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Old 2006-10-29, 16:16   #3
S485122
 
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Sep 2006
Brussels, Belgium

157410 Posts
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the easy part is x=y and needs no explaining

This simplifies the problem to find the integer solutions of x2+y2+xy-3x-3y=0 or y2+(x-3)y+x2-3x=0

Wich has two solutions:
y=(-x+3+sqrt((x-3)2-4(x2-3x)))/2
and
y=(-x+3+sqrt((x-3)2-4(x2-3x)))/2

the determinant must be positive, thus (x-3)2-4(x2-3x)=-3x2+2x+3 >= 0

Which implies that x is bounded by -1 and 3

The integer solutions are {-1,2}, {0,3}, {2,-1} and {3,0} and of course the solution [0,0]
I must learn to use tex :-(

Last fiddled with by S485122 on 2006-10-29 at 16:17
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