Mathematics: How many pairs of integers?

You want to find solutions x and y such that
0 = (x^2 - x) + xy +y^2 + y

x^2 - x >= 0, and = 0 iff x = 0 or x=1
(It's >= 0 since it = 0 if x is 0 or 1, and is >0 if x > 1 because x^2 > x)

xy+y^2 + y >= 0 and = 0 iff y=0 (all the terms are nonnegative, and we need this to be 0 also).
So there are two solutions:
x=0 and y=0
x=1 and y=0

plot it:

http://www.wolframalpha.com/input/?i=x-y+%3D+x%5E2+%2B+xy+%2By%5E2+

then mark points where x and y are integers
looks like we have only two that are non-negative:
(0,0)
(1,0)

x - y = x^2 + xy + y^2 ≥ 0
x ≥ y

If x = y, then
0 = 3x^2
x = 0 and y = 0.

If x = y + k where k is a positive integer, then
k = (y + k)^2 + y(y + k) + y^2
y^2 + 2ky + k^2 + y^2 + ky + y^2 - k = 0
3y^2 + 3ky + (k^2 - k) = 0.

b^2 - 4ac ≥ 0 as y is a non-negative integer, which is a real number.
9k^2 - 4(3)(k^2 - k) ≥ 0
9k^2 - 12k^2 + 12k ≥ 0
-3k^2 + 12k ≥ 0
3k^2 - 12k ≤ 0
k(k - 4) ≤ 0
k = 4 or 0 < k < 4.

Thus,
k = 4, y = -2 and x = 2 (neg.), or
k = 3, y = -2 and x = 1, or k = 3, y = -1 and x = 2 (neg.), or
k = 2, y = -1 - 1/√3 and x = 1 - 1/√3, or k = 2, y = 1/√3 - 1 and x = 1 + 1/√3 (neg.), or
k = 1, y = -1 and x = 0 (neg.), or k = 1, y = 0, x = 1.

Therefore, There are 2 pairs of non-negative integers, including (0, 0) and (1, 0).