math
posted by Steven .
How many ordered pairs of integers solutions (x,y) are there to
2x^2+3xy+y^2+2x+y+18=0?

2(x+1/2)^2 + (y+1/2)^2 + 3xy + 18 = 3/4
Obviously either x or y will be negative, since we need 3xy to be negative to wipe out all those other positive numbers.
I count 12 pairs:
(18,18)
(18,35)
(10,11)
(10,18)
(8,10)
(8,13)
(10,17)
(10,14)
(12,22)
(12,15)
(20,39)
(20,22)