# 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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References22
#1Yasutsugu Fujita (College of Industrial Technology)H-Index: 11
#2Takafumi Miyazaki (Tokyo Metropolitan University)H-Index: 9
#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.
#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.
#1Takafumi Miyazaki (Tokyo Metropolitan University)H-Index: 9
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...
#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).
#1Maohua LeH-Index: 1
#2Maurice MignotteH-Index: 14
We study the solutions of a Diophantine equation of the form a^x+b^y=c^z where a\equiv 2 \pmod 4 b\equiv 3 \pmod 4and \gcd (a,b,c)=1 The main result is that if there exists a solution (x,y,z)=(2,2,r)with r>1odd then this is the only solution in integers greater than 1, with the possible exception of finitely many values (c,r) We also prove the uniqueness of such a solution if any of a b cis a prime power. In a different vein, we obtain various inequalities that must...
#1Kalman Gyory (University of Debrecen)H-Index: 13
#2Lajos Hajdu (University of Debrecen)H-Index: 14
Last. N. Saradha (TIFR: Tata Institute of Fundamental Research)H-Index: 12
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We show that the product of four or five consecutive positive terms in arithmetic progression can never be a perfect power whenever the initial term is coprime to the common difference of the arithmetic progression. This is a generalization of the results of Euler and Oblath for the case of squares, and an extension of a theorem of Gyory on three terms in arithmetic progressions. Several other results concerning the integral solutions of the equation of the title are also obtained. We extend res...
In this paper, we establish a number of theorems on the classic Diophantine equation of S. S. Pillai, , where and are given nonzero integers with . In particular, we obtain the sharp result that there are at most two solutions in positive integers and and deduce a variety of explicit conditions under which there exists at most a single such solution. These improve or generalize prior work of Le, Leveque, Pillai, Scott and Terai. The main tools used include lower bounds for linear forms in the lo...
#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 $a = m^2-n^2 , b=2mn , and c=m^2+n^2 with m and n positive coprime integers, m>n and m \not \equiv n \pmod 2 . A famous conjecture of Jeśmanowicz asserts that the only positive integer solution to the Diophantine equation a^x+b^y=c^z is (x,y,z)=(2,2,2). A solution (x,y,z) \ne (2,2,2) of this equation is called an exceptional solution. In this note, we will prove that for any n>0 there exists an explicit... #1Jerome T. Dimabayao (UP: University of the Philippines Diliman)H-Index: 1 Let (a, b, c) be pairwise relatively prime integers such that$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.
In this paper, we consider the exponential Diophantine equation a^{x}+b^{y}=c^{z},where a, b, cbe relatively prime positive integers such that a^{2}+b^{2}=c^{r}, r\in Z^{+}, 2\mid rwith beven. That is $a=\mid Re(m+n\sqrt{-1})^{r}\mid, b=\mid Im(m+n\sqrt{-1})^{r}\mid, c=m^{2}+n^{2}, where , n are positive integers with >n, m-n\equiv1(mod 2), gcdm, n)=1. x, y, z)= (2, 2, r)$ is called the trivial solution of the equation. In this paper we prove that the equation has no no...
Let $m \ge 1 be a positive integer. We show that the exponential Diophantine equation (m^2+m+1)^x+m^y=(m+1)^z has no positive integer solutions other than (x,y,z)=(1,1,2) when m \not \in \{1, 2, 3 \} . 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... #1Haizhou Song (Huaqiao University)H-Index: 1 Here we study the character and expression of n-order Pythagorean matrix using number theory. Theories of Pythagorean matrix are obtained. Using related algebra skills, we prove that the set which constitutes all n-order Pythagorean matrices is a finitely generated group of matrix multiplication and gives a generated tuple of this finitely generated group ( ) simultaneously. #1Maohua LeH-Index: 1 #2Gökhan Soydan (Uludağ University)H-Index: 5 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... #1Nainrong Feng (Anhui Normal University) Let a,b,cbe relatively prime positive integers such that a^2+b^2=c^2, 2|b In this paper, we show that Pythagorean triples (a, b,c)must satisfy abc\equiv{0\; (\mod3\cdot{4}\cdot{5}})and c\neq{0\; (\mod{3}}) and we also prove that for (a,b,c)\in\{(a,b,c)|a\equiv{0\;(\mod{3}}),b\equiv{0\;(\mod{4}}),c\equiv{0\; (\mod{5}})\}\bigcup\{(a,b,c)|b\equiv{0\;(\mod{12}}),c\equiv{0\;(\mod{5}})\} the only solution of$a^x+b^y=c^z\qquad{z},y,z\in{N} in positive integers is x, y, z) = (2, ...