On Some Equations Related to Ma’s Conjecture

DOI：10.16205/j.cnki.cama.2021.0018

 作者 单位 罗家贵 西华师范大学数学与信息学院, 四川 南充 637009. 费双林 西华师范大学数学与信息学院, 四川 南充 637009. 李　垣 西华师范大学数学与信息学院, 四川 南充 637009.

设$p$是奇素数, $b,t,r\in{\rm N}$. 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$t\geq r$,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,$t\geq r$, 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$t=r=1, x+a\sqrt{2^{b+2}(2^b-1)}=(2^{b+1}-1+\sqrt{2^{b+2}(2^b-1)})^{n}$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.

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).