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

  • math -

    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)

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