由 ---- 於 星期一 五月 26, 2003 6:41 pm
My method:
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5x^2+16xy+14y^2=35
3(x+2y)^2+2(x+y)^2=35
(x+2y)^2<=35/3
-4<x+2y<4
x+2y is odd, since 2(x+y)^2 is even.
so x is odd.
x+2y=(+/-)3, (+/-)1
When x+2y=(+/-)3
3(-3)^2+2(x+y)^2=35
2(x+y)^2=8
x+y=2 or -2
y=-5 or -1 or 1 or 5
x=7 or -1 or 1 or -7
(x,y)=(7,-5),(-1,-1)
When x+2y=(+/-)1
3(-1)^2+2(x+y)^2=35
2(x+y)^2=32
x+y=4 or -4
y=-5 or 3 or 5 or -3
x=9 or -7 or -9 or 7
(x,y)=(9,-5),(-7,3)
So (x,y)=(7,-5),(-1,-1),(9,-5),(-7,3),(-7,5),(1,1),(-9,5),(7,-3)
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