幾類(lèi)丟番圖方程的研究
發(fā)布時(shí)間:2018-07-26 14:57
【摘要】:本文主要研究幾類(lèi)丟番圖方程.文章主要由三部分構(gòu)成.1.第一部分,我們研究了廣義費(fèi)馬方程,得到了下面幾個(gè)結(jié)論:(1.1)設(shè)素?cái)?shù)p滿(mǎn)足br+1 = 2pt,其中r,t,6為正整數(shù),且6 = 3,5 (mod 8),則丟番圖方程x2+bm=pn只有一組正整數(shù)解(x,m,n) = (pt-1,r,2t).進(jìn)而,如果6 = q是奇素?cái)?shù),并且r = t = 2,則上述方程滿(mǎn)足下述條件之一時(shí)只有一組正整數(shù)解,這里條件(i) q三3,5,7 (mod 8);條件(ⅱ)q ≡1 (mod 8)且d = 1或者d是偶數(shù),其中d為素理想p在虛二次域Q((?))的理想類(lèi)群中的階,p|p.(1.2)令p≠q為兩個(gè)素?cái)?shù),則除去q = 2的情況外,方程p2m-qn=z2至多有一組非平凡解(m,n,z).進(jìn)而,丟番圖方程x2m- yn=z2僅有有限組正整數(shù)解(x,y,z,m,n),其中xy是連續(xù)的兩個(gè)素?cái)?shù).2.在第二部分,我們研究了乘法元上的加法性質(zhì),并得到了以下結(jié)果:(2.1)設(shè)K為一個(gè)代數(shù)數(shù)域,OK為其整數(shù)環(huán).n是OK中的一個(gè)非零理想.任取剩余類(lèi)環(huán)OK/n中元素a,定義(OK/n)* · a為a的一個(gè)軌道.首先,我們給出了任意兩個(gè)軌道之和的軌道分解.其次,對(duì)兩個(gè)軌道之和中的任意元素,我們得到了它在這兩個(gè)軌道中不同分解個(gè)數(shù)的表達(dá)式.(2.2)我們給出了剩余類(lèi)環(huán)上例外單位之和的表達(dá)式,并得到一個(gè)恒等式.3.在第三部分,我們考慮一類(lèi)橢圓曲線E :y2 = x(x + 2tp)(x + 2tq),其中t為非負(fù)整數(shù),p q為兩個(gè)奇素?cái)?shù)且滿(mǎn)足q - p = 2s.我們給出了該橢圓曲線的秩與Shafarevich-Tate群(沙群)之間的關(guān)系.進(jìn)而在BSD猜想成立的條件下,得到了一類(lèi)橢圓曲線的秩.
[Abstract]:In this paper, we study several kinds of Diophantine equations. The article is mainly composed of three parts. In the first part, we study the generalized Fermat equation and obtain the following conclusions: (1) Let the prime number p satisfy br 1 = 2pt, where rn 6 is a positive integer, and 6 = 3N 5 (mod 8), then the Diophantine equation x 2 bm=pn has only one set of positive integer solutions (XM n) = (pt 1 n) 2t. Furthermore, if 6 = Q is an odd prime number and r = t = 2, then the above equation satisfies only one set of positive integer solutions for one of the following conditions, where (i) q 3 3 + 5N 7 (mod 8); condition (II) Q 鈮,
本文編號(hào):2146380
[Abstract]:In this paper, we study several kinds of Diophantine equations. The article is mainly composed of three parts. In the first part, we study the generalized Fermat equation and obtain the following conclusions: (1) Let the prime number p satisfy br 1 = 2pt, where rn 6 is a positive integer, and 6 = 3N 5 (mod 8), then the Diophantine equation x 2 bm=pn has only one set of positive integer solutions (XM n) = (pt 1 n) 2t. Furthermore, if 6 = Q is an odd prime number and r = t = 2, then the above equation satisfies only one set of positive integer solutions for one of the following conditions, where (i) q 3 3 + 5N 7 (mod 8); condition (II) Q 鈮,
本文編號(hào):2146380
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