On the Diophantine equation (8n)^{x}+(15n)^{y}=(17n)^{z}

Published on Oct 1, 2012in Bulletin of The Australian Mathematical Society0.63
· DOI :10.1017/S000497271100342X
Zhi-Juan Yang2
Estimated H-index: 2
(Anhui Normal University),
Min Tang5
Estimated H-index: 5
(Anhui Normal University)
Sources
Abstract
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
📖 Papers frequently viewed together
References7
Newest
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...
Source
Source
Let n be a positive integer, and let ( a, b, c ) be a primitive Pythagorean triple. In this paper we give certain conditions for the equation ( an ) x + ( bn ) y = ( cn ) z to have positive integer solutions ( x, y, z ) with ( x, y, z ) ≠ (2, 2, 2). In particular, we show that x, y and z must be distinct.
Source
#1Moujie DengH-Index: 1
#2Graeme L. Cohen (UTS: University of Technology, Sydney)H-Index: 9
Let a , b , c be relatively prime positive integers such that a 2 + b 2 = c 2 . Jeśmanowicz conjectured in 1956 that for any given positive integer n the only solution of ( an ) x + ( bn ) y = ( en ) z in positive integers is x = y = z = 2. Building on the work of earlier writers for the case when n = 1 and c = b + 1, we prove the conjecture when n > 1, c = b + 1 and certain further divisibility conditions are satisfied. This leads to the proof of the full conjecture for the five triples ( a , b...
Source
#1Kei TakakuwaH-Index: 1
Source
Source
#1Maohua LeH-Index: 1
Source
Cited By8
Newest
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, ...
#1Gökhan SoydanH-Index: 5
#2Musa DemirciH-Index: 3
Last. Alain TogbéH-Index: 13
view all 4 authors...
We give a survey on some results covering the last 60 years concerning Jeśmanowicz' conjecture. Moreover, we conclude the survey with a new result by showing that the special Diophantine equation $(20k)^x+(99k)^y=(101k)^z has no solution other than x,y,z)=(2,2,2)$.
In 1956, conjectured that, for any positive integer n and any primitive Pythagorean triple (a, b, c) with , the equation has the unique solution (x, y, z) = (2, 2, 2). In this paper, under some conditions, we prove the conjecture for the primitive Pythagorean triples .
Source
Let a, b, c be relatively prime positive integers such that a 2 + b 2 = c 2. Jeśmanowicz’ conjecture on Pythagorean numbers states that for any positive integer N, the Diophantine equation (aN) x + (bN) y = (cN) z has no positive solution (x, y, z) other than x = y = z = 2. In this paper, we prove this conjecture for the case that a or b is a power of 2.
Source
#1Cuifang SunH-Index: 1
#2Zhi ChengH-Index: 1
Let (a;b;c) be a primitive Pythagorean triple. Jesmanowicz conjectured in 1956 that for any positive integer n, the Diophantine equation (an) x + (bn) y = (cn) z has only the positive integer solution (x;y;z) = (2;2;2). Let p 3 (mod 4) be a prime and s be some positive integer. In the paper, we show that the conjecture is true when (a;b;c) = (4p 2s 1;4p s ;4p 2s + 1) and certain divisibility conditions are satised.
Let (a, b, c)be a primitive Pythagorean triple satisfying {a}^{2} + {b}^{2} = {c}^{2} . In 1956, Jeśmanowicz conjectured that for any given positive integer nthe only solution of \mathop{(an)}\nolimits ^{x} + \mathop{(bn)}\nolimits ^{y} = \mathop{(cn)}\nolimits ^{z} in positive integers is x= y= z= 2. In this paper, for the primitive Pythagorean triple (a, b, c)= (4{k}^{2} - 1, 4k, 4{k}^{2} + 1)with k= {2}^{s} for some positive integer s\geq 0, we prove the conjecture whe...
Source
#1Min Tang (Anhui Normal University)H-Index: 5
#2Zhi-Juan Yang (Anhui Normal University)H-Index: 2
Let a, b, cbe relatively prime positive integers such that {a}^{2} + {b}^{2} = {c}^{2} . In 1956, Jeśmanowicz conjectured that for any positive integer n, the only solution of \mathop{(an)}\nolimits ^{x} + \mathop{(bn)}\nolimits ^{y} = \mathop{(cn)}\nolimits ^{z} in positive integers is (x, y, z)= (2, 2, 2). In this paper, we consider Jeśmanowicz’ conjecture for Pythagorean triples (a, b, c)if a= c- 2and cis a Fermat prime. For example, we show that Jeśmanowicz’ conjectur...
Source
This website uses cookies.
We use cookies to improve your online experience. By continuing to use our website we assume you agree to the placement of these cookies.
To learn more, you can find in our Privacy Policy.