In this paper we examine single-step iterative methods for the solution of the nonlinear algebraic equation f (x) = x2 - N = 0 , for some integer N, generating rational approximations p/q that are optimal in the sense...In this paper we examine single-step iterative methods for the solution of the nonlinear algebraic equation f (x) = x2 - N = 0 , for some integer N, generating rational approximations p/q that are optimal in the sense of Pell’s equation p2 - Nq2 = k for some integer k, converging either alternatingly or oppositely.展开更多
Let D≠1 be a positive non-square integer and k≥2 be any fixed integer. Extending the work of A. Tek-can, here we obtain some formulas for the integer solutions of the Pell equation X2 - Dy2 = ± k2 .
Let P:=P(t) be a polynomial in Z[X]\{0,1} In this paper, we consider the number of polynomial solutions of Diophantine equation E:X2–(P2–P)Y2–(4P2–2)X+(4P2–4P)Y=0. We also obtain some formulas and recurrence rela...Let P:=P(t) be a polynomial in Z[X]\{0,1} In this paper, we consider the number of polynomial solutions of Diophantine equation E:X2–(P2–P)Y2–(4P2–2)X+(4P2–4P)Y=0. We also obtain some formulas and recurrence relations on the polynomial solution (Xn,Yn) of展开更多
This paper corresponds to the written versions of many lectures at several locations including the most recent one at Weinan Teachers University on June 8,2011.I would like to thank Professor Hailong Li for inviting m...This paper corresponds to the written versions of many lectures at several locations including the most recent one at Weinan Teachers University on June 8,2011.I would like to thank Professor Hailong Li for inviting me to publish this in the journal of his university.I wish also to express my deep gratitude to my friend Shigeru Kanemitsu,thanks to whom I could visit Weinan Teachers University,and who also came up with a written version of these notes. The topic is centered around the equation x2-dy2=±1,which is important because it produces the(infinitely many) units of real quadratic fields.This equation,where the unknowns x and y are positive integers while d is a fixed positive integer which is not a square,has been mistakenly called with the name of Pell by Euler.It was investigated by Indian mathematicians since Brahmagupta(628) who solved the case d=92,then by Bhaskara II(1150) for d=61 and Narayana(during the 14-th Century) for d=103.The smallest solution of x2-dy2=1 for these values of d are respectively 1 1512-92·1202=1, 1 766 319 0492-61·226 153 9802=1 and 227 5282-103·22 4192=1, and hence they could not have been found by a brute force search! After a short introduction to this long story of Pell's equation,we explain its connection with Diophantine approximation and continued fractions(which have close connection with the structure of real quadratic fields),and we conclude by saying a few words on more recent developments of the subject in terms of varieties.Finally we mention applications of continued fraction expansion to electrical circuits.展开更多
with elementary method of Pell’s equation, and the results we get is generalization of the studies of Ljunggren （a=b=1）and Sun Qi et al. （b=1）.Theorem 1. If a, b∈N, then the Diophantine equation （1） only has s...with elementary method of Pell’s equation, and the results we get is generalization of the studies of Ljunggren （a=b=1）and Sun Qi et al. （b=1）.Theorem 1. If a, b∈N, then the Diophantine equation （1） only has solutions in positive integers x=y=1 （when a】1, b=1） and m=4s+1, x=3, y=3<sup>2s</sup>+2 (when a=1/4 (3<sup>2s-1</sup>+1), b=1), where s is a positive integer.展开更多
文摘In this paper we examine single-step iterative methods for the solution of the nonlinear algebraic equation f (x) = x2 - N = 0 , for some integer N, generating rational approximations p/q that are optimal in the sense of Pell’s equation p2 - Nq2 = k for some integer k, converging either alternatingly or oppositely.
文摘Let D≠1 be a positive non-square integer and k≥2 be any fixed integer. Extending the work of A. Tek-can, here we obtain some formulas for the integer solutions of the Pell equation X2 - Dy2 = ± k2 .
文摘Let P:=P(t) be a polynomial in Z[X]\{0,1} In this paper, we consider the number of polynomial solutions of Diophantine equation E:X2–(P2–P)Y2–(4P2–2)X+(4P2–4P)Y=0. We also obtain some formulas and recurrence relations on the polynomial solution (Xn,Yn) of
文摘This paper corresponds to the written versions of many lectures at several locations including the most recent one at Weinan Teachers University on June 8,2011.I would like to thank Professor Hailong Li for inviting me to publish this in the journal of his university.I wish also to express my deep gratitude to my friend Shigeru Kanemitsu,thanks to whom I could visit Weinan Teachers University,and who also came up with a written version of these notes. The topic is centered around the equation x2-dy2=±1,which is important because it produces the(infinitely many) units of real quadratic fields.This equation,where the unknowns x and y are positive integers while d is a fixed positive integer which is not a square,has been mistakenly called with the name of Pell by Euler.It was investigated by Indian mathematicians since Brahmagupta(628) who solved the case d=92,then by Bhaskara II(1150) for d=61 and Narayana(during the 14-th Century) for d=103.The smallest solution of x2-dy2=1 for these values of d are respectively 1 1512-92·1202=1, 1 766 319 0492-61·226 153 9802=1 and 227 5282-103·22 4192=1, and hence they could not have been found by a brute force search! After a short introduction to this long story of Pell's equation,we explain its connection with Diophantine approximation and continued fractions(which have close connection with the structure of real quadratic fields),and we conclude by saying a few words on more recent developments of the subject in terms of varieties.Finally we mention applications of continued fraction expansion to electrical circuits.
文摘with elementary method of Pell’s equation, and the results we get is generalization of the studies of Ljunggren （a=b=1）and Sun Qi et al. （b=1）.Theorem 1. If a, b∈N, then the Diophantine equation （1） only has solutions in positive integers x=y=1 （when a】1, b=1） and m=4s+1, x=3, y=3<sup>2s</sup>+2 (when a=1/4 (3<sup>2s-1</sup>+1), b=1), where s is a positive integer.
