Displaying similar documents to “A diophantine equation involving special prime numbers”

The exceptional set for Diophantine inequality with unlike powers of prime variables

Wenxu Ge, Feng Zhao (2018)

Czechoslovak Mathematical Journal

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Suppose that λ 1 , λ 2 , λ 3 , λ 4 are nonzero real numbers, not all negative, δ > 0 , 𝒱 is a well-spaced set, and the ratio λ 1 / λ 2 is algebraic and irrational. Denote by E ( 𝒱 , N , δ ) the number of v 𝒱 with v N such that the inequality | λ 1 p 1 2 + λ 2 p 2 3 + λ 3 p 3 4 + λ 4 p 4 5 - v | < v - δ has no solution in primes p 1 , p 2 , p 3 , p 4 . We show that E ( 𝒱 , N , δ ) N 1 + 2 δ - 1 / 72 + ε for any ε > 0 .

A Diophantine inequality with four squares and one k th power of primes

Quanwu Mu, Minhui Zhu, Ping Li (2019)

Czechoslovak Mathematical Journal

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Let k 5 be an odd integer and η be any given real number. We prove that if λ 1 , λ 2 , λ 3 , λ 4 , μ are nonzero real numbers, not all of the same sign, and λ 1 / λ 2 is irrational, then for any real number σ with 0 < σ < 1 / ( 8 ϑ ( k ) ) , the inequality | λ 1 p 1 2 + λ 2 p 2 2 + λ 3 p 3 2 + λ 4 p 4 2 + μ p 5 k + η | < max 1 j 5 p j - σ has infinitely many solutions in prime variables p 1 , p 2 , , p 5 , where ϑ ( k ) = 3 × 2 ( k - 5 ) / 2 for k = 5 , 7 , 9 and ϑ ( k ) = [ ( k 2 + 2 k + 5 ) / 8 ] for odd integer k with k 11 . This improves a recent result in W. Ge, T. Wang (2018).

Mersenne numbers as a difference of two Lucas numbers

Murat Alan (2022)

Commentationes Mathematicae Universitatis Carolinae

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Let ( L n ) n 0 be the Lucas sequence. We show that the Diophantine equation L n - L m = M k has only the nonnegative integer solutions ( n , m , k ) = ( 2 , 0 , 1 ) , ( 3 , 1 , 2 ) , ( 3 , 2 , 1 ) , ( 4 , 3 , 2 ) , ( 5 , 3 , 3 ) , ( 6 , 2 , 4 ) , ( 6 , 5 , 3 ) where M k = 2 k - 1 is the k th Mersenne number and n > m .

The number of solutions to the generalized Pillai equation ± r a x ± s b y = c .

Reese Scott, Robert Styer (2013)

Journal de Théorie des Nombres de Bordeaux

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We consider N , the number of solutions ( x , y , u , v ) to the equation ( - 1 ) u r a x + ( - 1 ) v s b y = c in nonnegative integers x , y and integers u , v { 0 , 1 } , for given integers a &gt; 1 , b &gt; 1 , c &gt; 0 , r &gt; 0 and s &gt; 0 . When gcd ( r a , s b ) = 1 , we show that N 3 except for a finite number of cases all of which satisfy max ( a , b , r , s , x , y ) &lt; 2 · 10 15 for each solution; when gcd ( a , b ) &gt; 1 , we show that N 3 except for three infinite families of exceptional cases. We find several different ways to generate an infinite number of cases giving N = 3 solutions.

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.

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 perfect powers in k -generalized Pell sequence

Zafer Şiar, Refik Keskin, Elif Segah Öztaş (2023)

Mathematica Bohemica

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Let k 2 and let ( P n ( k ) ) n 2 - k be the k -generalized Pell sequence defined by P n ( k ) = 2 P n - 1 ( k ) + P n - 2 ( k ) + + P n - k ( k ) for n 2 with initial conditions P - ( k - 2 ) ( k ) = P - ( k - 3 ) ( k ) = = P - 1 ( k ) = P 0 ( k ) = 0 , P 1 ( k ) = 1 . In this study, we handle the equation P n ( k ) = y m in positive integers n , m , y , k such that k , y 2 , and give an upper bound on n . Also, we will show that the equation P n ( k ) = y m with 2 y 1000 has only one solution given by P 7 ( 2 ) = 13 2 .

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 .

A note on the weighted Khintchine-Groshev Theorem

Mumtaz Hussain, Tatiana Yusupova (2014)

Journal de Théorie des Nombres de Bordeaux

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Let W ( m , n ; ψ ̲ ) denote the set of ψ 1 , ... , ψ n –approximable points in m n . The classical Khintchine–Groshev theorem assumes a monotonicity condition on the approximating functions ψ ̲ . Removing monotonicity from the Khintchine–Groshev theorem is attributed to different authors for different cases of m and n . It can not be removed for m = n = 1 as Duffin–Schaeffer provided the counter example. We deal with the only remaining case m = 2 and thereby remove all unnecessary conditions from the Khintchine–Groshev theorem. ...

Bartz-Marlewski equation with generalized Lucas components

Hayder R. Hashim (2022)

Archivum Mathematicum

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Let { U n } = { U n ( P , Q ) } and { V n } = { V n ( P , Q ) } be the Lucas sequences of the first and second kind respectively at the parameters P 1 and Q { - 1 , 1 } . In this paper, we provide a technique for characterizing the solutions of the so-called Bartz-Marlewski equation x 2 - 3 x y + y 2 + x = 0 , where ( x , y ) = ( U i , U j ) or ( V i , V j ) with i , j 1 . Then, the procedure of this technique is applied to completely resolve this equation with certain values of such parameters.

On the exponential diophantine equation x y + y x = z z

Xiaoying Du (2017)

Czechoslovak Mathematical Journal

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For any positive integer D which is not a square, let ( u 1 , v 1 ) be the least positive integer solution of the Pell equation u 2 - D v 2 = 1 , and let h ( 4 D ) denote the class number of binary quadratic primitive forms of discriminant 4 D . If D satisfies 2 D and v 1 h ( 4 D ) 0 ( mod D ) , then D is called a singular number. In this paper, we prove that if ( x , y , z ) is a positive integer solution of the equation x y + y x = z z with 2 z , then maximum max { x , y , z } < 480000 and both x , y are singular numbers. Thus, one can possibly prove that the equation has no positive integer solutions...

Padovan and Perrin numbers as products of two generalized Lucas numbers

Kouèssi Norbert Adédji, Japhet Odjoumani, Alain Togbé (2023)

Archivum Mathematicum

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Let P m and E m be the m -th Padovan and Perrin numbers respectively. Let r , s be non-zero integers with r 1 and s { - 1 , 1 } , let { U n } n 0 be the generalized Lucas sequence given by U n + 2 = r U n + 1 + s U n , with U 0 = 0 and U 1 = 1 . In this paper, we give effective bounds for the solutions of the following Diophantine equations P m = U n U k and E m = U n U k , where m , n and k are non-negative integers. Then, we explicitly solve the above Diophantine equations for the Fibonacci, Pell and balancing sequences.

On a system of equations with primes

Paolo Leonetti, Salvatore Tringali (2014)

Journal de Théorie des Nombres de Bordeaux

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Given an integer n 3 , let u 1 , ... , u n be pairwise coprime integers 2 , 𝒟 a family of nonempty proper subsets of { 1 , ... , n } with “enough” elements, and ε a function 𝒟 { ± 1 } . Does there exist at least one prime q such that q divides i I u i - ε ( I ) for some I 𝒟 , but it does not divide u 1 u n ? We answer this question in the positive when the u i are prime powers and ε and 𝒟 are subjected to certain restrictions. We use the result to prove that, if ε 0 { ± 1 } and A is a set of three or more primes that contains all prime divisors of any...