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IRJET- On the Positive Pell Equation Y2=35x2+29

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International Research Journal of Engineering and Technology (IRJET)

e-ISSN: 2395-0056

Volume: 06 Issue: 03 | Mar 2019

p-ISSN: 2395-0072

www.irjet.net

On the Positive Pell Equation y 2  35 x 2  29 M.A. Gopalan1, T. Mahalakshmi2, K. sevvanthi3 1Professor,

Department of Mathematics, Shrimati Indira Gandhi College, Trichy-620 002, Tamil Nadu, India. Professor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-620 002, Tamil Nadu, India. 3PG Scholor, Department of Mathematics, Shrimati Indira Gandhi College, Trichy-620 002, Tamil Nadu, India. ----------------------------------------------------------------------***--------------------------------------------------------------------2Assistant

Abstract – The binary quadratic Diophantine equation represented by the positive Pellian

y 2  35x 2  29 is analyzed for its

non-zero distinct integer solutions. A few interesting relations among the solutions are given. Employing the solutions of the above hyperbola, we have obtained solutions of other choices of hyperbolas and parabolas.

Key Words: Binary quadratic, Hyperbola, Parabola, Pell equation, Integer solutions 1. INTRODUCTION The binary quadratic equation of the form y 2  Dx 2  1 where D is non-square positive integer, has been selected by various mathematicians for its non-trivial integer solutions when D takes different integral values [1-4]. For an extensive review of various problems, one may refer [5-17]. In this communication, yet another an interesting equation given by y 2  35x 2  29 is considered and infinitely many integer solutions are obtained. A few interesting properties among the solutions are presented.

2. METHOD OF ANALYSIS The Positive pell equation representing hyperbola under consideration is

y 2  35x 2  29

(1)

The smallest positive integer solutions of (1) are

x0  1 , y0  8 To obtain the other solutions of (1), consider the Pell equation

y 2  35x 2  1

(2)

whose initial solution is

~ x0  1, ~ y0  6 xn , ~ yn ) of (2) is given by The general Solution ( ~ ~ xn 

1

1 gn , ~ yn  f n 2 2 35

where

f n  (6  35 ) n1  (6  35 ) n1

g n  (6  35 ) n1  (6  35 ) n1 , n  1, 0 ,1,...

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