Browsing by Author "Yuan, Pingzhi"

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    On the exponential diophantine equation (n-1)(x) + (n+2)(y) = n(z)
    (Ars Polona-Ruch, 2020-03-30) Bai, Hairong; Yuan, Pingzhi; Kızıldere, Elif; Soydan, Gökhan; Bursa Uludağ Üniversitesi/Fen-Edebiyat Fakültesi/Matematik Bölümü.; 0000-0002-6321-4132; 57204173004; 23566953200
    Suppose that n is a positive integer. We show that the only positive integer solutions (n, x, y, z) of the exponential Diophantine equation (n - 1)(x) + (n + 2)(y) = nz, n >= 2, xyz not equal 0, are (3, 2, 1, 2), (3,1, 2, 3). The main tools in the proofs are Baker's theory and Bilu-Hanrot-Voutier's result on primitive divisors of Lucas numbers.
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    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.