On Jeśmanowicz' conjecture concerning Pythagorean numbers

Published on Jan 1, 1996
· DOI :10.3792/PJAA.72.97
Maohua Le1
Estimated H-index: 1
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Abstract
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1995
2 Authors (Yondong Guo, Maohua Le)
1 Author (Maohua Le)
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#1Kei TakakuwaH-Index: 1
#2You AsaedaH-Index: 1
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Let f, g be positive integers such that null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null null $f>g null null null null null null null null null null null null null null null null null null null null null null null null null , null null null null null null null ...
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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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#1Van Thien NguyenH-Index: 1
#2Viet Kh. NguyenH-Index: 1
Last. Pham Hung QuyH-Index: 1
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Let (a, b, c)be a primitive Pythagorean triple parameterized as a=u^2-v^2,\ b=2uv,\ c=u^2+v^2\ where u>v>0are co-prime and not of the same parity. In 1956, L. Je{\'s}manowicz conjectured that for any positive integer n the Diophantine equation (an)^x+(bn)^y=(cn)^zhas only the positive integer solution (x,y,z)=(2,2,2) In this connection we call a positive integer solution (x,y,z)\ne (2,2,2)with n>1exceptional. In 1999 M.-H. Le gave necessary conditions for the existence of...
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...
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...
#1Q. Han (Guangdong University of Foreign Studies)H-Index: 1
#2Pingzhi Yuan (SCNU: South China Normal University)H-Index: 16
Jeśmanowicz [9] conjectured that, for positive integers m and n with m > n, gcd(m,n) = 1 and \({m\not\equiv n\pmod{2}}\), the exponential Diophantine 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). We prove the conjecture for \({2 \| mn}\) and m + n has a prime factor p with \({p\not\equiv1\pmod{16}}\).
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#1M.-J. Deng (Haida: Hainan University)H-Index: 1
#2J. Guo (Haida: Hainan University)H-Index: 1
Let \({(m^2 - n^2, 2mn, m^2 + n^2)}\) be a primitive Pythagorean triple such that m, n are positive integers with \({ \gcd (m,n)=1}\), \({m > n}\), \({m\not\equiv n\pmod{2}}\). In 1956, Jeśmanowicz conjectured that the only positive integer solution to the exponential Diophantine equation \({(m^2-n^2)^x + (2mn)^y = (m^2+n^2)^z}\) is x = y = z = 2. Let \({(m,n)\equiv(u,v)\pmod{d}}\) denote \({m\equiv u\pmod{d}}\) and \({n\equiv v\pmod{d}}\). Using the theory of quartic residue character and eleme...
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#1Mou Jie Deng (Haida: Hainan University)H-Index: 3
#2Dong Ming Huang (Haida: Hainan University)H-Index: 1
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#1Takafumi Miyazaki (Nihon University)H-Index: 9
#2N. Terai (Oita University)H-Index: 1
Jeśmanowicz [7] conjectured that the exponential Diophantine equation \({{(m^2-n^2)}^x + {(2mn)}^y = {(m^2 + n^2)}^z}\) has only one 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. In [17], the second author showed that if n = 2, then Jeśmanowicz’ conjecture is true without any assumption on m. Let n be a positive integer satisfying \({n \equiv 2\mod 4}\) and m > 72n. We prove that if n satisfies at least one of thre...
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