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)