Displaying similar documents to “Effective results for Diophantine equations over finitely generated domains”

On systems of diophantine equations with a large number of solutions

Jerzy Browkin (2010)

Colloquium Mathematicae

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We consider systems of equations of the form x i + x j = x k and x i · x j = x k , which have finitely many integer solutions, proposed by A. Tyszka. For such a system we construct a slightly larger one with much more solutions than the given one.

Multiplicative relations on binary recurrences

Florian Luca, Volker Ziegler (2013)

Acta Arithmetica

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Given a binary recurrence u n n 0 , we consider the Diophantine equation u n 1 x 1 u n L x L = 1 with nonnegative integer unknowns n 1 , . . . , n L , where n i n j for 1 ≤ i < j ≤ L, m a x | x i | : 1 i L K , and K is a fixed parameter. We show that the above equation has only finitely many solutions and the largest one can be explicitly bounded. We demonstrate the strength of our method by completely solving a particular Diophantine equation of the above form.

Finiteness results for Diophantine triples with repdigit values

Attila Bérczes, Florian Luca, István Pink, Volker Ziegler (2016)

Acta Arithmetica

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Let g ≥ 2 be an integer and g be the set of repdigits in base g. Let g be the set of Diophantine triples with values in g ; that is, g is the set of all triples (a,b,c) ∈ ℕ³ with c < b < a such that ab + 1, ac + 1 and bc + 1 lie in the set g . We prove effective finiteness results for the set g .

A remark on a Diophantine equation of S. S. Pillai

Azizul Hoque (2024)

Czechoslovak Mathematical Journal

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S. S. Pillai proved that for a fixed positive integer a , the exponential Diophantine equation x y - y x = a , min ( x , y ) > 1 , has only finitely many solutions in integers x and y . We prove that when a is of the form 2 z 2 , the above equation has no solution in integers x and y with gcd ( x , y ) = 1 .

Diophantine equations involving factorials

Horst Alzer, Florian Luca (2017)

Mathematica Bohemica

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We study the Diophantine equations ( k ! ) n - k n = ( n ! ) k - n k and ( k ! ) n + k n = ( n ! ) k + n k , where k and n are positive integers. We show that the first one holds if and only if k = n or ( k , n ) = ( 1 , 2 ) , ( 2 , 1 ) and that the second one holds if and only if k = n .

On some Diophantine equations involving balancing numbers

Euloge Tchammou, Alain Togbé (2021)

Archivum Mathematicum

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In this paper, we find all the solutions of the Diophantine equation B 1 p + 2 B 2 p + + k B k p = B n q in positive integer variables ( k , n ) , where B i is the i t h balancing number if the exponents p , q are included in the set { 1 , 2 } .

A note on the article by F. Luca “On the system of Diophantine equations a ² + b ² = ( m ² + 1 ) r and a x + b y = ( m ² + 1 ) z ” (Acta Arith. 153 (2012), 373-392)

Takafumi Miyazaki (2014)

Acta Arithmetica

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Let r,m be positive integers with r > 1, m even, and A,B be integers satisfying A + B ( - 1 ) = ( m + ( - 1 ) ) r . We prove that the Diophantine equation | A | x + | B | y = ( m ² + 1 ) z has no positive integer solutions in (x,y,z) other than (x,y,z) = (2,2,r), whenever r > 10 74 or m > 10 34 . Our result is an explicit refinement of a theorem due to F. Luca.

Diophantine triples with values in binary recurrences

Clemens Fuchs, Florian Luca, Laszlo Szalay (2008)

Annali della Scuola Normale Superiore di Pisa - Classe di Scienze

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In this paper, we study triples a , b and c of distinct positive integers such that a b + 1 , a c + 1 and b c + 1 are all three members of the same binary recurrence sequence.

On x n + y n = n ! z n

Susil Kumar Jena (2018)

Communications in Mathematics

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In p. 219 of R.K. Guy’s , 3rd edn., Springer, New York, 2004, we are asked to prove that the Diophantine equation x n + y n = n ! z n has no integer solutions with n + and n > 2 . But, contrary to this expectation, we show that for n = 3 , this equation has infinitely many primitive integer solutions, i.e. the solutions satisfying the condition gcd ( x , y , z ) = 1 .

Complete solution of the Diophantine equation x y + y x = z z

Mihai Cipu (2019)

Czechoslovak Mathematical Journal

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The triples ( x , y , z ) = ( 1 , z z - 1 , z ) , ( x , y , z ) = ( z z - 1 , 1 , z ) , where z , satisfy the equation x y + y x = z z . In this paper it is shown that the same equation has no integer solution with min { x , y , z } > 1 , thus a conjecture put forward by Z. Zhang, J. Luo, P. Z. Yuan (2013) is confirmed.

On the diophantine equation x y - y x = c z

Zhongfeng Zhang, Jiagui Luo, Pingzhi Yuan (2012)

Colloquium Mathematicae

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Applying results on linear forms in p-adic logarithms, we prove that if (x,y,z) is a positive integer solution to the equation x y - y x = c z with gcd(x,y) = 1 then (x,y,z) = (2,1,k), (3,2,k), k ≥ 1 if c = 1, and either ( x , y , z ) = ( c k + 1 , 1 , k ) , k ≥ 1 or 2 x < y m a x 1 . 5 × 10 10 , c if c ≥ 2.

On the Diophantine equation ( 2 x - 1 ) ( p y - 1 ) = 2 z 2

Ruizhou Tong (2021)

Czechoslovak Mathematical Journal

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Let p be an odd prime. By using the elementary methods we prove that: (1) if 2 x , p ± 3 ( mod 8 ) , the Diophantine equation ( 2 x - 1 ) ( p y - 1 ) = 2 z 2 has no positive integer solution except when p = 3 or p is of the form p = 2 a 0 2 + 1 , where a 0 > 1 is an odd positive integer. (2) if 2 x , 2 y , y 2 , 4 , then the Diophantine equation ( 2 x - 1 ) ( p y - 1 ) = 2 z 2 has no positive integer solution.

On the Lebesgue-Nagell equation

Andrzej Dąbrowski (2011)

Colloquium Mathematicae

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We completely solve the Diophantine equations x ² + 2 a q b = y (for q = 17, 29, 41). We also determine all C = p a p k a k and C = 2 a p a p k a k , where p , . . . , p k are fixed primes satisfying certain conditions. The corresponding Diophantine equations x² + C = yⁿ may be studied by the method used by Abu Muriefah et al. (2008) and Luca and Togbé (2009).

On the Diophantine equation j = 1 k j F j p = F n q

Gökhan Soydan, László Németh, László Szalay (2018)

Archivum Mathematicum

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Let F n denote the n t h term of the Fibonacci sequence. In this paper, we investigate the Diophantine equation F 1 p + 2 F 2 p + + k F k p = F n q in the positive integers k and n , where p and q are given positive integers. A complete solution is given if the exponents are included in the set { 1 , 2 } . Based on the specific cases we could solve, and a computer search with p , q , k 100 we conjecture that beside the trivial solutions only F 8 = F 1 + 2 F 2 + 3 F 3 + 4 F 4 , F 4 2 = F 1 + 2 F 2 + 3 F 3 , and F 4 3 = F 1 3 + 2 F 2 3 + 3 F 3 3 satisfy the title equation.