OntheextendibilityoftheDiophantinetriple

1、 ontheextendibilityofthediophantinetriple1,5,cfadwa s.abumuriefah andamal al-rashedreceived 15may 2003we study the problem of extendibility of the triples of the form 1,5,c. we prove thatif c = s +1, where2(s ) is a binary recursive sequence,kis a positive integer, and thekkkstatement thatallsolutio

2、ns ofasystemofsimultaneous pellianequations z2c x2=c 1,kk5z2c y2 =c 5 are given by (x,y,z) =(0,2,s ), is valid for 2k31, then it iskkkvalid forallpositive integer k.2000 mathematics subject classication: 11d09, 11d25.1. introduction. letnbeaninteger.asetofpositiveintegersa1,a2,.,amissaidtohave the p

3、roperty d(n) ifa a +nisaperfect square forall1ijm;such aijsetiscalledadiophantine m-tupleorap setofsizem.theproblemofconstructionnof such sets was studied by diophantus (see 4). a famous conjecture related to thisproblemisasfollows.conjecture 1.1. there does not exist a diophantine quadruple with th

4、e propertyd(1).for certain triples a,b,c with 1a,b,c, the validity of this conjecture can beveriedbysimpleuseofcongruences(see5).thecasea=1ismoreinvolvedandtherst important result concerning thisconjecture was proved in1985 bymohanty andramasamy 8; they proved that the triple 1,5,10 cannot be extend

5、ed. also, brown5 proved the conjecture for the triples n2+1, (n+1)2+1, (2n+1)2+1 , wheren 0(mod4), for the triples 2, 2n2+2n+1, 2n2+6n+5, where n 1(mod4),and proved nonextendibility of triples 17,26,68 and 1,2,5. in 1998, kedlaya 7verieditforthetriples1,2,145,1,2,4901,1,5,65,1,5,20737,1,10,17 ,and1,

6、26,37.sincedujella6hasprovedtheconjectureforalltriplesoftheform1,2,c,theconsideration oftriplesoftheform1,5,cseemstobethenaturalnextstep.inthepresentpaper,wewillstudytheextendibilityofalltriplesoftheform1,5,c.inourproof,wewillfollowthestrategyof6.2. preliminaries. since the triple 1,5,c satises the

7、property d(1) , thereforethereexistintegerss,t suchthatc1=s2 and5c1=t2 whichimplyt25s2=4.(2.1)ifthistriplecanbeextendedtoadiophantine quadruple, thenthereareintegersd,x,y,zsuchthatd1=x2,5d1=y2,cd1=z2.(2.2) 1738f. s. abu muriefah and a. al-rashedeliminating d,wegetz2cx2=c1,z2cy2=c5;(2.3)itisobviousth

8、atifallthesolutionsofthissystemaregivenby(x,y,z)=(0,2,c1),then,from(2.2),wegetd=1,sothetriple1,5,ccannotbeextended.thepellequation (2.1)hasthreeclasses ofsolutions andallthesolutions aregivenby tk+sk = 3+59+4 5 k,5 9+4 5 k,(2.4)tk+sk= 3+ ktk+sk=2 9+4 5.hence,ifthetriple1,5,cisadiophantine triplewith

9、thepropertyd(1),thenthereexistsapositiveintegerksuchthattheintegerchasthefollowingthreeformulas(see3):1 7+3 5 161+72 5 k+ 73 5 16172 5 +6 ,22k(2.5)(2.6)(2.7)c=ck =c=ck=101 2k2k73 5 161+72 5 + 7+3 5 16172 5 +6 ,10c=ck= 15 kk161+72 5 + 16172 5 +3 .the main result of this paper is in the following theo

10、rem, where ck denotes one oftheformulas in(2.5),(2.6),and(2.7).theorem 2.1. let kbe apositive integer and let ck=sk2+1,where (sk)is abinaryrecursivesequence.ifthestatementthatallsolutionsofasystemofsimultaneouspellianequationsz2ckx2=ck1,5z2cky2=ck5(2.8)are given by (x,y,z) =(0,2,sk) is valid for k31

11、, then it is valid for all positiveintegerk.remark 2.2. thetheorem istruewhenk=05andk=1(see1,7,8).sowewillsupposethatk2.forsimplicity,wewillomittheindexkandwewilldividetheproofofthetheoremintomanylemmas.3. asystem ofpellianequations. therearenitesets+x0(i)c:i=1,2,.,i0,(i)z0(3.1)(j)5+y1(j) c:j=1,2,.,

12、j0z,1 on the extendibility of the diophantine triple 1,5,c1739ofelementsofzcandz5c,respectively,suchthatallsolutionsof(2.8)aregivenbyz+xc= z +x0(i)c 2c1+2scm, i=1,.,m0,(3.2)(3.3)(i)0 (j)5+y1(j)10ck1+2t 5c n, i=1,.,n0,z 5+y c= zc1respectively (see6).from(3.2),weconclude thatz=vm(i) forsomeindexiandin

13、tegerm ,where(i)(i)(i)mv0(i)=z0(i),v1(i)=(2c1)z0(i)+2scx0(i),v=(4c2)vv,(3.4)(3.5)m+2m+1andfrom(3.3),weconclude thatz=wn(j) forsomeindexj andintegern ,where1(j)=(10c1)z1(j)+2tcy1(j),w(j)n+2=(20c2)wn(j)+1wn(j).w0(j)=z1(j),wthuswereformulated system(2.8)tonitelymanydiophantineequationsoftheformvm(i)=w(

14、j).(3.6)nifwechooserepresentatives z0(i)+x0(i)candz1(j)5+y1(j)csuchthat|z0(i)|and|z1(j)|areminimal, then,by9,theorem 108a,wehavethefollowing estimates:0 z0(i) 212c(c1)c,(3.7)0 z0(i) c(c5)c.4. application of congruence relations. in the following lemma, we prove that if(2.2) has a nontrivial solution

15、, then the initial terms of sequences vm(i) and wn(j)arerestricted.lemma4.1. letk2betheleastpositiveinteger(ifitexists)forwhichthestatementof(i)(j)ntheorem 2.1isnotvalid.let1ii ,1jj ,andletv andw bethesequences00(j)nmdenedin(3.4)and(3.5).iftheequationv =w hasasolution,then|z0(i)|=|z1(j)|=s.(i)mproof

16、.from(3.4)and(3.5),itfollowseasilybyinduction thatv2(i)mz0(i)(mod2c),w2(j)n z1(j)(mod2c),z0(i)(mod2c),(4.1)(i)2m+1v(j)2n+1z1(j)(mod2c).w 1740f. s. abu muriefah and a. al-rashedtherefore,iftheequationv =w(j) hasasolutioninintegersmandn,thenwemust(i)mnhave|z0(i)|=|z1 |.now,letd0=(z(i)2+1)/c;thenwehave

17、(j)0222,d01= x0(i),5d01= y1(j)d0 c2c+1,cd01= z0(i)(4.2)1.itfollowsfrom(4.2)thatthereexistsapositiveintegerlv1.letwlvl,wherel0;thenwl+2(20c2)wl+1vl=(20c2)wl+1 (4c2)vl+1vl+2,(4.6)(4.7)hencewl+2vl+20,whichimplieswhasasolutionandn0,thenvnvm=wn.butthesequencev isincreasing,son. on the extendibility of th

18、e diophantine triple 1,5,c1741(4.8)now,from(3.4),wehaves 12c1+2scm,mmvm=2c1+2sc+ 2c12sc22andfrom(3.5),wehave1 10c1+2t 5c n+1/2. nnwn= 2s 52 c 10c1+2t 5c + 10c12t 5c5(4.9) s5+2 c+110c1+2t 5cn1522since k 2, therefore from (2.5), (2.6), and (2.7), we have c 3026. thus vm = wnimplies ln 10c1+2t5cmn+1/2

19、1.1712.(4.10)(4.11)ln 2c1+2scifn=0,thenm=0,andifn1,then(4.10)impliesm1.1712n+0.5856(1/2)4c.proof.weassumethatn(1/2)(1)thecasewheremandnarebotheven.4c.usinglemma4.2andfromvm=wn,weget2c(2m)2s+2cs(2m)x010c(2n)2s2ct(2n)y1 mod8c2 .(4.12)(4.13)(4.14)butx0=0andy1=2,so8cm2s40cn2s8ctn mod8c2 ,whichimpliess 5

20、n2m2 tn(modc).ontheotherhand,wehave,fromlemma4.5,1 4 cs5n2m2c4n2 5/44c3 ,thentn 5cn 5 c4c=c3c.(4.16)(4.17) 12544so,from(4.14),(4.15),and(4.16),wegets 5n2m2 =tn. 1742f. s. abu muriefah and a. al-rashedalso,from(4.14),wehaves2 5n2m2 t2n2(modc).2(4.18)(4.19)buts2t2(modc),so(4.18)becomesm25n22n2(modc).n

21、ow,since1 4 c5n2m242 4n2 =16n4161 2=c,2(4.20)(4.21)1cc,n4 c n2 42so,from(4.19),(4.20),weget5n2m22=n2.finally,from(4.17)and(4.21),wegett2=s2,whichisimpossible.(2)thecasewheremandnarebothodd.we assume that n(1/2)andy1=2,weget4c. using lemma4.2 and from vm =wn, where x0 =0s 5n2m2 2tn(modc).(4.22)(4.23)

22、(4.24)(4.25)(4.26)(4.27)asabove,andsinces5n2m2c,2tn2 5cn 5c4cc,therefore (4.22),(4.23),and(4.24)implys 5n2m2 =2tn.also,from(4.22),wehaves2 5n2m2 4t2n2(modc),2whichimplies(m25n2)24n2(modc).but(5n2m2)2c and4n2c ,so5n2m22=4n2.finally,from(4.25)and(4.27),wegett2=s2,whichisimpossible. on the extendibilit

23、y of the diophantine triple 1,5,c17435. linearformsinlogarithmslemma 5.1. ifvm=wn ,then0nlog 10c1+2t 5c mlog 10c1+2t 5c +logs52c(5.1)(5.2)(5.3)(5.4)1andq1;alsopq= c5q1s2p1q, then from (5.4), we get pq 1 and p(4s c)s =4s2 c=4(c1) ccc1.henceqp,andwemayassumethatm1.furthermore0log q1=logq=log 1qp.(5.5)

24、ppqsincelog(1x)x+x2,therefore, from(5.5),weget+qpqp qp20q(c1)p1q(c1) ,sop1 q(c1)1;hence,from(5.3)and(5.7),wegetqp(c1) q(c1)1c5q14cq+c2+54cq+c2+5.q5q(5q5c+5) (s52c)(4c) impliesc2qc2c2c4 s52 c (4c) 4c = ,(5.9)so(5.8)becomes4c+cq2+(5.10)qp 5qq1 4c+4c+5 q1=174c+5 q1.q 1744f. s. abu muriefah and a. al-ra

25、shedfrom(5.6)and(5.10),wegetc+5 q1+174 c+5 q2.0logpq20,so17417417517192c+5 q1c+5 c1=.(5.12)(5.13)(5.14)4 +c4 +4=thus(5.11)becomesc+5 q1+174 c+5 q20logpq=174 c+5 q172417112171 1+4 c+5 q14 c+5 q1n112174 c+5 s5=c 10c1+2t 5c52n11 5 178 c+2 10c1+2t 5c5 n5111153c+2 4 5c1 5c 53c+2 (4c)n54c(4c)n.butlogpq =n

26、log 10c1+2t 5c mlog 10c1+2t 5c +logs52c.5cso,(5.13)and(5.14)complete theproofofthelemma.now,toprovethetheorem, weapplythefollowing theorem.theorem 5.2 2. for a linear form = 0 in logarithms of l algebraic numbers1,., withrationalcoecients b ,.,b,l1llog|18(l+1)!ll+1(32d)l+2h 1 h log(2ld)logb,(5.15)lw

27、here b=max(|b1|,.,|bl|)and where d is the degree of the number eld generatedby ,.,.1lhereh()= 1rmax h(),|log|,1(5.16)andh()denotesthestandard logarithmic weilheightof. on the extendibility of the diophantine triple 1,5,c17456. proofoftheorem 2.1. (1)thecasewheremandnarebotheven.we consider the equat

28、ion v2m =w2n with n=0. we apply the above theorem andwehavel=3,d=4,b=2m ,where1=10c1+2t 5c,2=2c1+2sc,(6.1)5+2 c3= s.5stheequations satisedby1,2,3 are12(20c2)1+1=0,22(4c2)2+1=0,(6.2)(5c5)32(10c10)3+c5=03223+ c55 5 =0.chence1log1 12log20c,h 1 = 2h 2 = 12log2 12log4c,(6.3)(6.4)5+2ch 3 = 21logs 21log(1+

29、2c).5sfromlemma5.1,whereniseven,wehavelog(4c)12n=(2n1)log4c.so,fromtheorem 5.2,weget(2n1)log4c184!34(324)512log(20c)12log(4c)12log(2c+1)log24log2m.(6.5)now,usinglemmas4.5and4.6,weget(2n1)2.074311014log8000n4log 800n4+1 log2 5n ,(6.6)(6.7)(6.8)whichimpliesthatn21019,andnally,c256 1076 . 1746f. s. abu

30、 muriefah and a. al-rashedtondkintherstclass,substitute in(2.5);hence klog 161+72 5 log256+77log10log 73 5 ,(6.9)whichimpliesk31.similarly, wendthatintheothertwoclasses,k31.(2)thecasewheremandnarebothodd.inthiscase,usinglemma5.1,wherenisodd,relation(6.4)becomeslog(4c)2n=(2n1)log4c.(6.10)(6.11)hence(

31、6.6)becomes2n2.074311014log8000n4log 800n4+1 log2 5n ,whichimpliesthatn21019,andnallyc256(1076),hencek31.references1234f.s.abumuriefah anda.al-rashed, onthediophantine triple 1,5,442,toappear inarab.j.sci. eng. sect. asci.a.baker andg.wstholz, logarithmic forms andgroup varieties,j.reineangew. math.

32、 442(1993), 1962.e. j. barbeau, pells equation, problem books in mathematics, springer-verlag, new york,2003.i. g. bashmakova (ed.), diophantus of alexandria, arithmetics and the book of polygonalnumbers,nauka, moscow, 1974.56e.brown,setsinwhichxy+kisalwaysasquare,math.comp.45(1985),no.172,613620.a.

33、 dujella, complete solution of a family of simultaneous pellian equations, acta math. in-form. univ. ostraviensis 6(1998), no.1,5967.78k.s.kedlaya, solving constrained pellequations,math.comp. 67(1998), no.222,833842.s.p.mohanty and a.m.s.ramasamy, the simultaneous diophantine equations 5y220=x2 and

34、 2y2+1=z2,j.number theory 18(1984), no.3,356359.9t.nagell, introduction tonumber theory,john wiley &sons inc., new york, 1951.fadwa s.abu muriefah: mathematics department, girls college ofeducation, p.o. box 60561,riyadh 11555, saudi arabiae-mail address:amal al-rashed: mathematics department, girlscollege ofeducation, p.o.box60561, riyadh11555, saudi arabia advances inoperations researchadvances indecision sciencesjournal ofapplied mathematicsjournal ofprobability and statisticsalgebrahindawi publishing corporationhindawi publishing corporationhindawipublishingcorporationh