Person:
SOYDAN, GÖKHAN

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SOYDAN

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GÖKHAN

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Now showing 1 - 10 of 23
  • Publication
    On the diophantine equation (5 pn 2 - 1) x
    (Honam Mathematical Soc, 2020-03-01) Kızıldere, Elif; Soydan, Gökhan; SOYDAN, GÖKHAN; Kızıldere, Elif; Bursa Uludağ Üniversitesi/Fen-Edebiyat Fakültesi/Matematik Bölümü; 0000-0002-6321-4132; M-9459-2017; GRN-4828-2022
    Let p be a prime number with p > 3, p 3 (mod 4) and let n be a positive integer. In this paper, we prove that the Diophantine equation (5pn(2) - 1)(x) + (p(p - 5)n(2) + 1)(y) = (pn)(z) has only the positive integer solution (x; y; z) = (1; 1; 2) where pn +/- 1 (mod 5). As an another result, we show that the Diophantine equation (35n(2) - 1)(x) + (14n(2) + 1)(y) = (7n)(z) has only the positive integer solution (x, y, z) = (1; 1; 2) where n +/- 3 (mod 5) or 5 vertical bar n. On the proofs, we use the properties of Jacobi symbol and Baker's method.
  • Publication
    A note on the diophantine equation x2=4pn-4pm + l2
    (Indian Nat Sci Acad, 2021-11-11) Abu Muriefah, Fadwa S.; Le, Maohua; Soydan, Gokhan; SOYDAN, GÖKHAN; Bursa Uludağ Üniversitesi/Fen Edebiyat Fakültesi/Matematik Bölümü.
    Let l be a fixed odd positive integer. In this paper, using some classical results on the generalized Ramanujan-Nagell equation, we completely derive all solutions (p, x, m, n) of the equation x(2) = 4p(n)-4p(m)+l(2) with l(2) < 4p(m) for any l > 1, where p is a prime, x, m, n are positive integers satisfying gcd(x, l) = 1 and m < n. Meanwhile we give a method to solve the equation with l(2) > 4p(m). As an example of using this method, we find all solutions (p, x, m, n) of the equation for l is an element of {5, 7}.
  • Publication
    A note on the exponential diophantine equation A2n)x + (B2n)y = ((A2 + B2)n)z
    (Croatian Mathematical Society, 2020-12-01) Le, Maohua; Soydan, Gökhan; SOYDAN, GÖKHAN; Bursa Uludağ Üniversitesi/Fen-Edebiyat Fakültesi/Matematik Bölümü.; 0000-0002-6321-4132; M-9459-2017
    Let A, B be positive integers such that. inin{A, B} > 1, gcd(A, B) = 1 and 2 vertical bar B. In this paper, using an upper bound for solutions of ternary purely exponential Diophantine equations due to R. Scott and R. Styer, we prove that, for any positive integer n, if A > B-3/8, then the equation (A(2)n)(x) + (B(2)n)(y) = ((A(2) + B-2)n)(z) has no positive integer solutions (x, y, z) with x > z > y; if B > A(3)/6, then it has no solutions (x, y, z) with y > z > x. Thus, combining the above conclusion with some existing results, we can deduce that, for any positive integer n, if B 2 (mod 4) and A > B-3/8, then this equation has only the positive integer solution (x, y, z)= (1,1,1).
  • Publication
    The shuffle variant of a diophantine equation of miyazaki and togbe
    (Soc Matematice Romania, 2021-01-01) Kızıldere, Elif; Soydan, Gökhan; Han, Qing; Yuan, Pingzhi; Kızıldere, Elif; SOYDAN, GÖKHAN; Bursa Uludağ Üniversitesi/Fen-Edebiyat Fakültesi/Matematik Bölümü; 0000-0002-6321-4132; GRN-4828-2022; GEK-9891-2022
    In 2012, T. Miyazaki and A. Togbe gave all of the solutions of the Diophantine equations (2am - 1)(x) + (2m)(y) = (2am + 1)(z) and b(x) + 2(y) = (b + 2)(z) in positive integers x, y, z, a > 1 and b >= 5 odd. In this paper, we propose a similar problem (which we call the shuffle variant of a Diophantine equation of Miyazaki and Togbe). Here we first prove that the Diophantine equation (2am + 1)(x) + (2m)(y) = (2am - 1)(z) has only the solutions (a, m, x, y, z) = (2, 1, 2, 1, 3) and (2, 1, 1, 2, 2) in positive integers a > 1, m, x, y, z. Then using this result, we show that the Diophantine equation b(x) + 2(y) = (b - 2)(z) has only the solutions (b, x, y, z) = (5,2, 1, 3) and (5,1, 2, 2) in positive integers x, y, z and b odd.
  • Publication
    A note on terai's conjecture concerning primitive pythagorean triples
    (Hacettepe Üniversitesi, 2021-01-01) Le, Maohua; Soydan, Gökhan; SOYDAN, GÖKHAN; Bursa Uludağ Üniversitesi/Fen-Edebiyat Fakültesi/Matematik Bölümü; 0000-0002-6321-4132; M-9459-2017
    Let f,g be positive integers such that f > g, gcd(f,g) =1 and f not equivalent to g (mod 2). In 1993, N. Terai conjectured that the equation x(2) + (f(2) - g(2))(y) = (f(2) + g(2))(z) has only one positive integer solution (x, y, z) = (2 fg, 2, 2). This is a problem that has not been solved yet. In this paper, using elementary number theory methods with some known results on higher Diophantine equations, we prove that if f = 2(r)s and g = 1, where r, s are positive integers satisfying 2 inverted iota s, r >= 2 and s < 2(r-)(1), then Terai's conjecture is true.
  • Publication
    Rational sequences on different models of elliptic curves
    (Croatian Mathematical Soc, 2019-06-01) Sadek, Mohammad; Soydan, Gökhan; SOYDAN, GÖKHAN; Çelik, Gamze Savaş; Bursa Uludağ Üniversitesi/Fen Edebiyat Fakültesi/Matematik Anabilim Dalı.; 0000-0002-6321-4132; M-9459-2017
    Given a set S of elements in a number field k, we discuss the existence of planar algebraic curves over k which possess rational points whose x-coordinates are exactly the elements of S. If the size vertical bar S vertical bar of S is either 4, 5, or 6, we exhibit infinite families of (twisted) Edwards curves and (general) Huff curves for which the elements of S are realized as the x-coordinates of rational points on these curves. This generalizes earlier work on progressions of certain types on some algebraic curves.
  • Publication
    Rational points in geometric progression on the unit circle
    (Kossuth Lajos Tudomanyegyetem, 2021-01-01) Çelik, Gamze Savaş; Sadek, Mohammad; Soydan, Gökhan; Çelik, Gamze Savaş; SOYDAN, GÖKHAN; Bursa Uludağ Üniversitesi/Fen-Edebiyat Fakültesi/Matematik Bölümü; 0000-0002-6321-4132; EPC-6610-2022; GEK-9891-2022
    A sequence of rational points on an algebraic planar curve is said to form an r-geometric progression sequence if either the abscissae or the ordinates of these points form a geometric progression sequence with ratio r. In this work, we prove the existence of infinitely many rational numbers r such that for each r there exist infinitely many r-geometric progression sequences on the unit circle x(2) + y(2) = 1 of length at least 3.
  • Publication
    A note on the ternary purely exponential diophantine equation A x + Y = C z with A plus B = C 2
    (Akademiai Kiado Zrt, 2020-06-01) Kızıldere, Elif; le, Maohua; Soydan, Gökhan; Kızıldere, Elif; SOYDAN, GÖKHAN; Bursa Uludağ Üniversitesi/Fen-Edebiyat Fakültesi/Matematik Bölümü; 0000-0002-6321-4132; M-9459-2017; GRN-4828-2022
    Let l,m,r be fixed positive integers such that 2 vertical bar l, 3 lm, l > r and 3 vertical bar r. In this paper, using the BHV theorem on the existence of primitive divisors of Lehmer numbers, we prove that if min{rlm(2) - 1,(l-r)lm(2) + 1} >30, then the equation (rlm(2) - 1)(x) + ((l - r)lm(2) + 1)(y) = (lm)(z) only the positive integer solution (x,y,z) = (1,1,2).
  • Publication
    On triangles with coordinates of vertices from the terms of the sequences {u kn} and {vkn}
    (Croatian Acad Sciences Arts, 2020-01-01) Ömür, Neşe; Soydan, Gökhan; Ulutaş, Yücel Türker; Doğru, Yusuf; SOYDAN, GÖKHAN; Bursa Uludağ Üniversitesi/Fen-Edebiyat Fakültesi/Matematik Bölümü; 0000-0002-6321-4132; M-9459-2017
    In this paper, we determine some results of the triangles with coordinates of vertices involving the terms of the sequences {U-kn} and {V-kn} where U-kn are terms of a second order recurrent sequence and V-kn are terms in the companion sequence for odd positive integer k, generalizing works of Cerin. For example, the cotangent of the Brocard angle of the triangle Delta(kn) iscot(Omega(Delta kn)) = Uk(2n+3) V-2k - Vk(2n+3)Uk/(-1)U-n(2k).
  • Publication
    Resolution of the equation (3 x 1-1)(3x2-1) = (5y1-1)(5y2-1)
    (Rocky Mt Math Consortium, 2020-08-01) Liptai, Kalman; Nemeth, Laszlo; Soydan, Gökhan; Szalay, Laszlo; SOYDAN, GÖKHAN; Bursa Uludağ Üniversitesi/Fen-Edebiyat Fakültesi/Matematik Bölümü; 0000-0002-6321-4132; M-9459-2017
    Consider the diophantine equation (3(x1) - 1)(3(x2) - 1) = (5(y1) - 1)(5(y2) - 1) in positive integers x(1) <= x(2) and y(1) <= y(2). Each side of the equation is a product of two terms of a given binary recurrence. We prove that the only solution to the title equation is (x(1), x(2), y(1), y(2)) = (1, 2, 1, 1). The main novelty of our result is that we allow products of two terms on both sides.