2016
DOI: 10.1007/s10998-016-0133-4
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Generalized Fibonacci numbers of the form $$wx^{2}+1$$ w x 2 + 1

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Cited by 2 publications
(3 citation statements)
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“…In [1], the authors showed that when a ̸ = 0 and b ̸ = ∓2 are integers, the equation V n (P, ∓1) = ax 2 + b has only a finite number of solutions n. Moreover, the same authors showed that when a ̸ = 0 and b are integers, the equation U n (P, ±1) = ax 2 + b has only a finite number of solutions n. Keskin [14] solved the equation V n (P, −1) = wx 2 ∓ 1 for w = 1, 2, 3, 6 when P is odd. Keskin and Öğüt [15] solved the equations U n (P, −1) = wx 2 + 1 for w = 1, 2, 3, 5, 7, 10 when P is odd. Öğüt and Keskin [22] showed that only U 1 (P, −1) and U 2 (P, −1) may be of the form 11x 2 + 1 if P is odd.…”
Section: Introductionmentioning
confidence: 99%
“…In [1], the authors showed that when a ̸ = 0 and b ̸ = ∓2 are integers, the equation V n (P, ∓1) = ax 2 + b has only a finite number of solutions n. Moreover, the same authors showed that when a ̸ = 0 and b are integers, the equation U n (P, ±1) = ax 2 + b has only a finite number of solutions n. Keskin [14] solved the equation V n (P, −1) = wx 2 ∓ 1 for w = 1, 2, 3, 6 when P is odd. Keskin and Öğüt [15] solved the equations U n (P, −1) = wx 2 + 1 for w = 1, 2, 3, 5, 7, 10 when P is odd. Öğüt and Keskin [22] showed that only U 1 (P, −1) and U 2 (P, −1) may be of the form 11x 2 + 1 if P is odd.…”
Section: Introductionmentioning
confidence: 99%
“…Furthermore, the same authors [12] solved V n = kx 2 under some assumptions on k. In [1], when P is odd, Cohn solved V n = P x 2 and V n = 2P x 2 with Q = ±1. In [14], the authors determined, assuming Q = 1, all indices n such that V n (P, 1) = kx 2 when k | P and P is odd, where k is a square-free positive divisor of P. The values of n have been found for which V n (P, −1) is of the form kx 2 , 2kx 2 , kx 2 ± 1, and 2kx 2 ± 1 with k | P and k > 1 [4]. Moreover, the values of n have been found for which V n (P, −1) is of the form 2x 2 ± 1, 3x 2 − 1, and 6x 2 ± 1 [4] and the author give all integer solutions of the preceding equations.…”
Section: Introductionmentioning
confidence: 99%
“…In [14], the authors determined, assuming Q = 1, all indices n such that V n (P, 1) = kx 2 when k | P and P is odd, where k is a square-free positive divisor of P. The values of n have been found for which V n (P, −1) is of the form kx 2 , 2kx 2 , kx 2 ± 1, and 2kx 2 ± 1 with k | P and k > 1 [4]. Moreover, the values of n have been found for which V n (P, −1) is of the form 2x 2 ± 1, 3x 2 − 1, and 6x 2 ± 1 [4] and the author give all integer solutions of the preceding equations. Our results in this paper add to the above list the values of n for which V n (P, −1) is of the form 5kx 2 , 5kx 2 ± 1, 7kx 2 , and 7kx 2 ± 1 when k | P and k > 1.…”
Section: Introductionmentioning
confidence: 99%