv= bx28axy4by2 ,w=x2+4y2,wededucedthat,inf,v b(2iy)28a(2iy)y4by28y2(b2ai)0(mod),w0(mod),(2.13)thatis,isalsoadivisorofvandw,contradictingthefactthatgcd(v,w)=1according to(2.1).hence,wehavegcd(v+2i,)=b+2ai.(2.14)thenweshowthat(2.14)istrueforv0arisin

18、gfromtheoddminimalsolutionof(1.5).as(v,w)isanoddsolutionof(1.5),wehavebythetheoryofthepellianequation2n+1v0+w0,ifv>0,2v+w=(2.15)2n+12v0w0,ifv<0,2forsomeintegern0.developing (2.15),weobtain2n+12n1w +···,20vifv>0,024nv=(2.16)n+1v02n1w02···, ifv<0,2n+1v022where

19、thetermsarealldivisiblebyexceptv02n+1.hence,asv024(mod),wehave( )nv0(mod),1ifv>0,v(2.17)(2.18)(1)n+1v0(mod), ifv<0.from(2.14)and(2.17),wededucethatb+2ai=gcd(v+2i,)=gcd ±v0+2i, =gcd v0±2 i,asrequired. thisprovesthat(v0,w0)isamonicsolutionof(1.5). the diophantine equation ax2+2bxy4ay2=

20、±12245(iii)(i).supposethat(v0,w0)isamonicsolutionof(1.5).theequalityv02w02=4(2.19)maybeexpressed intheformv0+2iv02ib2ai=w ,2(2.20)b+2ai0where,from(2.3),(v0+2i)/(b+2ai)and(v02i)/(b2ai)arecoprimeinf.hence,forsomeunit=±1andintegersx,y,wehavev0+2ib+2ai =(x+2iy)2,w0=x2+4y2.(2.21)takingv0+2i=(b+

21、2ai)(x+2iy)2(2.22)(2.23)andequating coecients ofionbothsidesof(2.22),weobtainax2+2bxy4ay2=,showingthat(x,y)z2 isasolutionof(1.1).remark 2.4. theproofabovealsoconrms thefollowing result.theorem 2.5. ifa3andb1areoddintegers with gcd(a,2b)=1and=4a2+b2 nonsquare inz,thefollowing statements areequivalent

22、:(i) (1.1)hasasolution(x,y)z2;(ii) (1.5)hasanoddviolainsolution(v,w)z2;(iii) (1.5)hasanoddmonicsolution(v,w)z2.wehavealsothefollowing proposition.proposition 2.6. let a3and b1beodd integers with gcd(a,2b)=1and = 4a2+b2 nonsquare in z. if the diophantine equation (1.1) has anysolution(x,y)z2 ,then

23、77; 14n x + , µ, z.a=±n y +µinotherwords,±a(resp.,±4a)isthenormofanelementofz.or(2.24)proof.wesupposethat(x,y)z2isanysolutionof(1.1).thentheequa-tionat2+2bty4ay2=0 (=±1)(2.25)(2.26)hasanintegerroot,henceitsdiscriminant isasquareinz:b2y2+4a2y2a=µ2, µz, 2246lion

24、el bapounguÉwhencea=y2 b2+4a2 µ2=y2µ2, µz,(2.27)(2.28)(2.29)sothata=n y +µ , µz.exchanging therolesofx andy,weobtainalso4a=n x + , z.3. uniqueness ofthepair(a,b),given. weassumeinthissectionthatisgivenandcanbefactorized intoseveralsumsoftwosquaresinz.thenweuse theorem 2

25、.3 to show that among all the pairs of positive integers (a,b),thereisexactlyonepairforwhich(1.1)issolvable.theorem3.1. letbeanoddnonsquarepositiveintegerforwhich(1.5)hasanoddsolution (v,w)z2.then among allthepairs ofoddpositive coprimeintegers (a,b) satisfying =4a2+b2,there is exactly one pair (a,b

26、)=(a,b)suchthat(1.1)issolvable.proof.itiveintegersaandb asfollows:let(v,w)z2 beanyoddviolainsolutionof(1.5).wedenepos-a=|a|,b=b.(3.1)letg=gcd(a,b) .thenv+2i=g(+2i) 1=g,(3.2)henceg=1.since(v,w)z2isanoddsolutionof(1.5),wehave5(mod8),andthusaandbareodd.hence,wehavegcd(a,b)=1(3.3)withbothaandb allodd.th

27、enweshowthat=4a2+b2.fromthedenition ofaandb,weseethatb+2ai|v+2iorv2i .hence,wemayassume,forexample,thatb+2ai|v+2i.then(1.5)maybeexpressedintheformv+2i (v2i)=b+2aib+2aiw2,(3.4)where (v+2i)/(b+2ai)and/(b+2ai)arecoprime elements off(since v, and b are odd). equation (3.4) shows that /(b+2ai) divides v2

28、i , but the diophantine equation ax2+2bxy4ay2=±1/(b+2ai)alsodivides,therefore /(b+2ai)divides2247(3.5)gcd(v2i,)=b2ai,inf,andso|b2+4a2.(3.6)ontheotherhand,sinceb+2ai|,takingconjugates, weobtainb2ai|.letzibeanyprimefactorofb+2aiandb2ai.thenwehaveb2+4a2.(3.7)since(v,w) isanyoddviolainsolutionof(1.

29、5),wehave v+2i v2i,b2ai,(3.8)b+2aiandasvb(mod2),2,lemma2appliedtozishowsthat=1,thentherelation(3.7)becomesb2+4a2|.(3.9)thus,=4a2+b2followsfrom(3.6)and(3.9).hence,=4a2+b2isadecom-positionofwhichsatisesstatement(ii)oftheorem 2.3.so,bytheorem2.3,theequationax2+2bxy4ay2=±1(3.10)issolvable. aandb ar

30、eunique.applying theorems 2.3(or2.5)and3.1,weobtainthefollowing corollary.corollary 3.2. if 5(mod8) is a prime number for which (1.5) is solv-able, d, e denote integers such that = 4d2+e2 (they are odd, unique, andpositive), thenthediophantine equationdx2+2exy4dy2=±1(3.11)issolvable.proof.this

31、results from the fact that any prime number of the form4m+1mayberepresented asthesumoftwosquares(cf.5).4. construction of (a,b), given. we show in this section how the pair(a,b) canbeconstructed. 2248lionel bapounguÉtheorem 4.1. letbeanoddnonsquare positive integer suchthat(1.5)issolvableinoddi

32、ntegers(v,w)z2.thenthereexistsauniquepairofcoprimeoddintegers(a,b)satisfyingb±av0(mod),0<a<,0<b<,(4.1)=4a2+b2.then,forthatuniquepair(a,b),(1.1)issolvablein(x,y)z2.proof.andb>0suchthattaking=vintheorem 1.1,weseethatthereexistintegersa>0b±av0(mod), a< , b<.(4.2)(4.3)(

33、4.4)since(v,)=1,wehaveb2+4a2a2v2+4a24a2+4a20(mod).but0<b2+4a2<5,hencetheequationsb2+4a2=2,3,4are insolvable in z since, modulus 4, the rst and the second congruencesb22,3areimpossible andthethirdimposesbtobeeven.hence,wehave=4a2+b2,(4.5)which veries 5(mod8)since (v,w) z2 isan odd solution of(1

34、.5), andthusaandbarebothodd.next,weshowthatif(a,b)satises(4.2)and(4.5),thengcd(a,b)=1.letg=gcd(a,b),andseta=ga,b=gb.then(4.5)becomes(b)2+4(a)2=1(4.6)with 1 = /g2. relations (4.2) show that there exists z such that b =±av+,andthusb=±av+g .replacing b in(4.6),weobtain1±av+g2+4(a)2=1,(4.

35、7)(4.8)1and,using(1.5),wededucefromitthatg w2(a)2±2av+2g1 =1,provingthatg=1. the diophantine equation ax2+2bxy4ay2=±12249now, we show that (a,b) is unique. we suppose that (a1,b1) is anothersolution of(4.2).thenfromcongruencesb+avb1+a1v0(mod)bavb1a1v0(mod),(4.9)(4.10)(4.11)(4.12)(4.13)orwe

36、seethatbb1+4aa10(mod),ab1a1b0(mod).fromtheproductofthetwofollowing expressions:b2+4a2=,b12+4a21=,wededucethatbb1+4aa1 +4 ab1a1b =222sothat(dividing by2)bb1+4aa1ab1a1b22=1,+4(4.14)whichgivesbb1+4aa1=±,relations (4.15)imposeab1a1b=0.(4.15)(4.16)a1,b1 =±(a,b).thus,thereexistsauniquesolutionof

37、(4.2)satisfying a>0,b>0.finally, our last assertion is to prove that (a,b) dened by (4.1) satisestheorem 2.3(ii).wesupposethatb±av0(mod).(4.17)(4.18)(4.19)asv24(mod),multiplying (4.17)byv,weobtainbv±av2±4a(mod),andso2±b+aviv±2i = bv±4ab+2ai 2250lionel bapoungu

38、1;isanelementoff .thusb+2ai|v±2i.(4.20)this proves that (v,w) is an odd violain solution of (1.5). so, from theorem2.3,(1.1)issolvableinintegers(x,y)z2.remark 4.2. denoting by (v0,w0) the odd minimal solution of (1.5), wecan easily determine a and b, such that b+2ai is the gcd(v0± 2i,) , u

39、singthefollowing algorithm:(1)(i) factorize inz;(ii) calculate thenormofv0+2iandfactorize itinz;(2) factorize theprimefactorsobtained inf;(3) deducefrom(2)thecommon divisorsofandv0+2i.5. complete set of solutions of (1.1). first of all, we prove the followingtheorem.theorem 5.1. undertheconditions o

40、ftheorem 2.3,thediophantine equa-tion(1.1)hasaninnityofsolutions inz.proof.gcd(x0,y0)=1,hencethereexists(a,b)z2 suchthatwe assume that (1.1) has a solution (x0,y0)z2. then we haveax0+by0=1.(5.1)(5.2)weset= xyx0 xy,y0g(x,y)=f(x,y)=ax2+2bxy4ay2.thengandf aretwoequivalent quadratic forms.furtherg(x,y)=

41、x22bxy cy2, b,cz,(5.3)(5.4)(5.5)with=ax02+2bx0y04ay02,b=(ab)x0+(b+4a)y0,c=a2+2b+4a2.buttheequationsf(x,y)=,g(x,y)= the diophantine equation ax2+2bxy4ay2=±12251(5.6)areequivalent, henceasg(x,y)=n(xy),whereisarootoftheequationt2bt+c=0,(5.7)weconcludethatiff(x,y)=hasasolutioninz,ithasaninnityofsol

42、utionsinz.now,wedescribethefamilyofsolutions of(1.1).proposition 5.2. under the conditions of theorem 2.3, let (x0,y0) be aparticular solutionof(1.1).thenthesetofsolutions (x,y) of(1.1)isgivenby3nv0+w20ax+by+y =±ax0+by0+y0(5.8)inwhich(v0,w0)istheminimalsolutionof(1.5)andnz.let (x0,y0) be a part

43、icular solution of (1.1). we show how all theproof.solutions(x,y) of(1.1)maybeobtainedintermsof(x0,y0)andtheminimalsolution(v0,w0)of(1.5).if(x,y) isanysolutionof(1.1)andifwesetax+by+yj=,(5.9)ax +by +y000thenormofj is(ax+by)2y2a ax2+2bxy4ay22 =,(5.10)2a ax2 2bx y4ay2ax +byy0+000000thatis,±1.more

44、over, j isoftheformd+e,wheredande areintegersgivenbyd=axx0+b xy0+x0y 4ayy0,e=x yxy .(5.11)00hence,bythetheoryofthepellianequation, wehavej=±,(5.12)3nv0+w02wherenz.thus,wehaveshowntheexistence ofaninteger nsuchthatwehave(5.8).conversely, let x and y be dened by (5.8) for some nz. taking normsofb

45、othsidesof(5.8),weseethatxandy verify(1.1).itremainstoshowthattheyarebothintegers. 2252lionel bapounguÉdeneintegersm andn bym+n =±.(5.13)3nv0+w2 0thenequating coecients in(5.8),weobtainax+by =m ax0+by0 +ny0,(5.14)(5.15)y=my0+n ax0+by.0clearly,yz.using=4a2 +b2,weobtainx=(mbn)x0+4any0,sothat

46、x isalsoaninteger.6. numerical examplesexample6.1. ifa=19andb=71,then=4(19)2+712=64855(mod8)isnonsquaresuchthat(v ,w )=(1369,17)istheminimalsolutionof(1.5).in00thiscase,gcd(1369+2i,6485)=71+38iandtheorem 2.3showsthat(1.1)issolvable; infact,itis19x2+142xy76y2=1(x,y)=(1,2)isasolution).example 6.2. if

47、a = 3 and b = 5, then = 4(3)2+52 = 61 5(mod8)is prime such that (v,w) =(39,5) is a solution of (1.5); corollary 3.2 showsthat (1.1) is solvable; in fact, it is 3x2+10xy12y2=1 (x,y) =(1,1) is asolution).example 6.3. in case =2941=4(25)2+212=4(27)2+52, we have 5(mod8)nonsquare suchthat(v ,w )=(705,13)

48、istheminimal solution of00(1.5).asgcd(7054+2i,2941)=2150i,theorems2.3and3.1showthat(1.1)issolvable onlyinthecasewhen(a,b)=(25,21);infact,itis25x2+42xy100y2=1(x,y)=(3,1)isasolution).theequation 27x2+10xy108y2=±1isinsolvable.example 6.4. take =3077. we have 5(mod8) nonsquare such that(v ,w )=(943

49、,17) is the minimal solution of (1.5). the candidates for the00unique pair (a,b) satisfying (4.1) must be solutions of =4a2+b2. that is,(a,b) = ±(13,49),±(23,31). the only pair satisfying b+av 0(mod) is0(a,b)=(13,49)sothat(13,49)istheunique pairforwhich (1.1) issolvable;infact,itis13x2+98x

50、y52y2=1(x,y)=(1,2)isasolution).theequation 23x2+62xy92y2=±1isinsolvable.references1l. bapoungué, sur la résolubilité de léquation diophantienne ax2 +2bxy kay2 =±1 on the solvability of the diophantine equation ax2+2bxy the diophantine equation ax2+2bxy4ay2=±12253ka

51、y2 =±1, c. r. acad. sci. paris sér. i math. 309 (1989), no. 5, 235238(french).23, un critère de résolution pour léquation diophantienne ax2+2bxy kay2=±1 a criterion for the solution of the diophantine equation ax2+2bxykay2=±1,exposition. math. 16(1998), no.3,249262 (french).,factorisation dans unordre non maximal dun corps quadratique factor-ization in a nonmaximal order of a quadratic eld, exposition. math. 20(2002), no.1