THEDIOPHANTINEEQUATIONax2+2bxy4ay2=1LIONEL BAPOUNGUReceived 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 isrelative 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,(1.1)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 a1, b 0, 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 )isaunitofQ( ),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), is1)and,hence, a00unitofQ( ).Writing (1/2)(v +w )forthefundamental unitofQ( ),weseethatthesolutionsinpairsofnaturalnumbers (v,w) of(1.2)comprisethevaluesofthesequence (vn,wn)(n1)denedbysettingn 12v0+w0vn+wn=.(1.3)2 2242LIONEL BAPOUNGUHence,weremarkeasily(cf.2)thatif(x,y) isasolutionof(1.1),thenv=bx2+8axy+4by2,w=x2+4y2(1.4)verifywith=4a2+b2 (a1, b1, and gcd(a,2b)=1)thePellsequationv2w2=4 withv,w odd.(1.5)Hence, our study will 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)issolvable, thenaisthenormofanelement 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 with the aid of the followingresultprovedin5.Theorem1.1(Thue). Ifandareintegerssatisfying1,gcd(,)=1,andmtheleastintegergreaterthan,thereexistxandyin0,msuchthatyx(mod).When solutions exist, we show using any of them in Section5 that (1.1)possessesaninnityofsolutions(Theorem 5.1);afterwards,wedescribeusingitafamily (Proposition 5.2). Wegive theconclusion ofourpaper inSection6withsomenumerical examples.2. Thecasea1,oddnonsquare, and(1.5)solvable withv,w odd2.1. Preliminaries. Letv,wbeodd integersgreaterthanorequalto1suchthatv2w2=4.Itisclearthatgcd(v,w)=1(2.1)becauseifv andw haveacommonprimefactord,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) is 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-sition4appliedtoFenablesustostatethefollowing 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). Weprovethefollowingtheorem.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,w00)of(1.5)ismonic.000Proof.(i)(ii).Let(x,y)Z2 beasolutionof(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),weobtai