Results 11 to 20 of about 44 (43)
On the exponential Diophantine equation mx+(m+1)y=(1+m+m2)z
Analele Stiintifice ale Universitatii Ovidius Constanta: Seria Matematica, 2021 Let m > 1 be a positive integer. We show that the exponential Diophantine equation mx + (m + 1)y = (1 + m + m2)z has only the positive integer solution (x, y, z) = (2, 1, 1) when m ≥ 2.
Alan Murat
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An elementary proof of Fermat’s last theorem for all even exponents
Journal of Mathematical Cryptology, 2020 An elementary proof that the equation x2n + y2n = z2n can not have any non-zero positive integer solutions when n is an integer ≥ 2 is presented. To prove that the equation has no integer solutions it is first hypothesized that the equation has integer ...
Karmakar Sudhangshu B.
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On a variation of the Erdős–Selfridge superelliptic curve
Bulletin of the London Mathematical Society, Volume 51, Issue 4, Page 633-638, August 2019., 2019 Abstract In a recent paper by Das, Laishram and Saradha, it was shown that if there exists a rational solution of yl=(x+1)…(x+i−1)(x+i+1)…(x+k) for i not too close to k/2 and y≠0, then logl<3k. In this paper, we extend the number of terms that can be missing in the equation and remove the condition on i.
Sam Edis
wiley +1 more source
Repdigits as Euler functions of Lucas numbers
Analele Stiintifice ale Universitatii Ovidius Constanta: Seria Matematica, 2016 We prove some results about the structure of all Lucas numbers whose Euler function is a repdigit in base 10. For example, we show that if Ln is such a Lucas number, then n < 10111 is of the form p or p2, where p3 | 10p-1 -1.
Bravo Jhon J. +3 more
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On the equation x2 + 2a · 3b = yn
International Journal of Mathematics and Mathematical Sciences, Volume 29, Issue 4, Page 239-244, 2002., 2002 We find all positive integer solutions (x, y, a, b, n) of x2 + 2a · 3b = yn with n ≥ 3 and coprime x and y.
Florian Luca
wiley +1 more source
On the Diophantine equation x2 + p2k+1 = 4yn
International Journal of Mathematics and Mathematical Sciences, Volume 31, Issue 11, Page 695-699, 2002., 2002 It has been proved that if p is an odd prime, y > 1, k ≥ 0, n is an integer greater than or equal to 4, (n, 3h) = 1 where h is the class number of the field Q(−p), then the equation x2 + p2k+1 = 4yn has exactly five families of solution in the positive integers x, y. It is further proved that when n = 3 and p = 3a2 ± 4, then it has a unique solution k =
S. Akhtar Arif, Amal S. Al-Ali
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International Journal of Mathematics and Mathematical Sciences, Volume 2025, Issue 1, 2025.Let N,Z, and Q be the sets of natural, integers, and rational numbers, respectively. Our objective, involving a predetermined positive integer a, is to study a characterization of Diophantine equations of the form a + y2 = z2. Building on this result, we aim to obtain a characterization for Pythagorean n‐tuples.
Roberto Amato, Anwar Saleh Alwardi
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Narayana Numbers With Zeckendorf Partition in Two Terms
International Journal of Mathematics and Mathematical Sciences, Volume 2025, Issue 1, 2025.The Narayan’s cow sequence starts with the terms 1, 1, and 1. Each subsequent term is obtained as the sum of the previous term and the term three places before. A term of this sequence is called a Narayana number. The mathematician Zeckendorf proved that every positive integer has a unique decomposition into a sum of distinct and nonconsecutive ...
Japhet Odjoumani +2 more
wiley +1 more source
Tribonacci numbers that are concatenations of two repdigits. [PDF]
Rev R Acad Cienc Exactas Fis Nat A Mat, 2020 Ddamulira M.
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Repdigits as sums of three Padovan numbers. [PDF]
Bol Soc Mat Mex, 2020 Ddamulira M.
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