关于丢番图方程Ax^4+Bx^2y^2+Cy^4=z^2的解  

On Diophantine Equation Ax^4 + 6x^2y^2+ Cy^4 = z^2

作　　者：佟瑞洲[1] 

机构地区：[1]朝阳师范高等专科学校,辽宁朝阳122000

出　　处：《河南科技大学学报（自然科学版）》2006年第2期91-93,共3页Journal of Henan University of Science And Technology:Natural Science

基　　金：辽宁省教育厅科研立项课题(20401232)

摘　　要：证明了丢番图方程4x4-6x2y2+3y4=z2,(x,y)=1的全部正整数解为(x,y,z)=(x0／2,ab,(3a4+b4)／4), (Xn,2yn,2zn),认为仅有正整数解(x,y,z)=(1,1,1)是不妥的,它漏掉了(xn,2yn,2zn)及(x0／2,ab,(3a4+b4)／ 4);丢番图方程x4-6x2y2+12y4=z2,(x,y)=1的全部正整数解为(x,y,z)=(x0,ab,(3a4+b4)／2),(xn,yn, zn),认为仅有正整数解(xn,yn,zn),则漏掉了(x0,ab,(3a4+b4)／2)。It is proved that the positive integer solutions to the Diophantine equation 4x4 + 6x2y2 + 3y4 = z2, ( x , y) = 1 include (x , y, z) = (x0/2, ab, (3 a4 + b4)/4) , ( xn ,2yn,2zn) instead of an only (x, y, z) = (1,1,1). The traditional view left out (xn,2yn,2zn) and ( x0/2, ab , (3o4 + b4)/4).The positive integer solutions to the Diophantine equation x4 -6x2 y2 + 12y4 = z2 ,(x,y) = l include (x , y, z) = ( x0, ab , (3 a4 + b4 )/2) , ( xn , yn , zn) . The traditional view that the equation has only one positive integer solution ( xn , yn , zn ) is incorrect since it left out (x0,ab,(3a4+ b4)/2) .

关 键 词：丢番图方程 正整数解 递推序列 

分 类 号：O157.6[理学—数学]