Abstract:Let q be an odd prime. Let c > 1 and t be positive integers such that q t þ 1 ¼ 2c 2 . Using elementary method and a result due to Ljunggren concerning the Diophantine equation x n À1 xÀ1 ¼ y 2 , we show that the Diophantine equation x 2 þ q m ¼ c 2n has the only positive integer solution ðx; m; nÞ ¼ ðc 2 À 1; t; 2Þ. As applications of this result some new results on the Diophantine equation x 2 þ q m ¼ c n and the Diophantine equation x 2 þ ð2c À 1Þ m ¼ c n are obtained. In particular, we prove that Terai's c… Show more
“…Using these results, together with results of Ljunggren [5], Zhu [7] and Arif-Abu Muriefah [1], Terai showed that, apart from c = 12, 24, his conjecture holds for 2 ≤ c ≤ 30. The cases c = 12, 24 have been treated in [4]. In this paper, we show that Terai's conjecture is true under a wider range of conditions on c and 2c − 1.…”
Section: Introductionmentioning
confidence: 61%
“…and[4], we may suppose that 31 ≤ c ≤ 499 with c ≡ 3 (mod 4).For c = p 2s+1 , where p is a prime, p ≡ 3 (mod 4), s ≥ 0 and 31 ≤ p 2s+1 ≤ 499, that is, c ∈ {31,43, 47, 59, 67, 71, 79, 83, 103, 107, 127, 131, 139, 151, 163, 167, 179, 191, …”
Let $c\geq 2$ be a positive integer. Terai [‘A note on the Diophantine equation $x^{2}+q^{m}=c^{n}$’, Bull. Aust. Math. Soc.90 (2014), 20–27] conjectured that the exponential Diophantine equation $x^{2}+(2c-1)^{m}=c^{n}$ has only the positive integer solution $(x,m,n)=(c-1,1,2)$. He proved his conjecture under various conditions on $c$ and $2c-1$. In this paper, we prove Terai’s conjecture under a wider range of conditions on $c$ and $2c-1$. In particular, we show that the conjecture is true if $c\equiv 3\hspace{0.6em}({\rm mod}\hspace{0.2em}4)$ and $3\leq c\leq 499$.
“…Using these results, together with results of Ljunggren [5], Zhu [7] and Arif-Abu Muriefah [1], Terai showed that, apart from c = 12, 24, his conjecture holds for 2 ≤ c ≤ 30. The cases c = 12, 24 have been treated in [4]. In this paper, we show that Terai's conjecture is true under a wider range of conditions on c and 2c − 1.…”
Section: Introductionmentioning
confidence: 61%
“…and[4], we may suppose that 31 ≤ c ≤ 499 with c ≡ 3 (mod 4).For c = p 2s+1 , where p is a prime, p ≡ 3 (mod 4), s ≥ 0 and 31 ≤ p 2s+1 ≤ 499, that is, c ∈ {31,43, 47, 59, 67, 71, 79, 83, 103, 107, 127, 131, 139, 151, 163, 167, 179, 191, …”
Let $c\geq 2$ be a positive integer. Terai [‘A note on the Diophantine equation $x^{2}+q^{m}=c^{n}$’, Bull. Aust. Math. Soc.90 (2014), 20–27] conjectured that the exponential Diophantine equation $x^{2}+(2c-1)^{m}=c^{n}$ has only the positive integer solution $(x,m,n)=(c-1,1,2)$. He proved his conjecture under various conditions on $c$ and $2c-1$. In this paper, we prove Terai’s conjecture under a wider range of conditions on $c$ and $2c-1$. In particular, we show that the conjecture is true if $c\equiv 3\hspace{0.6em}({\rm mod}\hspace{0.2em}4)$ and $3\leq c\leq 499$.
“…The above conjecture has been verified in some special cases (see [1,[10][11][12][13]15], [23] and [26]). But, in general, it is far from solved problem.…”
Let k be a positive integer. In this paper, using the modular approach, we prove that if k ≡ 0 (mod 4), 30 < k < 724 and 2k − 1 is an odd prime power, then the equation ( * ) x 2 + (2k − 1) y = k z has only one positive integer solution (x, y, z) = (k − 1, 1, 2). The above results solve some difficult cases of Terai's conecture concerning the equation ( * ). 2020 Mathematics Subject Classification. 11D61.
“…For each set S = {2, 3, p}, we have tabulated these solutions in the file http://www.math.ubc.ca/~bennett/2-3-p.pdf. By quick examination of this table for p = 23 and p = 47, we immediately deduce the following result about equations that remained unsolved in Proposition 3.3 of Terai [24] (but have been recently solved via rather different methods by Deng [11]).…”
Section: 2mentioning
confidence: 89%
“…We now deal with equation (2) when n = 3 using a (p, p, 3) Frey-Hellegouarch curve approach. The corresponding equations to treat are of the shape (11) x + y = wz 3 where x, y and w are pairwise coprime S-units. We may assume, without loss of generality, that w is cubefree and positive and that we have x ≡ 0 (mod 3) and y ≡ 2 (mod 3).…”
In this paper, we develop a new method for finding all perfect powers which can be expressed as the sum of two rational S-units, where S is a finite set of primes. Our approach is based upon the modularity of Galois representations and, for the most part, does not require lower bounds for linear forms in logarithms. Its main virtue is that it enables to carry out such a program explicitly, at least for certain small sets of primes S; we do so for S = {2, 3} and S = {3, 5, 7}.
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