# 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
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#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...
#1Maohua LeH-Index: 1
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.
#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...
#1Kei TakakuwaH-Index: 1
#1Maohua LeH-Index: 1
#1Maohua LeH-Index: 1
Cited By8
#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 $a^x+b^y=c^z\qquad{z},y,z\in{N} in positive integers is x, y, z) = (2, ... #1Gökhan SoydanH-Index: 5 #2Musa DemirciH-Index: 3 Last. Alain TogbéH-Index: 13 view all 4 authors... We give a survey on some results covering the last 60 years concerning Jeśmanowicz' conjecture. Moreover, we conclude the survey with a new result by showing that the special Diophantine equation$(20k)^x+(99k)^y=(101k)^z has no solution other than x,y,z)=(2,2,2)\$.
#1Mi-Mi MaH-Index: 1
#2Jian-Dong WuH-Index: 1
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 .
#1Takafumi Miyazaki (Nihon University)H-Index: 1
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.
#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.
#1Mou Jie Deng (Haida: Hainan University)H-Index: 3
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...
#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...