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Int Jr. of Mathematical Sciences & Applications Vol. 5, No. 2, (July-December, 2015) Copyright ο Mind Reader Publications ISSN No: 2230-9888 www.journalshub.com NON β EXTENDABLE SPECIAL RATIONAL DIO TRIPLES M.A.Gopalan1,V.Sangeetha2,Manju Somanath3 1 Professor,Department of Mathematics,Srimathi Indira Gandhi College,Trichy-2,India. e-mail: [email protected] 2 Assistant Professor,Department of Mathematics,National College,Trichy-1,India. e-mail:[email protected] 3 Assistant Professor,Department of Mathematics,National College,Trichy-1,India email:[email protected] ABSTRACT In this paper, we present three non β extendable special rational Dio triples with suitable property. Keywords: Diophantine triples,specialDiotriples,special rational Dio triples. 2000 MSC Number: 11D09,11D99 1. INTRODUCTION A set of positive integers (π1 , π2 , β¦ , ππ ) is said to have the property π·(π), π β π§ β {0}, if ππ ππ + π is a perfect for all 1 β€ π < π β€ π and such a set is called a Diophantine m-tuple with property π·(π).Many mathematicians considered the problem of existence of Diophantine triples with the property π·(π) for any arbitrary integer n [1] and also for any linear polynomial in n. In this context, one may refer [2-21] for an extensive review of various problem on Diophantine triples. These results motivated us to search for non-extendable special rational Dio triples with suitable property, where the special mention is provided because it differs from the earlier one and special Dio triple is constructed 233 NON β EXTENDABLE SPECIAL RATIONAL DIO TRIPLES where the product of any two members of the triple with the addition of their sum and increased by the given property is a perfect square. 2. Section A: Non- extendable π« ( Let π = 1 π2 and π = 1 π2 +1 ππ +ππβπ ππ METHOD OF ANALYSIS ) β special rational Dio triple. be two rational numbers such that ππ + π + π + ( π 2 +2πβ1 π2 ) is a perfect square. Let cbe any rational number such that 1 π2 π+ 1 π2 +1 1 π2 π+ +π+( 1 π2 +1 π 2 +2πβ1 π2 +π+( ) = πΌ2 π 2 +2πβ1 π2 (1) ) = π½2 (2) Eliminating cfrom (1) and (2), we obtain β 1 π2 (π2 +1) ( π 2 +2πβπ2 β1 π2 )=( π2 +2 π2 +1 ) πΌ2 β ( π2 +1 π2 ) π½2 Using the linear transformation πΌ =π+( π2 +1 π2 )π (4) π½ =π+( π2 + 2 )π π2 + 1 in (3),it leads to the Pell equation π2 = ( π2 +1 π2 Let π0 = 1 ; π0 = )( π+1 π π2 +2 π2 +1 ) π2 + ( π 2 βπ2 +2πβ1 π2 ) (5) be the initial solution of (5).Thus (4) yields πΌ= π2 +(π+1)π+1 π2 and using (1), we get π= π2 +2(π+1)π+1 π2 234 (3) M.A.Gopalan, ,V.Sangeetha, Manju Somanath Hence (π, π, π) = ( 1 , π2 1 , π2 +1 π2 +2(π+1)π+1 π2 ) is a special rational Dio triple with property π· ( π 2 +2πβ1 We show that the above triple cannot be extended to a quadruple. Let d be any rational number such that 1 π2 π+ 1 π2 +1 ( 1 π2 π+ +π+( 1 π2 +1 π2 +2(π+1)π+1 π2 π 2 +2πβ1 π2 +π+( )π + ( ) = π2 π 2 +2πβ1 π2 (6) ) = π2 π2 +2(π+1)π+1 π2 (7) )+π+( π 2 +2πβ1 π2 ) = π2 (8) Eliminating d from (7) and (8), we obtain π2 β(π2 +1)(π2 +2(π+1)π+1) =( π2 (π2 +1) 2π2 +2(π+1)π+1 π2 ) π2 β ( π2 +2 π2 +1 ) π2 (9) Using the linear transformations π =π+( π =π+( π2 +2 π2 +1 )π (10) 2π2 +2(π+1)π+1 π2 )π in (9),it leads to the Pell equation π2 = ( π2 +2 π2 +1 )( 2π2 +2(π+1)π+1 π2 Let π0 = 1 and π0 = ) π2 β ( π2 βπ 2 β2π+1 π2 π3 +(π+1)π2 +2π+(π+1) π(π2 +1) π= ) (11) be the initial solution of (11).Thus (10) yields 2π3 + (π + 1)π2 + 4π + (π + 1) π(π2 + 1) and using (7),we get π= 4π6 + 4(π + 1)π5 + 17π4 + 12(π + 1)π3 + 19π2 + 8(π + 1)π + 2 π2 (π2 + 1)(π2 + 2) Verification for non-extendabiltiy of quadruple from the above triple is given below: 235 π2 ). NON β EXTENDABLE SPECIAL RATIONAL DIO TRIPLES Substituting the values of a and d in LHS of (6),we have 2 LHS of (6) = (2π2 +(π+1)π+2) β3 π4 Note that the RHS is not a perfect square. Section B: Non- extendable π« ( Let π = 1 2π2 and π = 1 2π2 +1 ππ +ππ ππ ) β special rational Dio triple. be two rational numbers such that ππ + π + π + ( π 2 +2π π2 ) is a perfect square. Let cbe any rational number such that 1 2π2 π+ 1 2π2 +1 1 2π2 π+ +π+( 1 2π2 +1 π 2 +2π π2 +π+( ) = πΌ2 π 2 +2π π2 (12) ) = π½2 (13) Eliminating cfrom (12) and (13), we obtain β 1 2π2 (2π2 +1) ( π 2 +2πβπ2 π2 )=( 2π2 +2 2π2 +1 ) πΌ2 β ( 2π2 +1 2π2 ) π½2 (14) Using the linear transformation πΌ =π+( 2π2 +1 2π2 )π (15) π½ =π+( 2π2 + 2 )π 2π2 + 1 in (14),it leads to the Pell equation π2 = ( 2π2 +1 2π2 Let π0 = 1 ; π0 = )( π+1 π 2π2 +2 2π2 +1 ) π2 + ( π 2 +2πβπ2 π2 ) (16) be the initial solution of (16).Thus (15) yields πΌ= 2π2 +2(π+1)π+1 2π2 236 M.A.Gopalan, ,V.Sangeetha, Manju Somanath and using (12), we get π= Hence (π, π, π) = ( π·( π 2 +2π π2 1 2π2 , 1 2π2 +1 , 4π4 +8(π+1)π3 +6π2 +4(π+1)π+1 2π2 (2π2 +1) 4π4 +8(π+1)π3 +6π2 +4(π+1)π+1 2π2 (2π2 +1) ) is a special rational Dio triple with property ). We show that the above triple cannot be extended to a quadruple. Let d be any rational number such that 1 2π2 π+ 1 2π2 +1 ( 1 2π2 +π+( 1 π+ 2π2 +1 π 2 +2π π2 +π+( ) = π2 π 2 +2π π2 4π4 +8(π+1)π3 +6π2 +4(π+1)π+1 2π2 (2π2 +1) (17) ) = π2 )π + ( (18) 4π4 +8(π+1)π3 +6π2 +4(π+1)π+1 2π2 (2π2 +1) )+π+( π 2 +2π π2 ) = π 2 (19) Eliminating d from (18) and (19), we obtain ( 4π4 +8(π+1)π3 +6π2 +4(π+1)π+1 2π2 (2π2 +1) )( π 2 +2πβπ2 π2 )=( 8π4 +8(π+1)π3 +8π2 +4(π+1)π+1 2π2 (2π2 +1) ) π2 β ( 2π2 +2 2π2 +1 ) π2 (20) Using the linear transformations π =π+( π =π+( π2 = ( 2π2 +2 2π2 +1 )π (21) 8π4 +8(π+1)π3 +8π2 +4(π+1)π+1 2π2 +2 2π2 +1 2π2 (2π2 +1) )( )π 8π4 +8(π+1)π3 +8π2 +4(π+1)π+1 Let π0 = 1 and π0 = 2π2 (2π2 +1) in (20) leads to the Pell equation ) π2 + ( 2π3 +2(π+1)π2 +2π+(π+1) π(2π2 +1) π 2 +2πβπ2 π2 ) (22) be the initial solution of (22).Thus (21) yields 4π3 +2(π+1)π2 +4π+(π+1) π(2π2 +1) and using (18),we get 237 π= NON β EXTENDABLE SPECIAL RATIONAL DIO TRIPLES π= 16π6 + 16(π + 1)π5 + 34π4 + 24(π + 1)π3 + 19π2 + 8(π + 1)π + 1 2π2 (2π2 + 1)(π2 + 1) Verification for non-extendabiltiy of quadruple from the above triple is illustrated below: Substituting the values of a and d in LHS of (17),we have 2 LHS of (17) = (4π2 +2(π+1)π+2) β3 4π4 Note that the RHS is not a perfect square. Section C: Non- extendable π« (π β Let π = 1 π and π = 1 π ππ ) β special rational Dio triple. be two rational numbers such that ππ + π + π + (1 β 4π 1 4π ) is a perfect square. Let cbe any rational number such that 1 π 1 1 π 4π π + + π + (1 β 1 4π π+ 1 4π + π + (1 β ) = πΌ2 1 4π (23) ) = π½2 (24) Eliminating cfrom (23) and (24), we obtain 3 16π2 =( 4π+1 4π ) πΌ2 β ( π+1 π ) π½2 (25) Using the linear transformation πΌ =π+( π+1 π )π (26) π½ =π+( in (25),it leads to the Pell equation 238 4π + 1 )π 4π M.A.Gopalan, ,V.Sangeetha, Manju Somanath π2 = ( π+1 π Let π0 = 1 ; π0 = )( 4π+1 4π 2π+1 2π ) π2 β 1 (27) 4π be the initial solution of (27).Thus (26) yields πΌ= 4π+3 2π and using (23), we get π= 1 1 π 4π Hence (π, π, π) = ( , , 12π+9 4π 12π+9 4π ) is a special rational Dio triple with property π· (1 β We show that the above triple cannot be extended to a quadruple. Let d be any rational number such that 1 1 4π ( 1 1 π 4π π + + π + (1 β π π+ 12π+9 4π 1 4π + π + (1 β )π + ( 12π+9 4π ) = π2 1 4π (28) ) = π2 (29) ) + π + (1 β 1 4π ) = π 2 (30) Eliminating d from (29) and (30), we obtain ( 3π+2 π β1 16π+9 4π 4π )( ) = ( ) π2 β ( 4π+1 4π ) π2 (31) Using the linear transformations π =π+( π =π+( 4π+1 4π )π 16π+9 4π (32) )π in (31),it leads to the Pell equation π2 = ( 4π+1 4π )( 16π+9 4π ) π2 β Let π0 = 1 and π0 = 8π+3 4π 1 4π (33) be the initial solution of (33).Thus (32) yields 239 1 4π ). NON β EXTENDABLE SPECIAL RATIONAL DIO TRIPLES π= 3π + 1 π π= 8π + 4 π and using (29),we get Verification for non-extendabiltiy of quadruple from the above triple is given below: Substituting the values of a and d in LHS of (28),we have LHS of (28) = (6π+4)2 +3 4π2 Note that the RHS is not a perfect square. 3. CONCLUSION To conclude, one may search for other non-extendable special rational Dio triples with suitable properties. 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