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    Please use this identifier to cite or link to this item: http://ir.lib.ncu.edu.tw/handle/987654321/7819


    Title: 關於方程式2x^2+1=3^n的研究;The Diophantine Equation 2x^2+1=3^n
    Authors: 李冠葦;Guan-Wei Li
    Contributors: 數學研究所
    Keywords: 方程式;Binary Recurrent Sequence;Diophantine Equatoin
    Date: 2002-06-04
    Issue Date: 2009-09-22 11:06:16 (UTC+8)
    Publisher: 國立中央大學圖書館
    Abstract: 約在1913年左右 Ramanujan 猜想方程式x^2+7=2^n只有五組正整數解,此猜想在1948年由Nagell首先給出證明,其後有釵h數學家以不同的方法再一次證明此猜想。在1980年以後,釵h數學家開始研究方程式D_1x^2+D_2=λ^2k^n 的正整數解。在余茂華一系列的論文中證明除了方程式x^2+7=2^(n+2)有五組正整數解及方程式 3x^2+5=2^(n+2)、x^2+11=2^2×3^n、x^2+19=2^2×5^n各有三組正整數解外,方程式D_1x^2+D_2=λ^2k^n 的正整數解個數最多二組。此篇論文主要討論方程式2x^2+1=3^n的正整數解,由余茂華之前的結果得到方程式2x^2+1=3^n的正整數解個數最多二組,而在Begeard 和 Sheory 的論文中給定此方程式的正整數解分別為(x,n)=(1,1)及(2,2),但是除了n=1,2,我們發現n=5也是此方程式的一個解。在此篇論文中,我們將求正整數解的問題轉換成二次遞迴數列(binary recurrent sequence)的問題並且利用Beukers論文中的相關結果來證明方程式2x^2+1=3^n僅有三組正整數解(x,n)=(1,1),(2,2)以及(11,5)。 Ramanujan observed in 1913 that 1^2 + 7 = 2^3 , 3^2 + 7 = 2^4 , 5^2 + 7 = 2^5 , 11^2 + 7 = 2^7 , (181)^2 + 7 = 2^15 .(1.1) When looking for the solutions of the equation x^2 + 7 = 2^n in integers x > 0, n > 0,(1.2) Ramanujan conjectured in 1913 that all the solutions of equation (1.2) are given by (1.1). This was proved by Nagell in 1948 and by others, using several different proofs. The equation (1.2) is called the Ramanujan-Nagell equation and has applications to binary error-correcting codes. After 1980’s, many mathematicians concentrated on studying the equation D_1x^2 + D_2 =λ^2 k^n, (1.3) We denote by N(λ, D_1, D_2, k) the number of solutions(x, n) of the equation (1.3). For D_1 = 1, equation (1.3) is usually called the generalized Ramanujan-Nagell equation. In a series of papers, Le proved that N(λ,D_1, D_2,p)≦2 except for N(2, 1, 7, 2) = 5 and N(2, 3, 5, 2) = N(2, 1, 11, 3) = N(2, 1, 19, 5) =3. Bugeaud and Shorey offer the newest results and related ref-erences about equation (1.3). In this paper, we are looking for solutions of the equation 2x^2 + 1 = 3^n in integer x≧1, n≧1. (1.4) The previous result of Le implies N(1, 2, 1, 3) ≦2, and Bugeaud and Shorey refer that equation (1.4) has two solutions which are given by n = 1, 2. But except for n = 1, 2, we find n = 5 is also a solution of equation (1.4). In this paper we will reduce the problem of determining N(1, 2, 1, 3) to a binary recurrent sequence and use the results of Beukers [1] to prove that the equation (1.4) has only the solutions (x, n) = (1, 1), (2, 2) and (11, 5).
    Appears in Collections:[數學研究所] 博碩士論文

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