Generalizations of classical results on Jeśmanowicz' conjecture concerning Pythagorean triples II

Published on Feb 1, 2013in Journal of Number Theory0.72
· DOI :10.1016/J.JNT.2014.01.011
Takafumi Miyazaki9
Estimated H-index: 9
,
Pingzhi Yuan16
Estimated H-index: 16
(SCNU: South China Normal University),
Danyao Wu1
Estimated H-index: 1
(SCNU: South China Normal University)
Sources
Abstract
Abstract In 1956 L. Jeśmanowicz conjectured, for any primitive Pythagorean triple ( a , b , c ) satisfying a 2 + b 2 = c 2 , that the equation a x + b y = c z has the unique solution ( x , y , z ) = ( 2 , 2 , 2 ) in positive integers x , y and z . This is a famous unsolved problem on Pythagorean numbers. In this paper we broadly extend many of classical well-known results on the conjecture. As a corollary we can verify that the conjecture is true if a − b = ± 1 .
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#1Yasutsugu Fujita (College of Industrial Technology)H-Index: 11
#2Takafumi Miyazaki (Tokyo Metropolitan University)H-Index: 9
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#1Takafumi MiyazakiH-Index: 1
Let (a,b,c)be a primitive Pythagorean triple such that a^2+b^2=c^2with even b In 1956, L. Jeśmanowicz conjectured that the equation a^x+b^y=c^zhas only the solution (x,y,z)=(2,2,2)in positive integers. In this paper, we give various new results on this conjecture. In particular, we prove that if the equation has a solution (x,y,z)with even x,zthen x/2and z/2are odd.
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#1Takafumi Miyazaki (Tokyo Metropolitan University)H-Index: 1
In 1956, Leon Jeśmanowicz conjectured that, for any primitive Pythagorean triple (a,b,c) with a2+b2 = c2, the equation ax+by = cz in positive integers has only the solution (x,y,z) = (2,2,2). There are some classical and celebrated results on this conjecture. In this paper, we broadly generalize many of them. As a corollary, we can verify that the conjecture is true when |b−a| = 1.
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Let a , b , c be relatively prime positive integers such that a 2 + b 2 = c 2 with b even. In 1956 Jeśmanowicz conjectured that the equation a x + b y = c z has no solution other than ( x , y , z )=(2,2,2) in positive integers. Most of the known results of this conjecture were proved under the assumption that 4 exactly divides b . The main results of this paper include the case where 8 divides b . One of our results treats the case where a has no prime factor congruent to 1 modulo 4, which can b...
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#1Bo He (SICNU: Sichuan Normal University)H-Index: 1
#2Alain Togbé (Purdue University North Central)H-Index: 13
Let n be a positive integer. In this paper, we consider the diophantine equation We prove that this equation has only the positive integer solutions ( n , x , y , z ) = (1, t , 1, 1), (1, t , 3, 2), (3, 2, 2, 2). Therefore we extend the work done by Leszczynski ( Wiadom. Mat ., vol. 3, 1959, pp. 37–39) and Makowski ( Wiadom. Mat ., vol. 9, 1967, pp. 221–224).
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#1Mihai Cipu (Romanian Academy)H-Index: 9
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#1Kalman Gyory (University of Debrecen)H-Index: 13
#2Lajos Hajdu (University of Debrecen)H-Index: 14
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#1Trygve NagellH-Index: 3
A specific feature of this text on number theory is the rather extensive treatment of Diophantine equations of second or higher degree. A large number of non-routine problems are given. The book is intended for graduate students and research mathematicians.
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Let (a, b, c) be a primitive Pythagorean triple. Set $a = m^2-n^2 , b=2mn , and c=m^2+n^2 with m and n positive coprime integers, m>n and m \not \equiv n \pmod 2 . A famous conjecture of Jeśmanowicz asserts that the only positive integer solution to the Diophantine equation a^x+b^y=c^z is (x,y,z)=(2,2,2). A solution (x,y,z) \ne (2,2,2) of this equation is called an exceptional solution. In this note, we will prove that for any n>0 there exists an explicit...
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#1Jerome T. Dimabayao (UP: University of the Philippines Diliman)H-Index: 1
Let (a, b, c) be pairwise relatively prime integers such that $a^2 + b^2 = c^2 . In 1956, Jeśmanowicz conjectured that the only solution of a^x + b^y = c^z in positive integers is (x,y,z)=(2,2,2). In this note we prove a polynomial analogue of this conjecture.
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Let (a,b,c)be a primitive Pythagorean triple. Set a=m^2-n^2b=2mn and c=m^2+n^2with mand npositive coprime integers, m>n and m \not \equiv n \pmod 2 A famous conjecture of Je\'{s}manowicz asserts that the only positive solution to the Diophantine equation a^x+b^y=c^zis (x,y,z)(2,2,2).In this note, we will prove that for any n>0there exists an explicit constant c(n)>0such that if m> c(n) then the above equation has no exceptional solution when all xyan...
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Let m, n be positive integers such that \(m>n\), \(\gcd (m,n)=1\) and \(m \not \equiv n \pmod {2}\). In 1956, L. Jeśmanowicz conjectured that the equation \((m^2 - n^2)^x + (2mn)^y = (m^2+n^2)^z\) has only the positive integer solution \((x,y,z) = (2,2,2)\). This conjecture is still unsolved. In this paper, combining a lower bound for linear forms in two logarithms due to M. Laurent with some elementary methods, we prove that if \(mn \equiv 2 \pmod {4}\) and \(m > 30.8 n\), then Jeśmanowicz’ con...
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Let m nbe positive integers such that m>n \gcd(m,n)=1and m \not\equiv n \bmod 2 In 1956, L. Je\'smanowicz \cite{Jes} conjectured that the equation (m^2 - n^2)^x + (2mn)^y = (m^2+n^2)^zhas only the positive integer solution (x,y,z) = (2,2,2) This problem is not yet solved. In this paper, combining a lower bound for linear forms in two logarithms due to M. Laurent \cite{Lau} with some elementary methods, we prove that if mn \equiv 2 \bmod 4and m > 30.8 n then Je\'smanowi...
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