1) infinite recursive
无穷递回;无限递归
3) infinite recursion
无限递回
4) method of infinite descent
无穷递降法
1.
In this paper,we use Fermat method of infinite descent which shows that the Diophantine equations x 3±y 6=z 2 has no positive integer solutions when (x,y)=1 and y>1.
利用Fermat无穷递降法 ,证明了丢番图方程x3 ±y6=z2 ,(x ,y) =1仅有整数解 2 3 +16=32 。
5) recursive
[英][ri'kə:siv] [美][rɪ'kɝsɪv]
递回;递归
6) Fermat method of infinite descent
Fermat无穷递降法
1.
We use elementary theory of number and Fermat method of infinite descent,some necessary conditions if the Diophantine equations x 4+mx 2y 2+ny 4=z 2 has positive integer solutions that fit (x,y) =1m.
利用Fermat无穷递降法 ,证明了方程x4 +mx2 y2 +ny4 =z2 在 (m ,n) =(± 18,5 4 ) ,(36 ,- 10 8) ,(± 36 ,10 8) ,(± 18,- 10 8) ,(- 18,10 8) ,(± 36 ,75 6 )时均无正整数解 ,并且获得了方程在 (m ,n) =(± 6 ,-2 4 ) ,(± 12 ,132 ) ,(- 36 ,- 10 8) ,(18,10 8)时无穷多组正整数解的通解公式 。
2.
We make use elementary theory of number and Fermat method of infinite descent,somenecessary conditions if the diophantine equations x 4+mx 2y 2+ny 4=z 2 has positive Integer solutions that fir (x,y) =1
Fermat无穷递降法 ,证明了方程x4 +mx2 +ny4 =z2 =z2 在 (m ,n) =± (6,-3 3 ) ,(6,3 3 ) ,(-3 ,-6) ,(± 1 2 ,1 68) ,(-6,-1 2 ) ,(1 2 ,84)均无正整数解 ,并且获得了方程在 (-3 ,6) ,(6,-1 5 ) ,(± 3 ,-3 )时的无穷多组正整数解的通解公式 ,从而完善了Aubry等人的结
3.
With the help of the elementary theory of number and Fermat method of infinite descent,some necessary conditions have been proved provided that the Diophantine equations x 4+mx 2y 2+ny 4=z 2 has positive Integer solutions that fit (x,y) =1 m.
利用数论方法及Fermat无穷递降法 ,证明了丢番图方程x4 +mx2 y2 +ny4 =z2 在 (m ,n) =(± 6,-3 ) ,(6,3 ) ,(± 3 ,3 ) ,(-12 ,2 4) ,(± 12 ,-2 4) ,(± 6,15 ) ,(-6,-15 ) ,(3 ,6)仅有平凡整数解 ,并且获得了方程在 (-6,3 ) ,(12 ,2 4) ,(3 ,-6) ,(-6,3 3 )时的无穷多组正整数解的通解公式 ,从而完善了Aubry等人的结
补充资料：递降
1.依次降低;逐渐降低。
说明：补充资料仅用于学习参考，请勿用于其它任何用途。
参考词条