# Generator of Pythagorean triples and Je\acute{s}anowicz conjecture

Published on Jan 23, 2020in arXiv: Number Theory
Nainrong Feng (Anhui Normal University)
Sources
Abstract
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 $a^x+b^y=c^z\qquad{z},y,z\in{N} in positive integers is x, y, z) = (2, 2,2)$.
📖 Papers frequently viewed together
2 Authors
1 Author
1 Author
References7
#1Mou Jie Deng (Haida: Hainan University)H-Index: 3
#2Dong Ming Huang (Haida: Hainan University)H-Index: 1
#1Nobuhiro Terai (Ashikaga Institute of Technology)H-Index: 6
Abstract In 1956, Jeśmanowicz conjectured that the exponential Diophantine equation ( m 2 − n 2 ) x + ( 2 m n ) y = ( m 2 + n 2 ) z has only the positive integer solution ( x , y , z ) = ( 2 , 2 , 2 ) , where m and n are positive integers with m > n , gcd ( m , n ) = 1 and m ≢ n ( mod 2 ) . We show that if n = 2 , then Jeśmanowicz' conjecture is true. This is the first result that if n = 2 , then the conjecture is true without any assumption on m .
#1Min TangH-Index: 1
#2Jian-Xin WengH-Index: 1
Abstract. Let a, b, c be relatively prime positive integers such that a + b = c. In 1956, Jeśmanowicz conjectured that for any positive integer n, the only solution of (an) + (bn) = (cn) in positive integers is (x, y, z) = (2, 2, 2). Let k ≥ 1 be an integer and Fk = 2 k + 1 be k-th Fermat number. In this paper, we show that Jeśmanowicz’ conjecture is true for Pythagorean triples (a, b, c) = (Fk − 2, 22+1, Fk).
#1Takafumi MiyazakiH-Index: 9
#2Pingzhi Yuan (SCNU: South China Normal University)H-Index: 16
Last. Danyao Wu (SCNU: South China Normal University)H-Index: 1
view all 3 authors...
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 .
#1Zhi-Juan Yang (Anhui Normal University)H-Index: 2
#2Min Tang (Anhui Normal University)H-Index: 5
Let a,b,cbe relatively prime positive integers such that a^{2}+b^{2}=c^{2}.In 1956, Je\'{s}manowicz conjectured that for any given positive integer nthe only solution of (an)^{x}+(bn)^{y}=(cn)^{z}in positive integers is (x,y,z)=(2,2,2) In this paper, we show that (8n)^{x}+(15n)^{y}=(17n)^{z}has no solution other than (x,y,z)=(2,2,2)in positive integers. DOI: 10.1017/S000497271100342X
#1Maohua LeH-Index: 1
#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
view all 3 authors...
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...
Cited By0