| 罗家贵,费双林,李 垣.与马猜想有关的一类不定方程[J].数学年刊A辑,2021,42(2):229~236 |
| 与马猜想有关的一类不定方程 |
| On Some Equations Related to Ma’s Conjecture |
| Received:August 13, 2018 |
| DOI：10.16205/j.cnki.cama.2021.0018 |
| 中文关键词: McFarland's猜想, 丢番图方程, 基本解 |
| 英文关键词:McFarland’conjecture, Diophantine equations, Fundamental solution |
| 基金项目:国家自然科学基金 (No.,10571180) 和四川省教育厅重大培育项目(No.,16ZA0173) |
| Author Name | Affiliation | | LUO Jiagui | School of Mathematics and Information, China West Normal University, Nanchong 637009, Sichuan, China. | | FEI Shuanglin | School of Mathematics and Information, China West Normal University, Nanchong 637009, Sichuan, China. | | LI Yuan | School of Mathematics and Information, China West Normal University, Nanchong 637009, Sichuan, China. |
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| 中文摘要: |
| 设p是奇素数, b,t,rin{rm N}. 1992 年, 马少麟猜想丢番图方程 x^2=2^{2b+2}p^{2t}-2^{b+2}p^{t+r}+1有唯一的正整数解(x,b,p,t,r)=(49,3,5,1,2), 并且证明了这个猜想蕴含McFarland关于乘子为-1 的阿贝尔差集的猜想.在[Ma S L, MaFarland'conjecture on Abelian difference sets with multiplier-1[J]. {it Designs, Codes and Cryptography,} 1992, 1:321--332.]中, 马少麟证明了: 若tgeq r,则丢番图方程x^2=2^{2b+2}p^{2t}-2^{b+2}p^{t+r}+1没有正整数解. 本文证明了: 若a>1是奇数,tgeq r, 那么丢番图方程x^2=2^{2b+2}a^{2t}-2^{b+2}a^{t+r}+1的正整数解由t=r=1, x+asqrt{2^{b+2}(2^b-1)}=(2^{b+1}-1+sqrt{2^{b+2}(2^b-1)})^{n}给出, 其中n为奇数.作者也证明了: 若p是奇素数, 则(x,b,p,t,r)=(7,3,5,1,2)是丢番图方程x^4=2^{2b+2}p^{2t}-2^{b+2}p^{t+r}+1的唯一正整数解. |
| 英文摘要: |
| Let p be an odd prime and b, t, r ∈ N. In 1992, Ma conjectured that (x, b, p, t, r) =(49, 3, 5, 1, 2) is the only positive integer solution of equation x2 = 22b+2p2t ? 2b+2pt+r + 1.And Ma proved that the conjecture implies McFarland’s conjecture on Abelian difference sets with multiplier-1. In [Ma S L, MaFarland’conjecture on Abelian difference sets with multiplier-1 [J]. Designs, Codes and Cryptography, 1992, 1:321–332.], Ma proved that equation x2 = 22b+2p2t ? 2b+2pt+r + 1 had no positive integer solution if t > r. In the present paper, the authors prove that the positive integer solutions of Diophantine equation x2 = 22b+2a2t ? 2b+2at+r + 1 with a is an odd > 1 and t > r are given by t = r = 1 and x + ap2b+2(2b ? 1) = (2b+1 ? 1 + p2b+2(2b ? 1))n for some odd positive integer n. They also prove that the only positive integer solution of Diophantine equation x4 = 22b+2p2t ? 2b+2pt+r + 1 with p is an odd prime and x, b, t, r ∈ N is given by (x, b, p, t, r) = (7, 3, 5, 1, 2). |
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