关于丢番图方程|6x^2y^2-x^4+3y^4|=2z^2  

Stuely on Diophantine Equation｜6x^2y^2-x^4+3y^4｜=2z^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：《辽宁大学学报（自然科学版）》2006年第2期163-165,共3页Journal of Liaoning University：Natural Sciences Edition

基　　金：辽宁省教育厅科研项目(20401232)

摘　　要：证明了丢番图方程|-x4+6x2y2+3y4|=2z2,(x,y)=1的全部正整数解为:(Ⅰ)若z>2y2,则x=|m21n21-6m22n22|,y=m21m22+2n21n22,z=z=[24m21m22n21n22-2(|m21m22-2n21n22|±2m1m2n1n2)2],其中m2,n1满足-n41+6m22n21+3m42=2(D2)2,2n1m1m2;z=z-时,n2,m1满足(D-4m2m1)n2=m1(m22-n21)和(D+4m2n1)m1=2n2(n21+3m22),z=z+时,n2,m1满足n2(D±4m2n1)=(m22-n21)m1和m1(D4m2n1)=2n2(3m22+n21).(Ⅱ)若z<2y2,则x=|m21n21-6m22n22|,y=m21m22+2n21n22,z=±z0,z0=24m21m22n21n22-2(|m21m22-2n21n22|±2m1m2n1n2)2,其中m2,n1满足-n41+6m22n21+3m42=2(D2)2,2n1m1m2;z=z0时,n2,m1满足n2(D±4m2m1)=(m22-n21)m1和m1(D4m2n1)=2n2(3m22+n21),z=-z0时,n2,m1满足(D4m2n1)n2=m1(m22-n21)和(D±4m2n1)m1=2n2(n21+3m22).从而更正了梁莉莉,王云葵[1]关于上述方程仅有正整数解(x,y,z)=(1,1,2)的结果.We have proved that all positive integer solutions to the Diophantine equation ｜6x^2y^2 - X^4 ＋ 3y^4 ｜= 2z^2（x,y）=1 which are（Ⅰ）If z〉2y^2 then X=｜m1^2n1^2-6m2^2n2^2｜,Y=m1^2m2^2＋2n^2n2^2,z=z-＋=-＋[24m1^2m2^2n1^2n2^2-2（｜m1^2m2^2-2n1^2n2^2｜±2m1m2n1n2）^2]where m2, n1 satisfies-n1^4＋6m2^2n1^2＋3m2^4=2（D/2）^2,2χn1m1m2 when z=z-,n2m1 satisfies （D-4m2m1）m2=m1（m2^2-n1^2） and （D＋4m2n1）m1=2n2（m1^2＋3m2^2）when z=z＋,n2,m1 satisfies n2 （D±4m2n1）=（m2^2-n1^2）m1和m1（D-＋4m2n1）=2n2（3m2^2＋n1^2）（Ⅱ）If;z〈2y^2,then x=｜m1^2n1^2-6m2^2n2^2｜,y=m1^2m2^2＋2n1^2n2^2,z=±z0,z0=24m1^2m2^2n1^2n2^2-2（｜m1^2m2^2-2n1^2n2^2｜±2m1m2n1n2）^2 where m2,n1 satisfies-n1^4＋6m2^2n1^2＋3m2^4=2（D/2）^2,2χn1m1m2 when z=z0,n2 m1 satisfiesn2（D±4m2m1）=（m2^2-n1^2）m1 and m1（D-＋4m2n1）=2n2（3m2^2＋n1^2）,when z=-z0 ,n2 m1 satisfies （D-＋4m2n1）n2=m1（m2^2-n1^2）and（D±4m2n1）m1=2n2（n1^2＋3m2^2）The result that the Diophantine equation above has only positive integer solution （x, y, z） = （1,1,2） acquired by Liang Lili and Wang Yunkui is corrected.

关 键 词：丢番图方程 正整数解 本原解 

分 类 号：O157.6[理学—数学]