问题
解答题
(1)求满足y4+2x4+1=4x2y的所有整数对(x,y);
(2)求出所有满足5(xy+yz+zx)=4xyz的正整数解.
答案
(1)∵y4+2x4+1=4x2y⇒(2x4-4x2y+2y2)+(y4-2y2+1)=0⇒2(x2-y)2+(y2-1)2=0,
∴x2-y=0,y2-1=0,
∴原方程的所有整数解为(1,1)、(-1,1);
(2)∵5(xy+yz+zx)=4xyz⇒
+1 x
+1 y
=1 z
,4 5
∵4/5=(1/5)+(3/5)=(1/5)+(6/10)=(1/5)+(1/10)+(5/10)=(1/2)+(1/5)+(1/10).
⇒x,y,z的值循环为2,5,10,
∵
=4 5
+1 5
=3 5
+1 5
=6 10
+1 5
+1 10
=5 10
+1 5
+1 10
,1 2
=4 5
+1 4
=11 20
+1 4
+1 20
=10 20
+1 4
+1 20
,1 2
∴所有满足5(xy+yz+zx)=4xyz的正整数解为(5,10,2)、(5,2,10)、(10,5,2)、(10,2,5)、(2,10,5)、(2,5,10)、(4,20,2)、(4,2,20)、(2,4,20)、(2,20,4)、(20,2,4)、(20,4,2).